tag:blogger.com,1999:blog-90589308657208227262018-04-20T15:58:01.074+01:00algebradabraAlgebran for digesting the notion of variable ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-9058930865720822726.post-7322817712360039722018-04-15T19:55:00.000+01:002018-04-20T15:58:01.021+01:00ALG 10This week we describe various situations using symbolic algebra. We then 'play' with the algebra by choosing values that take us beyond the immediate situation: can we relate the algebra back to the situation? Does the situation still make sense?<br />The situations we've chosen are fairly straightforward, but the game we're playing is mathematically quite sophisticated. It's similar to starting with a familiar statement like 8 + <i>x</i> = 10, which works in natural numbers, and asking what happens in a case like 8 + <i>x</i> = 5: this can be made to work if we stretch our ideas about number, ie if invent new ones - the integers.<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">Note</span>: From an RME perspective, we could say that we are engaged in horizontal mathematisation (expressing a 'real' situation mathematically) and then in vertical mathematisation (developing the maths). <br />- <br /><b>MONDAY</b>: Here we play with an area formula, for a shape that can vary in size. We start with a straightforward application of the formula but then consider a case which only works if we allow (or invent) edges with negative lengths.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-UuaY1YDHkVQ/WtOdoOTrkFI/AAAAAAAAAWU/3Nj-3YhbZ7AAoEUs1zW5FhNYkzAqSMs-ACLcBGAs/s1600/Algrabya150-10A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-UuaY1YDHkVQ/WtOdoOTrkFI/AAAAAAAAAWU/3Nj-3YhbZ7AAoEUs1zW5FhNYkzAqSMs-ACLcBGAs/s1600/Algrabya150-10A.jpg" /></a></div>Some students might feel that negative lengths are simply <i>not allowed</i>. That is a perfectly defensible position, but it would restrict the mathematics that we are able to do. A simple response is to say we are going to enter (or invent) a new (mathematical) world where negative lengths <i>are allowed</i>. So there!<br />This is what the shapes in parts i and ii look like (if you allow negative lengths in part ii):<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-cBF4QcTiN5k/WtTDwma5IbI/AAAAAAAAAWk/0kjnv2rHXJMZM454JTTIWiFd9RPlrC-kgCLcBGAs/s1600/Algrabya150-10A-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://3.bp.blogspot.com/-cBF4QcTiN5k/WtTDwma5IbI/AAAAAAAAAWk/0kjnv2rHXJMZM454JTTIWiFd9RPlrC-kgCLcBGAs/s400/Algrabya150-10A-ans.jpg" width="400" /></a></div> -<br /><b>TUESDAY</b>: We look at another familiar area scenario, namely area of a trapezium (and, to keep it simple, a trapezium that is right-angled).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-L3MBR9svQKg/WtULi9UnleI/AAAAAAAAAW0/kCnXw58_3Z48kQeFuscvUKzsZe3v8rtbACLcBGAs/s1600/Algrabya150-10B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-L3MBR9svQKg/WtULi9UnleI/AAAAAAAAAW0/kCnXw58_3Z48kQeFuscvUKzsZe3v8rtbACLcBGAs/s1600/Algrabya150-10B.jpg" /></a></div>Here we need to accept the idea of a negative length again, but also the idea of a negative area (which we could sidestep or leave implicit in ALG 10A).<br />One way to find the area geometrically is to divide the trapezium into two triangles. For part i we can, for example, divide the trapezium into two triangles of area 30 and 15 square units (top-left diagram, below).<br />The top-right diagram shows what happens to the trapezium as point P moves until it is 5 units to the <i>left</i> of the formerly top-left vertex. The trapezium 'twists' over itself to form two triangles whose areas, we can argue, are 20 and -5 square units.<br />The two diagrams at the bottom of the slide, below, show an alternative interpretation for part ii. The yellow triangle corresponds to the 30 square units triangle in the top-left diagram. The green triangle corresponds to the 15 square units triangle in the same diagram, except its base has changed from 5 units to -5 units. If we 'cancel' the region where the yellow and green triangles overlap, we are left with the regions with area 20 and -5 square untis shown in the top-right diagram.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-V_84tkuMr6M/WtZOoIGnW0I/AAAAAAAAAXE/UYy8Xy9PSbgjBnHh3WQDWuuGyk_QsXotQCLcBGAs/s1600/Algrabya150-10B-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://1.bp.blogspot.com/-V_84tkuMr6M/WtZOoIGnW0I/AAAAAAAAAXE/UYy8Xy9PSbgjBnHh3WQDWuuGyk_QsXotQCLcBGAs/s400/Algrabya150-10B-ans.jpg" width="400" /> </a></div>- <br /><div class="separator" style="clear: both; text-align: left;"><b>WEDNESDAY</b>: Things get really interesting.... What's a 2-and-a-half sided regular polygon?<i>!</i> </div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-MRIOoO9utjQ/WtZaOOc0s6I/AAAAAAAAAXk/Mol0vqMo3GESTiehX7YZ2tqPOLXqWL0-wCLcBGAs/s1600/Algrabya150-10C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-MRIOoO9utjQ/WtZaOOc0s6I/AAAAAAAAAXk/Mol0vqMo3GESTiehX7YZ2tqPOLXqWL0-wCLcBGAs/s1600/Algrabya150-10C.jpg" /></a></div>I came across this beautiful idea, of replacing the whole number of sides, <i>n</i>, with a fraction, in one of David Fielker's articles in <i>Mathematics Teaching</i>, many years ago. For me, the idea is almost on a par with inventing negative or fractional indices. A simple but brilliant mathematical act!<br />You may recognise the form of the instructions if you are familiar with LOGO and Turtle Geometry. If you don't have a turtle to hand I hope you will have enacted the instructions yourself and traced the resulting paths on paper or in your head! This is what they turn out to look like:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-uSsWfBA9PdQ/Wteln6-zVvI/AAAAAAAAAX0/ojGcYSiU484cPJTAjJ7N-rNeES7COkktwCLcBGAs/s1600/Algrabya150-10C-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://3.bp.blogspot.com/-uSsWfBA9PdQ/Wteln6-zVvI/AAAAAAAAAX0/ojGcYSiU484cPJTAjJ7N-rNeES7COkktwCLcBGAs/s400/Algrabya150-10C-ans.jpg" width="400" /></a></div>- <br /><b>THURSDAY</b>: We again make the shift from whole numbers to fractions, this time on a familiar number grid.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-rpLEiiLO73U/WtemBJk96GI/AAAAAAAAAX4/lf49sJn5cQ8GF5Ix8jxsJJyA7_7z3e2XgCLcBGAs/s1600/Algrabya150-10D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-rpLEiiLO73U/WtemBJk96GI/AAAAAAAAAX4/lf49sJn5cQ8GF5Ix8jxsJJyA7_7z3e2XgCLcBGAs/s1600/Algrabya150-10D.jpg" /></a></div> It can be useful to consider what happens to the sum, <i>S</i>, when the T-shape moves across the grid (eg 1 square to the right, or 1 square up). We can think about this spatially (What happens to each of the numbers in the T-shape?) or algebraically (What happens to <i>S</i> when <i>n</i> in the expression 6<i>n</i>+120 is increased by 1, say, or by 10?). In the case of part i, <i>S</i> has increased by 240–210 = 30, which can be achieved by moving the T-shape 5 squares to the right... [Are there other ways?]<br />Here are positions for the T-shape for parts i and ii.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-17wTp3NvZI8/Wtj7t_RpbvI/AAAAAAAAAYM/JcuWpv2u9uYkyMl0U08gSAkK_uv7JdrogCLcBGAs/s1600/Algrabya150-10D-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://1.bp.blogspot.com/-17wTp3NvZI8/Wtj7t_RpbvI/AAAAAAAAAYM/JcuWpv2u9uYkyMl0U08gSAkK_uv7JdrogCLcBGAs/s400/Algrabya150-10D-ans.jpg" width="400" /></a></div><span style="font-family: "arial" , "helvetica" , sans-serif;">Note</span>: if we accept the principle behind the part ii answer, of allowing <i>fractions of a square</i>, we can find infinitely many positions for the T-shape, for parts i and ii, by moving the shape vertically (maybe just a tiny bit) as well as horizontally. What we're doing here, in effect, is to change the discrete 2-D grid into a continuous Cartesian plane.<br />- <br /><b>FRIDAY</b>: Here we consider square grids made of matchsticks - or parts of matchsticks.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-Nvy8eDwffYY/Wtj8Twma0RI/AAAAAAAAAYU/8cQ5DFD6ECQHtRShgMOR2CR52-ii7E9LQCLcBGAs/s1600/Algrabya150-10E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-Nvy8eDwffYY/Wtj8Twma0RI/AAAAAAAAAYU/8cQ5DFD6ECQHtRShgMOR2CR52-ii7E9LQCLcBGAs/s1600/Algrabya150-10E.jpg" /></a></div>I'm particularly fond of this pattern - it's one whose structure is fairly easy to discern <i>generically</i>, even though the relation between the dimension of the square and the number of matchsticks is <i>quadratic</i> rather than linear.<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">Note</span>: The slide below shows some interesting attempts to structure the grid by three Year 7 students (from a 'low attaining' set: set 3 of 4). I've written this up in chapter 3 of the Proof Materials Project <a href="http://www.mathsmedicine.co.uk/ioe-proof/PMPintro2.html">report</a>, <i>Looking for Structure</i>.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-wnp5rFUlfC8/Wtn-q1QP7II/AAAAAAAAAYk/gvBS-Q7IGWU5rUWdLNX1EE8l2KP3_735wCLcBGAs/s1600/y7sticks.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="408" data-original-width="849" height="305" src="https://4.bp.blogspot.com/-wnp5rFUlfC8/Wtn-q1QP7II/AAAAAAAAAYk/gvBS-Q7IGWU5rUWdLNX1EE8l2KP3_735wCLcBGAs/s640/y7sticks.png" width="640" /></a></div>xProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-71792221130308100772018-04-08T23:47:00.000+01:002018-04-12T23:32:07.894+01:00ALG 9As it's <i>not quite</i> the summer term yet (some schools go back on Monday, some on Monday week), the theme this week is <i>not quite</i> algebra. We are going to look at some tasks from well-known (or well-publicised) textbooks that involve faux, fake, phony, manqué, mock, pretend, pseudo, quasi, cod algebra.<br /><b>MONDAY</b>: a peach from Singapore (or several peaches, if you can afford them).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-MFtfHjadYi4/WsqdY0-u5_I/AAAAAAAAAVE/02OUg8eEa9kZTjWBUbcq53T0W1r7EqnpACLcBGAs/s1600/Algrabya150-9A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-MFtfHjadYi4/WsqdY0-u5_I/AAAAAAAAAVE/02OUg8eEa9kZTjWBUbcq53T0W1r7EqnpACLcBGAs/s1600/Algrabya150-9A.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>What is going on here? Does the story make sense?<br />-<br /><b>TUESDAY</b>: We're still in Singapore, with a task where <i>x</i> is not so much an unknown as an utter mystery. What on earth might it stand for?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-Lyg8T4GLGoo/WsvkPN-hChI/AAAAAAAAAVU/Z0lFyCK8oA89NvQgzOjgIctdAnv88mNHwCLcBGAs/s1600/Algrabya150-9B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-Lyg8T4GLGoo/WsvkPN-hChI/AAAAAAAAAVU/Z0lFyCK8oA89NvQgzOjgIctdAnv88mNHwCLcBGAs/s1600/Algrabya150-9B.jpg" /></a></div>Mei Heng has unusual powers. The longer she works, the faster she works. However, if she works for less that 3 hour 40 minutes she seems to destroy rather than make T-shirts. Perhaps the Singapore government has created a mechanism that discourages part-time employment...<br />What of her pay, I wonder. Is it per hour or per T-shirt?<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">NOTE</span>: I've written about this and a few other tasks from the same textbook in the journal <i>Mathematics in School</i> (November 2013, Volume 42, Issue 5, p25).<br />-<br /><b>WEDNESDAY</b>: The context here involves 'grids' made of rods joined by a variety of links. We are shown two specific grids and for each we are presented with a 'rule' stating how many rods and various links it consists of.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-q9zSAC9I2wg/Ws1Bqf8g6MI/AAAAAAAAAVk/8EGBCkhuB1sUS8s8fzMDDziW4mwNh4WsgCLcBGAs/s1600/Algrabya150-9C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-q9zSAC9I2wg/Ws1Bqf8g6MI/AAAAAAAAAVk/8EGBCkhuB1sUS8s8fzMDDziW4mwNh4WsgCLcBGAs/s1600/Algrabya150-9C.jpg" /></a></div>The 'rules' look 'algebraic' in that they contain letters. However, they are not general statements, nor do they involve any unknowns. And the letters that appear in the 'rules' don't stand for numbers. Instead, what we have here is the equivalent of <i>fruit salad algebra</i>, or what I have dubbed elsewhere as <i>letter as object</i>.<br />-<br /><b>THURSDAY</b>: Here we look at a task from an earlier edition of Wednesday's UK textbook series. The exercise is designed purely as a device for practising algebraic manipulation. But what does it convey to students about the purpose and utility of algebra - or, indeed, of geometry? <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-qGJOIU87d5o/Ws6FP9L8msI/AAAAAAAAAV0/nYSKUmbFYUcIMCClua4L1Ykl82EBAX_nwCLcBGAs/s1600/Algrabya150-9D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-qGJOIU87d5o/Ws6FP9L8msI/AAAAAAAAAV0/nYSKUmbFYUcIMCClua4L1Ykl82EBAX_nwCLcBGAs/s1600/Algrabya150-9D.jpg" /></a></div>It turns out that, treated with any kind of common sense, shape <b>e</b> collapses into nothing. Curiously, exercises of this sort, which reduce algebra to an exercise in manipulation, and which abuse geometry by treating it merely as a means to this end, are commonplace in UK textbooks.<br />-<br /><b>FRIDAY</b>: Finally, we consider a task from a recent UK adaptation of a Singapore textbook, which describes itself as <i>The Mastery Course for Key Stage 3</i>.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-yRxz-vzZJYA/Ws_Y6KuUHoI/AAAAAAAAAWE/S51s9sbwkRQuOCRLqyLXiSJ1FzhC9XHWgCLcBGAs/s1600/Algrabya150-9E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-yRxz-vzZJYA/Ws_Y6KuUHoI/AAAAAAAAAWE/S51s9sbwkRQuOCRLqyLXiSJ1FzhC9XHWgCLcBGAs/s1600/Algrabya150-9E.jpg" /></a></div>It helps if the algebra tasks we give students demonstrate the <i>purpose and utility</i> of algebra (to use a phrase coined by Janet Ainley and Dave Pratt). However, this is not always easy to bring about; in the case of this task, its absurd nature suggests that the exact opposite has been achieved!<br />-<br /><b>NEXT WEEK</b>: algebra takes control and we try to make sense of the consequences...ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-53850774196629855412018-03-25T23:46:00.002+01:002018-03-29T16:42:10.595+01:00ALG 8This week's set of tasks use a mapping diagram to represent liner functions. This model may not be familiar to some teachers and students but this can be advantage - it allows us to take a fresh look at the characteristics of linear functions and (though we don't do it here) can help us look at the Cartesian graph of linear functions with fresh eyes.<br /><b>MONDAY</b>: Here we are asked to read a mapping diagram and to determine the function that the given information represents. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-nKF8wWslT5Y/Wrgjznj0cVI/AAAAAAAAAS0/FkSQx-fpsVUuliBYjN7IQkdrN9CJrZuCwCLcBGAs/s1600/Algrabya150-8A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-nKF8wWslT5Y/Wrgjznj0cVI/AAAAAAAAAS0/FkSQx-fpsVUuliBYjN7IQkdrN9CJrZuCwCLcBGAs/s1600/Algrabya150-8A.jpg" /></a></div><b>TUESDAY</b>: Here we take a more detailed look at Monday's mapping diagram (which was for the function <i>y</i> = 3<i>x</i> – 2) and consider how its features can be used to identify the linear function. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-8LA__W1ZKLc/WroV27PI27I/AAAAAAAAATE/uP1XbhvT2yM0hatJZcTagqGYKY0iUDU9ACLcBGAs/s1600/Algrabya150-8B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-8LA__W1ZKLc/WroV27PI27I/AAAAAAAAATE/uP1XbhvT2yM0hatJZcTagqGYKY0iUDU9ACLcBGAs/s1600/Algrabya150-8B.jpg" /></a></div><b>WEDNESDAY</b>: I was amazed by this phenomenon when I first came across it (thanks to JH). Of course, as with many mathematical phenomena, the more you think about it the more obvious it becomes. But this one also becomes richer. All in all, an exciting and lovely result!<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-O2l57aFZraY/WrqM_oHXWmI/AAAAAAAAATU/M8Z0awoOKnQWBrYxNIol86xKzjiZ8C4LwCLcBGAs/s1600/Algrabya150-8C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-O2l57aFZraY/WrqM_oHXWmI/AAAAAAAAATU/M8Z0awoOKnQWBrYxNIol86xKzjiZ8C4LwCLcBGAs/s1600/Algrabya150-8C.jpg" /></a></div>We can visualise parts i and ii by translating point P <i>and</i> the arrows one unit to the right (part i) or one unit down (part ii). Alternatively, move both axes one unit to the left (part i) or one unit up (part ii). Which approach do you prefer?<br /><b>PS</b>: You will probably have noticed that P can be thought of as the <i>centre of enlargement </i>for the linear function. Image points (ie points on the <i>y</i> axis) are 3 times the distance from P as object points (ie the corresponding points on the <i>x</i> axis). So the scale factor of the 'enlargement' is ×3. Similarly, the distance between a pair of points on the <i>y</i> axis is 3 times the distance of corresponding points on the <i>x</i> axis.<br />In part i, the same distance-relationships hold, ie the scale factor is still ×3. However, in part ii the distance-relationships change, and therefore so does the scale factor. Has it got smaller or larger?<br /><b>THURSDAY</b> <b>1</b>: Here we extend the notion that the arrows on mapping diagrams of linear functions emanate from a 'centre of enlargement', by considering centres that lie between the two axes. What does that tell us about the 'scale factor', ie the value of <i>m</i> in <i>y = mx + c</i> ?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-CNPH_dh0E4g/WrwHlOz3EGI/AAAAAAAAATk/cxuQ8u9QMh4ncfDjuyWDur9rFb-l7DDGwCLcBGAs/s1600/Algrabya150-8D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-CNPH_dh0E4g/WrwHlOz3EGI/AAAAAAAAATk/cxuQ8u9QMh4ncfDjuyWDur9rFb-l7DDGwCLcBGAs/s1600/Algrabya150-8D.jpg" /></a></div> -<br /><b>THURSDAY 2</b> (instead of Good Friday): Here we consider the inverse of our original function.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-DElAe9QohUw/Wr0Doxp6fFI/AAAAAAAAAT0/TZBHB3V445MwyMTb9Jg-Vr6yHVqafFPUQCLcBGAs/s1600/Algrabya150-8E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-DElAe9QohUw/Wr0Doxp6fFI/AAAAAAAAAT0/TZBHB3V445MwyMTb9Jg-Vr6yHVqafFPUQCLcBGAs/s1600/Algrabya150-8E.jpg" /> </a></div><div class="separator" style="clear: both; text-align: left;">The inverse takes us back to where we started. One way of representing this there-and-back journey is by adding a second mapping diagram that is a reflection of the original, as here:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-aZXP7IunGq0/Wr0DtE3nf4I/AAAAAAAAAT4/RqZRZpHMo_UNfSa6vL3iGhP_8cWfMHnawCLcBGAs/s1600/Algrabya150-8E-flip.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="353" data-original-width="570" src="https://4.bp.blogspot.com/-aZXP7IunGq0/Wr0DtE3nf4I/AAAAAAAAAT4/RqZRZpHMo_UNfSa6vL3iGhP_8cWfMHnawCLcBGAs/s1600/Algrabya150-8E-flip.jpg" /></a></div><b>Note</b>: the inverse function itself is represented by just the second mapping diagram, but with the axes re-labeled (<i>x</i> axis at the top, <i>y</i> axis below), and with the arrows pointing downwards. As can be seen, the 'common point' (or 'centre of enlargement') is vertically below P, two units below the <i>y</i> axis. The scale factor here is 1/3, and if we look carefully, we can see that 0 is mapped onto 2/3, so the inverse function is <i>y</i> = 1/3 <i>x</i> + 2/3, or <i>y</i> = 1/3 (<i>x</i> + 2). We can verify this algebraically, by transforming the original function <i>y</i> = 3<i>x</i> – 2 into <i>x</i> = .... (and then transposing <i>x</i> and <i>y</i>).<br /><br />ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-30363788415916639932018-03-18T11:43:00.003+00:002018-03-30T12:08:29.520+01:00ALG 7This week we look at two tasks of classic form, both of which are rather contrived but which are also quite engaging if simply treated as a puzzles. The aim is to show the power of using a symbolic algebra approach, although in the case of the first task, below, we have deliberately devised a task which is also amenable to informal methods.<br />- <br /><div class="separator" style="clear: both; text-align: left;"><b>MONDAY</b>: This first task is of the classic 'How many cows and chickens' form, although we have tried to make it a bit more challenging by involving three rather than just two sets of 'elements', in this case, bikes, trikes and quadracycles.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-96Z6_z11ecI/Wq5GlRvB_1I/AAAAAAAAARk/YmmmD4dy0OUzocLZ_iDNhAIpcH0Ez785ACLcBGAs/s1600/Algrabya150-7A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-96Z6_z11ecI/Wq5GlRvB_1I/AAAAAAAAARk/YmmmD4dy0OUzocLZ_iDNhAIpcH0Ez785ACLcBGAs/s1600/Algrabya150-7A.jpg" /></a></div>As we spell-out in Tuesday's variant, this <i>Cycles</i> task can be solved algebraically by creating equations such as<br />T = Q + 5<br />31 = B + T + Q<br />76 = 2B + 3T + 4Q.<br />We can then substitute Q for T. Also, if we double the second equation, giving 64 = 2B + 2T + 2Q, we can eliminate B by subtracting this equation from the third equation, leaving us with 14 = T+ 2Q. <br />In Tuesday's variant, we also indicate that the task can be solved informally with this sort of argument:<br />Imagine that the cycles are all bikes;<br />we would then have 64 tyres (or wheels) which is 14 fewer than the actual number;<br />so we would need to replace some of the bikes with trikes and/or quads;<br />each trike would increase the number of tyres by 1, each quad would increase the number of tyres by 2;<br />so .... <br />There is a clear link between this narrative approach and the process that gave us the equation 14 = T + 2Q. Of course, students might not see this link immediately, though Freudenthal (in his <i>China Lectures</i> published in 1991) suggests that students can achieve this by solving a range of similar tasks in this informal way and generalising.<br />There is an interesting tension here. One of the strengths of a symbolic approach is that it 'frees' us from thinking about the context, once we have expressed the given information symbolically, and assuming we are fluent in manipulating the symbols.<br />On the other hand, the act of keeping hold of the context, or of referring back to it periodically, can help us (or more to the point, our students) to form the symbolisations and to check that these, and the subsequent manipulations, are valid.<br />- <br /><b>TUESDAY</b>: Here we present two methods for solving yesterday's <i>Cylces</i> task, one informal the other involving symbolic algebra, which we ask students to complete. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-FVz9oD2u9p8/WrBGcP758vI/AAAAAAAAAR0/oWBu2OFSHA8iXHqHObWiTz9GfO4QeOQXACLcBGAs/s1600/Algrabya150-7B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-FVz9oD2u9p8/WrBGcP758vI/AAAAAAAAAR0/oWBu2OFSHA8iXHqHObWiTz9GfO4QeOQXACLcBGAs/s1600/Algrabya150-7B.jpg" /></a></div>How students respond to these methods will depend on their knowledge and experience. Some students might gain a better appreciation of the power and efficiency of symbolic algebra, while others might gain some understanding of the algebraic approach by linking it to the more concrete, informal approach.<br />-<br /><b>WEDNESDAY</b>: Here we present two new tasks, both set in the context of 'sharing acorns' but where the given information is structured differently. We consider how amenable the tasks are to informal and formal approaches in the light of this difference in structure.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-DpffZOQ1fwY/WrGSJJxMWcI/AAAAAAAAASE/RtOsf5ut30UIxpzuc2dBJuJkY8oUDngQACLcBGAs/s1600/Algrabya150-7C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-DpffZOQ1fwY/WrGSJJxMWcI/AAAAAAAAASE/RtOsf5ut30UIxpzuc2dBJuJkY8oUDngQACLcBGAs/s1600/Algrabya150-7C.jpg" /></a></div>The first task can be solved in informal and formal ways that are very similar: the collection of 60 acorns can be partitioned into 1 plus 4 plus 5 equal shares, ie into 10 equal shares altogether; algebraically, we can construct something like this: <i>g</i> + 4<i>g</i> + 5<i>g</i> = 60, so 10<i>g</i> = 60.<br />The second task <i>can</i> be solved informally but it probably requires quite a lot of insight to do this successfully. On the other hand, it can be solved in a fairly routine way using symbolic algebra, assuming the solver has adequate technical skills.<br />Thus, we would argue that the second task is that elusive entity: a task (albeit a mere pointless puzzle!) where <i>the use of symbolic algebra comes into its own</i>. Of course, in the first task, we could also strengthen the need to record and symbolise, by increasing the number of recipients and by relating their shares to George's share in more varied ways.<br /><b>Note 1</b>: You may have noticed that the methods of sharing in the two tasks turn out to be equivalent. The second task has been made more complex by using what Dettori et al (2006) call 'circular references'. I came across this work in the excellent revue by Mason & Sutherland (2002), <a href="https://www.researchgate.net/profile/John_Mason3/publication/42788782_Key_aspects_of_teaching_algebra_in_schools/links/53f1f5750cf272810e4c7a23.pdf"><i>Key Aspects of Teaching Algebra in Schools</i></a>. You might like to devise similar tasks to challenge your students, by taking a straightforward sharing task and then using circular references to recast the description of some of the shares.<br /><b>Note 2</b>: The 'circular references' task used by Dettori (and reproduced in Mason & Sutherland) is this:<br /><a href="https://3.bp.blogspot.com/-G9eVEtEOOf0/Wr4WEXQd4RI/AAAAAAAAAUM/uERU8a7PMV0EAsqGAPKFkMc6VQSIIV70wCLcBGAs/s1600/dettori-fish.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="239" data-original-width="906" height="168" src="https://3.bp.blogspot.com/-G9eVEtEOOf0/Wr4WEXQd4RI/AAAAAAAAAUM/uERU8a7PMV0EAsqGAPKFkMc6VQSIIV70wCLcBGAs/s640/dettori-fish.png" width="640" /></a>Interestingly, an almost identical task occurs in an English translation (1822, p204) of a French translation of Leonhard Euler's <i>Vollständige Anleitung Zur Algebra</i> (1771). <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-pwgNMP0_dhw/Wr4XQFWPFKI/AAAAAAAAAUU/He5NMXygKB081dPItsKniRQVHRkT8NQpwCLcBGAs/s1600/euler-de-lalgrange-fish.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="112" data-original-width="665" height="107" src="https://4.bp.blogspot.com/-pwgNMP0_dhw/Wr4XQFWPFKI/AAAAAAAAAUU/He5NMXygKB081dPItsKniRQVHRkT8NQpwCLcBGAs/s640/euler-de-lalgrange-fish.png" width="640" /></a></div>The English book (below) contains 'Additions of M. De La Grange' and it is possible that the problem stems from him rather than Euler, as I can't find it in the original German edition.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-SXb9wpTL7Nc/Wr4ZqA8iMRI/AAAAAAAAAUo/LCqi91yJ6gk9mBrVdKhayc6mTlDb1_yugCLcBGAs/s1600/euler-elements-front.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="860" data-original-width="490" height="640" src="https://1.bp.blogspot.com/-SXb9wpTL7Nc/Wr4ZqA8iMRI/AAAAAAAAAUo/LCqi91yJ6gk9mBrVdKhayc6mTlDb1_yugCLcBGAs/s640/euler-elements-front.png" width="363" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><b>THURSDAY</b>: Here we introduce Auntie's way of sharing acorns. It is not equivalent to Grandpa's method, but it turns out to produce the same result for a particular number of acorns.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-DhE6gwoYFk0/WrLXMn7dR1I/AAAAAAAAASU/GXwCwIs1ui0_Ml0cB8w0mOaqkyocv8q7QCLcBGAs/s1600/Algrabya150-7D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-DhE6gwoYFk0/WrLXMn7dR1I/AAAAAAAAASU/GXwCwIs1ui0_Ml0cB8w0mOaqkyocv8q7QCLcBGAs/s1600/Algrabya150-7D.jpg" /></a></div>We can find the particular number of acorns using symbolic algebra, by solving the equation resulting from equating Peppa's shares (ie 4<i>g</i> and 2<i>g</i> + 20) or Chloe's shares (5<i>g</i> and 2<i>g</i> + 30) under the two ways of sharing. Both give the value <i>g</i> = 10, so that the total number of acorns is 100.<br />We have obviously carefully contrived the two sharing methods for them to produce the same outcome for all three recipients for a particular number (100) of acorns. This wouldn't happen by chance. However it raises an interesting (although rather subtle) question: given that the choice of Grandpa's or Auntie's sharing method makes no difference for 100 acorns, does it make a difference, for any individual recipient, for other numbers of acorns?<br /><b>FRIDAY</b>: Having established that Grandpa's and Auntie's method happen to produce identical shares when there are 100 acorns, today we consider what happens when there are more than 100 acorns: for each individual recipient, is one sharing method more favourable than the other?<br /><b>Note</b>: This task is quite complex and is probably more suitable as a challenge for teachers than as an activity in class. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-qk0IZlKHm0c/WrQ-d_o6V6I/AAAAAAAAASk/B1dh3vCEES42ov2tersvFiq30DLJiMfeQCLcBGAs/s1600/Algrabya150-7E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-qk0IZlKHm0c/WrQ-d_o6V6I/AAAAAAAAASk/B1dh3vCEES42ov2tersvFiq30DLJiMfeQCLcBGAs/s1600/Algrabya150-7E.jpg" /></a></div>In the case of George, he always gets one tenth of the acorns when Grandpa shares them out. This is not a large proportion, but still larger than what he gets from Auntie when there are less than 100 acorns. In the extreme case, when there are only 50 acorns, George gets none! George is better off with Auntie when there are more than 100 acorns.<br />The situation with Peppa turns out to be rather surprising. When there are 100 acorns, George's share is 10 acorns by either sharing method and she gets 40 acorns by either sharing method (ie 4×10 or 2×10 + 20). Now, when Georges share (<i>g</i>) is greater than 10, 4<i>g</i> will be greater than 2<i>g</i> + 20, which suggests that Grandpa's method will be more favourable than Auntie's when there are more than 100 acorns. However, it turns out that this is not the case! Why? Because Peppa's share depends on George's (as it's a function of <i>g</i>), but George's share differs depending on whether Grandpa or Auntie are sharing out the acorns. It turns out that Peppa's share comes to the same amount, whether it's Grandpa or Auntie doing the sharing, regardless of the number of acorns being shared out! For exmple, for 50 acorns she gets 20 under both methods, for 110 acorns she gets 44 under both methods. We leave it to the reader to verify/prove this.<br />For Chloe, it looks like Grandpa's method is more favourable than Auntie's when there are more than 100 acorns, as 5<i>g</i> > 2<i>g</i> + 30 when <i>g</i> > 10. But can we be sure? In the case of Grandpa's method, Chloe's share is 5/10 or 1/2 of the total number of acorns. In the case of Auntie's method, Chloe's share is 2/5 of the total plus 10 (you might like to verify this). When the total is more than 100, 1/2 of the total is greater than 2/5 of the total plus 10.<br /><br />ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-28657217223322757722018-03-11T22:58:00.003+00:002018-03-16T11:46:10.547+00:00ALG 6In this week's set of tasks we look at figurative patterns and express their structure in generic terms by using a <i>quasi variable</i> (in this case, 20). We look at equivalent ways of construing/expressing the structure.<br />Some of the week's patterns turn out to have a linear structure and some a quadratic structure - which raises the question, <i> </i><br /><i>How can we recognise this in the pattern itself, as well as in its symbolic representation</i>?<br />- <br /><b>MONDAY</b>: Here is the week's root task:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-257J4l3K0A4/WqWzh9BdcsI/AAAAAAAAAP8/mlMB4Io8i2UPFSXH44n4XPtoP5RcqMDsgCLcBGAs/s1600/Algrabya150-6A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-257J4l3K0A4/WqWzh9BdcsI/AAAAAAAAAP8/mlMB4Io8i2UPFSXH44n4XPtoP5RcqMDsgCLcBGAs/s1600/Algrabya150-6A.jpg" /></a></div><b>Note</b>: We have chosen 20 here (and in later tasks) on the basis that it is big enough to deter most students from simply counting each individual dot. Also, if students can find an efficient method for counting a 20-chain, it is likely that they can do so for any length of chain, perhaps even a chain of <i>n</i> Y-shapes. So, for many students, 20 is taking on the role of a variable.<br /><b>Note 2</b>: Another reason for going for such a 'far generalisation' early on, is to encourage students to look at the structure of the 20-chain as a whole. Of course, it is still possible to use an 'incremental' (or scalar or term-to-term) approach by noting that the 2-chain has 4 more dots than the 1-chain (and so the 20-chain has 19×4 more dots than the 1-chain). It would be interesting to see how the task would work without showing the 2-chain. We decided to put it in to avoid students thinking that there might be a hidden dot underneath the dot where the Ys join (in which case a 2-chain would simply have 2×5 dots and a 20-chain would simply have 20×5 dots).<br /><b>Note 3</b>: The presence of the 2-chain does open up other interesting possibilities: some students might argue that the 20-chain will have 10 times as many dots as the 2-chain; others might simply focus on the given numerical information, 1→5 and 2→9, and look for a mapping-rule that fits both, which can then be applied to 20.<br />-<br /><b>TUESDAY</b>: Here we are given two expressions for finding the number of dots in the 20-chain. The expressions are deliberately left open so as to show how the chain of Y-shapes can be structured. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-mTVDgwMsub4/Wqb71Mvq_vI/AAAAAAAAAQM/9M2KhdjnF-EB1xUu8x1p-BF1poO7DEMHQCLcBGAs/s1600/Algrabya150-6B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-mTVDgwMsub4/Wqb71Mvq_vI/AAAAAAAAAQM/9M2KhdjnF-EB1xUu8x1p-BF1poO7DEMHQCLcBGAs/s1600/Algrabya150-6B.jpg" /></a></div>It may be helpful to make an annotated sketch, as in the examples below, when describing each structure (ie when explaining Alma's and Bojan's methods).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-nN0PBVnDYKc/WqutYT4G8DI/AAAAAAAAARM/un9-W7efovsTlL8Py3JDjYUq8rqNYE10ACLcBGAs/s1600/Algrabya150-6B-structures.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://2.bp.blogspot.com/-nN0PBVnDYKc/WqutYT4G8DI/AAAAAAAAARM/un9-W7efovsTlL8Py3JDjYUq8rqNYE10ACLcBGAs/s400/Algrabya150-6B-structures.jpg" width="400" /></a></div>The given expressions are of course equivalent and you might want to challenge students to show how one expression can be transformed into the other. It is possible to structure the pattern in other (equivalent) ways, resulting in other (equivalent) expressions 'in 20'. How many can you find?<br />-<br /><b>WEDNESDAY</b>: Here we consider two further chains of Y-shapes, but this time the size of the Ys increases as the number of Ys in the chain increases. What does this do to the relationship between the number of Ys and the number of dots in a chain? Is it still linear? Can you tell from the pattern? Can you tell from the generic expression for the total number of dots?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-MNMkZhr1fDM/Wqgw-e8a9DI/AAAAAAAAAQc/qqt_d6k2o94fSx2qYdXfw7q0hTQWTDnYgCLcBGAs/s1600/Algrabya150-6C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-MNMkZhr1fDM/Wqgw-e8a9DI/AAAAAAAAAQc/qqt_d6k2o94fSx2qYdXfw7q0hTQWTDnYgCLcBGAs/s1600/Algrabya150-6C.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;"><b> Note</b>: Here we deliberately consider a <i>far generalisation</i> again, rather than ask for the number of dots in the 5-chain, say. The hope is that this will focus students' attention on the overall structure of the pattern rather than on the change from one chain to the next. Put another way, we are trying to see whether students will consider the pattern <i>generically</i>. Of course, it is perfectly legitimate to look at differences, ie to consider the pattern 'incrementally', but it's not particularly efficient here.</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;"><b>THURSDAY</b>: Here we look at some more dot patterns, but this time the Y-chains have morphed into single trees.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-QLUOTBuScm4/WqmiSAcEzJI/AAAAAAAAAQs/qBGkzPJKrYQEJmXXn9jxWXvnKtlx1U9ZgCLcBGAs/s1600/Algrabya150-6D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-QLUOTBuScm4/WqmiSAcEzJI/AAAAAAAAAQs/qBGkzPJKrYQEJmXXn9jxWXvnKtlx1U9ZgCLcBGAs/s1600/Algrabya150-6D.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">Yesterday, the relation between chain-length and the total number of dots turned out to be quadratic, in both cases. What about today? Can you tell just by looking at the pattern? Do the generic expressions for the 20-brown-dots trees confirm this?</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;"><b>FRIDAY</b>: Here we consider equivalent generic expressions for the number of dots in Fynn's 20-brown-dots tree. We try to relate them to the dot pattern - the first one is relatively straightforward, the second less so. We then try to transform one expression into the other.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-PD5Kiwkj7W8/WqsLY-7gA_I/AAAAAAAAAQ8/QbImIfe170g6ua06CZltdSpf-C50ix3ugCLcBGAs/s1600/Algrabya150-6E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-PD5Kiwkj7W8/WqsLY-7gA_I/AAAAAAAAAQ8/QbImIfe170g6ua06CZltdSpf-C50ix3ugCLcBGAs/s1600/Algrabya150-6E.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">We can show the structure contained in the expressions with annotated sketches like these:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-4WDJ_a0-hIY/Wqut0MirAEI/AAAAAAAAARQ/J9XHxcMzyeoHzyAxt6ynnH-fiS6znlH1QCLcBGAs/s1600/Algrabya150-6E-structures.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://4.bp.blogspot.com/-4WDJ_a0-hIY/Wqut0MirAEI/AAAAAAAAARQ/J9XHxcMzyeoHzyAxt6ynnH-fiS6znlH1QCLcBGAs/s400/Algrabya150-6E-structures.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">-</div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-80465657175364962412018-03-04T23:55:00.001+00:002018-03-09T18:23:40.662+00:00ALG 5In this week's set of ALG 5 tasks, we find specific unknown values by using the so-called bar model (curved bars in our case...) and by forming and equating algebraic expressions - and we look at alternative ways of forming the expressions.<br /><b>MONDAY</b>: The root ALG 5 task. A sketch might help or you might find another way to visualise or represent the situation - or you could use trial and improvement. You can view a dynamic version of the queue <a href="https://www.youtube.com/watch?v=6FwQhK6H85I">here</a>.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-a7_ueZwQC-M/Wp1ngM0fmeI/AAAAAAAAAN8/NozZUzh8jDcqtRev0MBk9gI2ia_ghquLACLcBGAs/s1600/Algrabya150-5A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-a7_ueZwQC-M/Wp1ngM0fmeI/AAAAAAAAAN8/NozZUzh8jDcqtRev0MBk9gI2ia_ghquLACLcBGAs/s1600/Algrabya150-5A.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>Note: there's plenty of scope for varying the task.<br />Here's an easy variant:<br /><i>When has Deka queued for 11 times as long as Eric?</i><br /><i>- </i><br />Note 2: The bar model (or in this case, 'bent rod model') is a powerful device for visualising quantities, but what if you decide to make a sketch and the proportions in your drawing turn out to be not that good? That's probably OK, as long as you don't expect instant answers from the model and are prepared to slow down and modify the drawing (on paper or in your head).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-HggbXLV7rCg/Wp71M1rbGBI/AAAAAAAAAOc/Sg9BLBS1AWQG857c7M4ME-16QRsZcy-tQCLcBGAs/s1600/ALG5-baaaaa.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="297" data-original-width="1600" height="118" src="https://2.bp.blogspot.com/-HggbXLV7rCg/Wp71M1rbGBI/AAAAAAAAAOc/Sg9BLBS1AWQG857c7M4ME-16QRsZcy-tQCLcBGAs/s640/ALG5-baaaaa.png" width="640" /></a></div>-<br /><b>TUESDAY</b>: Here we ask for a symbolic algebra approach. Constructing the expressions can be quite challenging, though we can check to see whether they make sense by making use of the values found on Monday, ie by checking to see whether we get a value of 30 when <i>e</i> = 10. Then, having got the expressions, we can consider how they could themselves be used to derive the value <i>e</i> = 10, eg by forming an equation and solving it in some way ....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NarKCpPE5TE/Wp3dLpboAKI/AAAAAAAAAOM/OSCGDKJjv80-MnM5Tcok4nCnPmSWp_v5ACLcBGAs/s1600/Algrabya150-5B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-NarKCpPE5TE/Wp3dLpboAKI/AAAAAAAAAOM/OSCGDKJjv80-MnM5Tcok4nCnPmSWp_v5ACLcBGAs/s1600/Algrabya150-5B.jpg" /></a></div>-<br /><b>WEDNESDAY</b>: In this variant, we solve the same problem by forming expressions in <i>d</i>, the time taken by Deka, instead of <i>e</i>, the time taken by Eric.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-r0dqEP8ojUM/Wp8nbJkTiuI/AAAAAAAAAOs/47vSwkdMTx8WOlX8bQaegk5qLqo1bE_kwCLcBGAs/s1600/Algrabya150-5C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-r0dqEP8ojUM/Wp8nbJkTiuI/AAAAAAAAAOs/47vSwkdMTx8WOlX8bQaegk5qLqo1bE_kwCLcBGAs/s1600/Algrabya150-5C.jpg" /></a></div>As well as solving the problem itself (whose solution we already know, of course), this gives us the opportunity to compare the two sets of expressions and to consider how an expression for <i>e</i> in terms of <i>d</i> is related to, and can be transformed into, the corresponding expression for <i>d</i> in terms of <i>e</i>.<br />-<br /><b>THURSDAY</b>: Here we use algebra to solve a slightly more complex problem which isn't quite as amenable to a bar-model approach as the previous task. So where previously we tried to give meaning to an algebraic approach, here we try to show its utility.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-H7HeuZ9RGZo/WqBcQxD-LjI/AAAAAAAAAO8/NqIfa_BbNIYplCItsVN0X4TxFl3UYBudwCLcBGAs/s1600/Algrabya150-5D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-H7HeuZ9RGZo/WqBcQxD-LjI/AAAAAAAAAO8/NqIfa_BbNIYplCItsVN0X4TxFl3UYBudwCLcBGAs/s1600/Algrabya150-5D.jpg" /></a></div> -<br /><b>FRIDAY</b>: Our final version of the ALG 5 task involves the same context but has a different structure. We again represent/solve it using the clock-diagram (or bar model) and in symbolic algebra - which approach do you find easier here?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-aEpeS8Pzd3Y/WqJi5s-25QI/AAAAAAAAAPc/CPL5tMICKl87VLOnC7Urg3R3F82VOWMngCLcBGAs/s1600/Algrabya150-5E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-aEpeS8Pzd3Y/WqJi5s-25QI/AAAAAAAAAPc/CPL5tMICKl87VLOnC7Urg3R3F82VOWMngCLcBGAs/s1600/Algrabya150-5E.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: left;">We can also approach the task using a mix of trial and improvement and analysis: </div><div class="separator" style="clear: both; text-align: left;">At present (with <i>d</i> = 20 and <i>e</i> = 10) Deka and Eric have queued for a total of 30 minutes. If we wind forward 5 minutes, say, this will add 10 minutes to the total, so if we wind forwards a total of 15 minutes (making <i>d</i> = 20+15 and <i>e</i> = 10+15) they will have queued for a total of 60 minutes. This movie might help students see when to 'stop' the clock: <a href="https://www.youtube.com/watch?v=5J7dkwWxC7A">Long lunch with Deka and Eric</a>. </div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">Another approach would be to argue like this:</div><div class="separator" style="clear: both; text-align: left;">30+30 = 60</div><div class="separator" style="clear: both; text-align: left;">31+29 = 60</div><div class="separator" style="clear: both; text-align: left;">....</div><div class="separator" style="clear: both; text-align: left;">.... </div><div class="separator" style="clear: both; text-align: left;">35+25 = 60.</div><div class="separator" style="clear: both; text-align: left;">- </div><div class="separator" style="clear: both; text-align: left;">These approaches are not purely empirical - they make use of the task's structure so they can be classed as algebraic, I think. But they don't make use of symbolic algebra, even though the last method could be said to embody <i>d</i>+<i>e</i>=30 and <i>d</i>–<i>e</i> = 10. This highlights a real dilemma: <b>it is not that easy to devise accessible tasks where the need for symbolic algebra is compelling</b>. However, as we shall see in later weeks, one strategy is to make the task more complex, so that symbolising provides a way of keeping track of the information.</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">Notice: There's an interesting symmetry in the clock-diagram when <i>d</i> + <i>e</i> = 60:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-rb0rh1p6R40/WqLIZmmRqEI/AAAAAAAAAPs/1DJfXRGQaL42-DrFMRmjF9WW_b-t6QWSQCLcBGAs/s1600/de60.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="718" data-original-width="750" height="306" src="https://2.bp.blogspot.com/-rb0rh1p6R40/WqLIZmmRqEI/AAAAAAAAAPs/1DJfXRGQaL42-DrFMRmjF9WW_b-t6QWSQCLcBGAs/s320/de60.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-27536211761509517292018-02-25T22:30:00.001+00:002018-03-01T22:39:49.048+00:00ALG 4This set of ALG 4 tasks might require a somewhat greater understanding of algebra than the previous three ALG tasks, if students are to grasp some of the algebraic relations, especially in their more general form. However, even relative novices should be able to engage with the tasks and construe patterns even if this might be at a more empirical level.<br /><b>MONDAY</b>: This is the root ALG 4 task for this week. It can be approached in quite an exploratory way - later variants become more focused.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-LhEQ-VcTzN4/WpM13dri2dI/AAAAAAAAAMM/F6eHBGgNyfcobRVMgPMXWh8A81r4Ts5AwCLcBGAs/s1600/Algrabya150-4A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-LhEQ-VcTzN4/WpM13dri2dI/AAAAAAAAAMM/F6eHBGgNyfcobRVMgPMXWh8A81r4Ts5AwCLcBGAs/s1600/Algrabya150-4A.jpg" /></a></div>Here it is worth dwelling for a while on the form of the three equations, before sketching any lines. Students who know about the form of linear functions may notice that the greater the slope of the corresponding lines, the higher the point of intersection with the <i>y</i>-axis (ie as <i>m</i> increases, so does <i>c</i>, in <i>y</i> = <i>mx</i> + <i>c</i>). In turn this might suggest that the lines are 'hinged' about a common point, which turns out to be the case.<br />Students without this kind of knowledge can still engage meaningfully with the task, by sketching the lines given by the equations and examining the result.<br />- <br /><b>TUESDAY</b>: In this variant we show the three lines and ask students to explain the fact that they go through a common point, using both algebraic and geometric arguments.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-qZYUsLMfVvo/WpSQ1mWbN5I/AAAAAAAAAMc/G-BPxosmCQE5MxMsRfhpzVlSCYzb3d4RACLcBGAs/s1600/Algrabya150-4B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-qZYUsLMfVvo/WpSQ1mWbN5I/AAAAAAAAAMc/G-BPxosmCQE5MxMsRfhpzVlSCYzb3d4RACLcBGAs/s1600/Algrabya150-4B.jpg" /></a></div> As can be seen, the lines all go through the point (-3, 0). We can verify this by letting <i>y</i> = 0 in each of the equations and then solving for <i>x</i>, or by solving for <i>x</i> in the general equation <i>y</i> = <i>ax</i> + 3<i>a</i> or <i>y</i> = <i>a</i>(<i>x</i> + 3) after letting <i>y</i> = 0.<br />We can also explain this in terms of the slope of the lines and the triangle formed by the points with coordinates (-3, 0), (0, 0) and (0, <i>c</i>). We can think of the slope as being the change in <i>y</i> resulting from a change of +1 in the value of <i>x</i> (below). This means if we increase <i>x</i> by +3, in moving from (-3, 0) to the origin, we increase <i>y</i> by +3×<i>m</i>, and hence cut the <i>y</i> axis at (0, 3<i>m</i>).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-xgubWiKxwIc/WpapEacBYOI/AAAAAAAAAM8/AumtSoH0lOwM8fDWm2-TwLDITHrGqsivwCLcBGAs/s1600/Algrabya150-4B-geo.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="358" data-original-width="556" height="257" src="https://4.bp.blogspot.com/-xgubWiKxwIc/WpapEacBYOI/AAAAAAAAAM8/AumtSoH0lOwM8fDWm2-TwLDITHrGqsivwCLcBGAs/s400/Algrabya150-4B-geo.jpg" width="400" /></a></div>-<br /><b>WEDNESDAY</b>: Here we find the common point for the lines given by another family of equations. But this time, rather than listing individual equations, the family is expressed in a general form. This might help students express the previous family in a similar form if they have not already done so (ie, as <i>y</i> = <i>ax</i> + 3<i>a</i>, as mentioned above). Also, as the two forms are very similar, it might help students see how, and perhaps why, the general form relates to the lines' common point.<br />[It might be argued that the form <i>y</i> = <i>a</i>(<i>x</i> – 2) is more salient than the given form <i>y</i> = <i>ax</i> – 2<i>a</i>, even though it diverges from the familiar <i>y</i> = <i>mx</i> + <i>c</i> form, as it immediately becomes apparent that <i>x</i> = 2 when <i>y</i> = 0, regardless of the value of <i>a</i>.] <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-SAqngCl_kCo/WpXr9UhWYuI/AAAAAAAAAMs/Fcj9UdzuehIfKc9Q5FXl2PLDrocBjF2VgCLcBGAs/s1600/Algrabya150-4C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-SAqngCl_kCo/WpXr9UhWYuI/AAAAAAAAAMs/Fcj9UdzuehIfKc9Q5FXl2PLDrocBjF2VgCLcBGAs/s1600/Algrabya150-4C.jpg" /></a></div>-<br /><b>THURSDAY</b>: Here we consider another family of equations whose general form, at least at first sight, seems rather different from the form of the two families we've considered so far. It turns out that the common point is not on the <i>x</i>-axis this time. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-lnhLmmzSUfE/Wpct-Hc72QI/AAAAAAAAANM/0XFTgP6KtCANSFdTZ-zdvdwpbZ2B1rwhQCLcBGAs/s1600/Algrabya150-4D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-lnhLmmzSUfE/Wpct-Hc72QI/AAAAAAAAANM/0XFTgP6KtCANSFdTZ-zdvdwpbZ2B1rwhQCLcBGAs/s1600/Algrabya150-4D.jpg" /></a></div>There are several ways of writing the general form of the family of equations, for example these:<br /><i>y</i> = <i>ax</i> + (3 – <i>a</i>), <i>y</i> = <i>a</i>(<i>x</i> – 1) + 3, <i>y</i> – 3 = <i>a</i>(<i>x</i> – 1).<br />What are the affordances of each?<br />-<br /><b>FRIDAY</b>: Here we reverse the process, by asking for the family of lines that goes through a specific point.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-QdcG-wfYmhs/Wph-YWw5fKI/AAAAAAAAANc/TkppB5_5rM8EKKhiACcT38IfchvX6yFdgCLcBGAs/s1600/Algrabya150-4E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-QdcG-wfYmhs/Wph-YWw5fKI/AAAAAAAAANc/TkppB5_5rM8EKKhiACcT38IfchvX6yFdgCLcBGAs/s1600/Algrabya150-4E.jpg" /></a></div> We can approach this in different ways. One way is to find the equations of some specific lines through (2, 5) and look for a pattern. Another would be to look back at Thursday's task and see what's general about it, ie how does the general equation (in whatever form) relate to the common point?<br />Another way of thinking about this is to consider lines through the origin (ie of the form <i>y</i> = <i>ax</i>) and to think of translating the lines 2 across and 3 up - how do these steps transform the equation?<br />An obvious extension of the task would be to consider a general point, with coordinates (<i>u</i>, <i>v</i>), say. If you've managed to solve the task for (2, 5), this should be fairly easy ....ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-32734979430119391802018-02-18T22:56:00.001+00:002018-02-23T16:31:44.696+00:00ALG 3In this week's set of tasks we try to get a better feel for a linear function by comparing it with a non-linear function. We see that as the independent variable changes at a steady rate, the values of the functions change at different rates, with one of them changing steadily.<br />The functions are set in a geometric context and we compare them geometrically (in two different ways, one of which is much more salient than the other), by expressing them symbolically, and by looking at numerical values in an ordered table.<br />-<br /><b>MONDAY</b>: This is the root task for the week. It can be quite tricky ....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-yvrdB4LW8Aw/WooDnjab81I/AAAAAAAAAI4/EojKe0bMS_oxYdqNvPvVy1lgtqllXrbpQCLcBGAs/s1600/Algrabya150-3A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-yvrdB4LW8Aw/WooDnjab81I/AAAAAAAAAI4/EojKe0bMS_oxYdqNvPvVy1lgtqllXrbpQCLcBGAs/s1600/Algrabya150-3A.jpg" /></a></div>After students have drawn the shape for <i>u</i> = 3, you might want to show this:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-x9lIpavfSNI/WosSjE0ISnI/AAAAAAAAAJI/n2q0WxHiJEIyxcM6WXzmI9vbTJY_1BdEACLcBGAs/s1600/Algrabya150-3A-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://3.bp.blogspot.com/-x9lIpavfSNI/WosSjE0ISnI/AAAAAAAAAJI/n2q0WxHiJEIyxcM6WXzmI9vbTJY_1BdEACLcBGAs/s400/Algrabya150-3A-ans.jpg" width="400" /></a></div>-<br /><b>TUESDAY</b>: Here we compare the area of our cross-shape with the area of another cross-shape:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-7VLPI6TcBWM/WotXyU-t-2I/AAAAAAAAAJY/lUhN5TAHLe0b9Wu4JmkN4Rmr50OOnob1gCLcBGAs/s1600/Algrabya150-3B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-7VLPI6TcBWM/WotXyU-t-2I/AAAAAAAAAJY/lUhN5TAHLe0b9Wu4JmkN4Rmr50OOnob1gCLcBGAs/s1600/Algrabya150-3B.jpg" /></a></div>You might want to present this on two separate slides, starting with the edited version below, so that students are not channeled too soon into answering the more structured second part. It is worth giving students plenty of time to explore the question, "Which shape has the larger area?".<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-r-x_uWoS3O0/WovVikpctxI/AAAAAAAAAJo/5XuxNW2ex0EWkCP0JJuzfll6corAzTVYgCLcBGAs/s1600/Algrabya150-3Bi.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://2.bp.blogspot.com/-r-x_uWoS3O0/WovVikpctxI/AAAAAAAAAJo/5XuxNW2ex0EWkCP0JJuzfll6corAzTVYgCLcBGAs/s400/Algrabya150-3Bi.jpg" width="400" /></a></div>After students have explored the task, they can check their hunches and confirm their insights with this <i>Dynamic crosses</i> <a href="https://www.youtube.com/watch?v=ucaukhX5Lgs">movie</a>. Some clips from the movie are shown below:<br /><a href="https://3.bp.blogspot.com/-Upz88ZKI4t4/WoxLUxTRF8I/AAAAAAAAAKA/hm1yhbHrQTkwcBPUlB_C9436y4Am3mu0gCLcBGAs/s1600/clips-u2468-noQ.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="1000" height="249" src="https://3.bp.blogspot.com/-Upz88ZKI4t4/WoxLUxTRF8I/AAAAAAAAAKA/hm1yhbHrQTkwcBPUlB_C9436y4Am3mu0gCLcBGAs/s640/clips-u2468-noQ.png" width="640" /></a>-<br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><b>WEDNESDAY</b>: Here we repeat Tuesday’s task but we treat the cross-shapes as the nets of two open boxes. </span><br /><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"></span></div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-DhRER02XuHc/WoxvLSlcWdI/AAAAAAAAAKQ/DGMM_0ny2ZcwLmZxbaVIGWNaGSYUFJzXQCLcBGAs/s1600/Algrabya150-3C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-DhRER02XuHc/WoxvLSlcWdI/AAAAAAAAAKQ/DGMM_0ny2ZcwLmZxbaVIGWNaGSYUFJzXQCLcBGAs/s1600/Algrabya150-3C.jpg" /></a></div><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">For one of the boxes, as <i>u</i> increases, the height of the box increases but the (square) base stays the same; for the other box, its base gets larger (in both dimensions) but the height stays the same. Thus one box changes in two dimensions simultaneously, while the other changes in only one. </span></div><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">These changes are far less obvious when one just looks at the 2D nets, unless one re-configures them. For example, the first cross-shape is changing in two directions as <i>u</i> changes. However, if one cuts and re-joins the net like this (below), the change can be seen to occur in just one dimension.</span></div><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"></span></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-g0Kn7yZNhYo/WoxvvatuCzI/AAAAAAAAAKY/gZ_AX8z9YOYJfUwATJ8Pqt2lPKjmrP6rgCLcBGAs/s1600/Algrabya150-3C-net-jig.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="143" data-original-width="467" src="https://2.bp.blogspot.com/-g0Kn7yZNhYo/WoxvvatuCzI/AAAAAAAAAKY/gZ_AX8z9YOYJfUwATJ8Pqt2lPKjmrP6rgCLcBGAs/s1600/Algrabya150-3C-net-jig.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">We can do the same for the net of the other box. As <i>u</i> increases, the net expands in two dimensions, because of the <i>u</i> by <i>u</i> square (tinted).</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-0klSPMxhfdo/Wo1uGWsfxaI/AAAAAAAAAKo/Tx2nobeYP2oVKB-7_AQKLZJHqeU2SSuowCLcBGAs/s1600/Algrabya150-3C-net-jig2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="164" data-original-width="620" src="https://4.bp.blogspot.com/-0klSPMxhfdo/Wo1uGWsfxaI/AAAAAAAAAKo/Tx2nobeYP2oVKB-7_AQKLZJHqeU2SSuowCLcBGAs/s1600/Algrabya150-3C-net-jig2.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">-</span><br /><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><b>THURSDAY</b>:This is Thursday's variant on the ALG 3 task. By now, students will, hopefully, have developed a good feel for how the areas of the two cross-shapes change with </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>: eg, that the first shape's area increases steadily for steady increases in </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>; and that the second shape's area changes more and more rapidly as </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span> increases, and so, though its area is smaller for values like <i>u</i> = 2, it overtakes the other area at some point, namely when <i>u</i> gets past 5.</span><br /><div class="separator" style="clear: both; text-align: center;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><a href="https://2.bp.blogspot.com/-2PPqv4PV-js/Wo3mHciAP0I/AAAAAAAAAK4/RdqvjZ3NvikWx42NY0cmlHFmHmehV-a8gCLcBGAs/s1600/Algrabya150-3D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-2PPqv4PV-js/Wo3mHciAP0I/AAAAAAAAAK4/RdqvjZ3NvikWx42NY0cmlHFmHmehV-a8gCLcBGAs/s1600/Algrabya150-3D.jpg" /></a></span></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">So in today's task, we represent the areas symbolically, with the aim of making these ideas about the changing areas more explicit and, conversely, giving meaning to the resulting algebraic expressions. This may also help students to get a better sense of the general form of what turn out to be linear and quadratic expressions.</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">- </span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><b>FRIDAY</b>: Here we try to consolidate some of the burgeoning ideas by revisiting the numerical values found in Tuesday's task, and judiciously extending these with the aid of an ordered table.</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-a3ZwmetcVVw/Wo9QMTtvXtI/AAAAAAAAALI/Yie6paLe4KoZzWHgdh6W9YVKzA1GtQltwCLcBGAs/s1600/Algrabya150-3E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-a3ZwmetcVVw/Wo9QMTtvXtI/AAAAAAAAALI/Yie6paLe4KoZzWHgdh6W9YVKzA1GtQltwCLcBGAs/s1600/Algrabya150-3E.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"> A possible next step would be to graph these sets of values - both as a way of illuminating the nature of the (linear and quadratic) relations, and as a way of giving meaning to graphs. And it would be worth relating this to the <i>Dynamic crosses</i> <a href="https://www.youtube.com/watch?v=ucaukhX5Lgs">movie</a> (with the mouse over the play/pause button).</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The completed table looks like this:</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-qyPDO2pSvkY/WpAJK1ibxJI/AAAAAAAAALY/0TOxE3ieTxcQtq1AJ1Kselg3971YCGLtQCLcBGAs/s1600/Algrabya150-3E-table.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="232" data-original-width="316" src="https://4.bp.blogspot.com/-qyPDO2pSvkY/WpAJK1ibxJI/AAAAAAAAALY/0TOxE3ieTxcQtq1AJ1Kselg3971YCGLtQCLcBGAs/s1600/Algrabya150-3E-table.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"> As we can see, as <i>u</i> increases by 1, 20</span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>+25 increases by the same amount, whether we start from </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>=5 or </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>=10, whereas 20</span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>+</span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>2 increases by an increasing amount ....</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">We can illustrate this with a graph, such as this off-the-peg version from Excel.</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-rBqPk-tW5lM/WpAKqBqaBJI/AAAAAAAAALk/3LR81Xs0pF87XMC83h0iqdbgzVTAAW4xgCLcBGAs/s1600/ALG-3E-graph.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="846" data-original-width="610" src="https://3.bp.blogspot.com/-rBqPk-tW5lM/WpAKqBqaBJI/AAAAAAAAALk/3LR81Xs0pF87XMC83h0iqdbgzVTAAW4xgCLcBGAs/s1600/ALG-3E-graph.png" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The graph raises some interesting questions:</span><br /><ul><li><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The blue line appears to be straight. Is this really the case and what would this mean (algebraically, and in the context of the cross-shape's area)? </span></li><li><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The red line seems to consist of ever-steeper line segments. Does this make sense? What if I had chosen different data-points - would I get different line-segments?? </span></li><li><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">What happens to the two graphs for negative values of <i>u</i> (algebraically, and in terms of the cross-shapes and their areas)?</span></li></ul><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">It can also be illuminating to see how equivalent forms of the algebraic expressions relate to (and can be derived from) the shapes of the crosses. This is for the first cross-shape:</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-jgOay4obLZk/WpAceBOUoCI/AAAAAAAAAL0/siUb3tP50Ekhk86rUvpw0A6f6FAper3rgCLcBGAs/s1600/Algrabya150-3C-net-jig-express.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="169" data-original-width="600" src="https://1.bp.blogspot.com/-jgOay4obLZk/WpAceBOUoCI/AAAAAAAAAL0/siUb3tP50Ekhk86rUvpw0A6f6FAper3rgCLcBGAs/s1600/Algrabya150-3C-net-jig-express.jpg" /> </a></div><div class="separator" style="clear: both; text-align: left;"> And this is for the second cross-shape: </div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-szqH1l_aaBE/WpAcvYApKhI/AAAAAAAAAL4/bNkEmZJZKQcpyF3v3YZIpHky6imC6mNPgCLcBGAs/s1600/Algrabya150-3C-net2-jig-express.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="299" data-original-width="525" src="https://3.bp.blogspot.com/-szqH1l_aaBE/WpAcvYApKhI/AAAAAAAAAL4/bNkEmZJZKQcpyF3v3YZIpHky6imC6mNPgCLcBGAs/s1600/Algrabya150-3C-net2-jig-express.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><br /></span></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-44054447094164285672018-02-15T12:56:00.002+00:002018-02-18T17:52:14.856+00:00ALG 2<div class="separator" style="clear: both; text-align: center;"></div>Here is an open version of our ALG 2 root task - a slightly more structured version will appear on Monday, with other variants later in the week.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-CWkxhu1Zf18/WoAjoTEpk6I/AAAAAAAAAGA/llLfnibVIagjGix-m4kQqnLqXjiSfLvxQCLcBGAs/s1600/Algrabya150-2Aopenopen.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-CWkxhu1Zf18/WoAjoTEpk6I/AAAAAAAAAGA/llLfnibVIagjGix-m4kQqnLqXjiSfLvxQCLcBGAs/s1600/Algrabya150-2Aopenopen.jpg" /></a></div>The task starts with the idea that an unknown can vary, rather than the more common idea in school mathematics that it has a specific value that we want to find.<br />Also, rather than immediately asking students to perform various actions on the unknown (such as collecting like terms, manipulating or evaluating expressions, forming and solving equations), the task may help students realise that they can sometimes make sense of expressions from the outset - especially in this case, where the given expressions (<i>x</i> and 2<i>x</i>) are fairly simple, and thus might fairly readily lead to an interpretation like 'one angle in the triangle is twice the size of another angle'.<br />A movie of the triangle can be found <span style="color: #3d85c6;"><a href="https://www.youtube.com/watch?v=9Ua6wIwRwcA">here</a></span>. <br />-<br />MONDAY: Here is this week's root task. It is more structured than the version above but it should still help students discern the general relationship between two of the triangle's angles (ie between the expressions <i>x</i> and 2<i>x</i>). Part b) provides a shift towards the use of algebraic procedures, by asking students to find a particular (critical) value of <i>x</i>. We can solve this algebraically by forming the equation <i>x</i> + 2<i>x</i> = 180. However, it is important to acknowledge that we can also find <i>x</i> quite easily by other means, eg trial and improvement. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-11znlvWKxGc/WoFs_cMiGOI/AAAAAAAAAGQ/SqgkK7GN_NUVGR7ibhI9pi3a2-WejO-MwCLcBGAs/s1600/Algrabya150-2A.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-11znlvWKxGc/WoFs_cMiGOI/AAAAAAAAAGQ/SqgkK7GN_NUVGR7ibhI9pi3a2-WejO-MwCLcBGAs/s1600/Algrabya150-2A.jpg" /></a></div> We have made two movies showing the triangle as <i>x</i> varies. In <a href="https://www.youtube.com/watch?v=9Ua6wIwRwcA">one </a><a href="https://www.youtube.com/watch?v=9Ua6wIwRwcA">movie</a> the value of <i>x</i> goes up to the 'permissible' value. In the <a href="https://www.youtube.com/watch?v=PK2_6S1VTwY">other movie</a> the value of <i>x</i> goes beyond this value: what happens to the triangle when this occurs, and should such values be 'allowed'?!<br />-<br />TUESDAY: Like Monday's part b), we don't need formal algebra to solve this variant of ALG 2, but it gives us the opportunity to construct two more equations in <i>x</i>, so it becomes interesting to compare them and to discuss ways in which they can be solved.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-oNvjNibJWhk/WoIcqtLRv9I/AAAAAAAAAGg/JcugcRYfR9s_YPSMhxzDfHFli1N96SJdACLcBGAs/s1600/Algrabya150-2B.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-oNvjNibJWhk/WoIcqtLRv9I/AAAAAAAAAGg/JcugcRYfR9s_YPSMhxzDfHFli1N96SJdACLcBGAs/s1600/Algrabya150-2B.jpg" /></a></div> It is also worth engaging with the geometry of the situation. For example, why is the triangle always weighted to the right in the given diagrams? Also, we can see from the given triangles that the value of <i>x</i> will be close to 35 for one of the isosceles triangles, and somewhat less than 50 for the other - do our algebraic approaches confirm this?<br />-<br />WEDNESDAY: Having looked for isosceles triangles, this variant involves finding values of <i>x</i> for which the triangle is right-angled:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-2ikf-I8l2uk/WoMljrt48fI/AAAAAAAAAGw/81W3xPQdgtcK_UxvHkDx8tDJRFbNgskGwCLcBGAs/s1600/Algrabya150-2C.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-2ikf-I8l2uk/WoMljrt48fI/AAAAAAAAAGw/81W3xPQdgtcK_UxvHkDx8tDJRFbNgskGwCLcBGAs/s1600/Algrabya150-2C.jpg" /></a></div>We can find <i>x</i> by using trial and improvement, say, but we can also proceed algebraically by forming equations to represent the geometric relations resulting from angle B or angle C being a right angle. This gives 2<i>x</i> = 90 and <i>x</i> + 2<i>x</i> = 90 (or <i>x</i> + 2<i>x</i> + 90 = 180). Similarly, we can form equations for the earlier variants: <i>x</i> + 2<i>x</i> = 180 (ALG 2A), <i>x</i> + 2<i>x</i> + 2<i>x</i> = 180 and <i>x</i> + <i>x</i> + 2<i>x</i> = 180 (ALG 2B). These equations are all fairly easy to solve, but what is particularly nice about the more complex ones is that we may be able to persuade students of their <i>utility</i>: by forming and writing down equations we can ease the strain on our working memory.<br />It can also be useful to discuss and compare different ways of solving the equations, eg by inspection, by trial and improvement, or by transforming the equations.<br />-<br />THURSDAY: Here we change the relationship between the base angles of the triangle from <i>x</i>˚, 2<i>x</i>˚ to <i>x</i>˚, 4<i>x</i>˚. This gives students a chance to recap on earlier ideas and methods, as well as perhaps starting to think about how changes in the angle relationship might affect the size of the various angles that we are trying to find.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-R8KIm_S3kXU/WoS8oYzF2cI/AAAAAAAAAHE/c6-j1yaYfogXlExWDv4pVwmLuRjJOrqlQCLcBGAs/s1600/Algrabya150-2D.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-R8KIm_S3kXU/WoS8oYzF2cI/AAAAAAAAAHE/c6-j1yaYfogXlExWDv4pVwmLuRjJOrqlQCLcBGAs/s1600/Algrabya150-2D.jpg" /></a></div>Thursday enjoyed a couple of bonus tasks (below). The first can be varied in quite nice ways. The other [based on a seminal idea from David Fielker, writing some years ago in <i>Mathematics Teaching</i>] says something about the freedom that symbolic algebra can give us.<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-TkH5UxByp2k/WoXsF4VLyJI/AAAAAAAAAHg/vxaAOMpWMZUZ8rcOiygnJWqwplXdvfBnACLcBGAs/s1600/2A-bonus.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="295" data-original-width="888" height="212" src="https://1.bp.blogspot.com/-TkH5UxByp2k/WoXsF4VLyJI/AAAAAAAAAHg/vxaAOMpWMZUZ8rcOiygnJWqwplXdvfBnACLcBGAs/s640/2A-bonus.png" width="640" /></a></div>And here's another bonus (actually posted on Friday) which continues the regular polygon theme:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-3Fvs1r68BTE/WomqtsKAaFI/AAAAAAAAAIA/clE-6AZgn4ACbpduhbVeWjay4Q2N9Dt3wCLcBGAs/s1600/poly-exterior2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="208" src="https://3.bp.blogspot.com/-3Fvs1r68BTE/WomqtsKAaFI/AAAAAAAAAIA/clE-6AZgn4ACbpduhbVeWjay4Q2N9Dt3wCLcBGAs/s320/poly-exterior2.jpg" width="320" /></a></div>-<br />FRIDAY: This is our final (for now?) ALG 2 variant. It gives plenty of scope for consolidation and for exploring the relationship between the base angles, which is expressed here in a more general way (with the aid of parameters <i>a</i> and <i>b</i>). Let us know what you or your students find.<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-g3a5oRZSCmk/WoXuyDQYd1I/AAAAAAAAAHw/B40bLKnx_DsVdIOi8zDKfM6uHcZbIgk-QCLcBGAs/s1600/Algrabya150-2E.jpg" /></div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">SUNDAY Supplement: Here's a set of three tasks still involving the base angles of a triangle but where there's an inverse proportional relationship. Some of the resulting triangle-families can be quite surprising.</div><div class="separator" style="clear: both; text-align: left;">Here's the first task (below). Part <b>a)</b> is to get students to start thinking about the scenario. The given information means that <i>a</i>/20 = 50, so <i>a</i> = 1000. It might seem surprising that a relation involving '1000' works in the context of triangles, where the interior angle sum is a mere 180˚. Of course, the relation ∠A = <i>x</i>˚, ∠B = 1000˚/<i>x</i> doesn't work for all values of <i>x</i>. For example, if <i>x</i> = 1, then ∠B would be 1000˚. However it turns out the relation works for this range of values (approximately): </div><div class="separator" style="clear: both; text-align: left;">6 ≤ <i>x</i> ≤ 174.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-G_lAUGFu8CY/WomsoX7qHmI/AAAAAAAAAIM/Wszw82YKruYIRyZMFIs0yheOPg9HNtsPgCLcBGAs/s1600/Algrabya150-2Asunday01.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-G_lAUGFu8CY/WomsoX7qHmI/AAAAAAAAAIM/Wszw82YKruYIRyZMFIs0yheOPg9HNtsPgCLcBGAs/s1600/Algrabya150-2Asunday01.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">It is worth spending quite a lot of time on part <b>b)i</b>, to get a feel for how the base angles, and the overall shape of the triangle, change. The triangle turns out to be fairly 'flat'. When <i>x</i> is 'small', a change in <i>x</i> produces a rapid change in ∠B; when <i>x</i> is fairly 'large' ∠B seems to change hardly at all. A movie of the changing triangle can be found <a href="https://www.youtube.com/watch?v=ThoSbTK03jo">here</a>.</div><div class="separator" style="clear: both; text-align: left;">A good way to start part <b>b)ii</b> is to adopt a visual approach, eg by running the aforementioned movie and pausing it when the triangle looks to be isosceles. This happens in three places. A more precise estimate can then be found by using a numerical approach, which can be made more and more precise with the aid of a spreadsheet (with a column for <i>x</i>, for 1000/<i>x</i> and for 180 – <i>x</i> – 1000/<i>x</i>).</div><div class="separator" style="clear: both; text-align: left;">In the isosceles case where ∠A = ∠B, <i>x</i> = √1000. The value of <i>x</i> for the other two cases (∠A = ∠C and ∠C = ∠B) can be found algebraically by forming an equation in <i>x</i> and writing it in the usual quadratic form. However, it doesn't factorise....</div><div class="separator" style="clear: both; text-align: left;">- </div><div class="separator" style="clear: both; text-align: left;">Here's the second Sunday Supplement task. This is very exploratory. You might want to start by seeing what happens with some fairly 'random' values of <i>a</i>, and then try some more 'informed' values.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-2N4fuIglEwU/Wom3tSNntBI/AAAAAAAAAIc/1jXcL07jPWUXVeovyTdmT5IVSnQUmJsSACLcBGAs/s1600/Algrabya150-2Asunday02.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-2N4fuIglEwU/Wom3tSNntBI/AAAAAAAAAIc/1jXcL07jPWUXVeovyTdmT5IVSnQUmJsSACLcBGAs/s1600/Algrabya150-2Asunday02.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">After some explorations, you might want to look at two further movies, where <a href="https://www.youtube.com/watch?v=ouPjdKXrJPg"><i>a</i> = 100</a> and <a href="https://www.youtube.com/watch?v=Aonr9c4xcf0"><i>a</i> = 1</a>.</div><div class="separator" style="clear: both; text-align: left;">The choice of '100' is very deliberate and you might like to predict what effect this has on the shape of the triangle. The effect of choosing <i>a</i> = 1 is very strange.... - but the movie won't win any Oscars.</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">The third Sunday Supplement:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-5tN1a93-zxA/Wom5-stMPPI/AAAAAAAAAIo/Ey_g3w0wGbY2Pxbywh9AXqSAtus-yFGLgCLcBGAs/s1600/Algrabya150-2Asunday03.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-5tN1a93-zxA/Wom5-stMPPI/AAAAAAAAAIo/Ey_g3w0wGbY2Pxbywh9AXqSAtus-yFGLgCLcBGAs/s1600/Algrabya150-2Asunday03.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">In this task the numbers are nicer. You might be able to see the value of <i>x</i> for which ∠A = ∠B. The value of <i>x</i> for the other two isosceles cases (∠A = ∠C and ∠C = ∠B) can be found by forming a quadratic equation which this time does factorise. Or you could adopt a visual/numeric approach again, perhaps with the aide of <a href="https://www.youtube.com/watch?v=aJLsk3zv5rI">this</a> movie, or <a href="https://www.youtube.com/watch?v=VlSDZm09cg4">this</a> second version which provides a trace of vertex C.</div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-15759552069684742432018-02-11T13:46:00.001+00:002018-02-11T13:49:44.447+00:00SingalongAll together now ....<br />- <br /><span style="font-size: x-small;">[with thanks to Steve Miller and The Steve Miller Band]</span><br /><span style="font-size: x-small;">[and to Wren]</span><br /><span style="font-size: x-small;">- </span><br /><div class="separator" style="clear: both; text-align: left;"><iframe allowFullScreen='true' webkitallowfullscreen='true' mozallowfullscreen='true' width='320' height='266' src='https://www.blogger.com/video.g?token=AD6v5dzpDgCOxbaxuAUFtaqkMxZrSMXOeQxcvWGl71S5F5nP6y47djx34D7K0yem25QLGqqufqS6cSa9kNF5xrmPsg' class='b-hbp-video b-uploaded' FRAMEBORDER='0' /></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-7951505657704461822018-02-04T23:17:00.003+00:002018-02-08T23:47:21.071+00:00ALG 1<style><!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:Calibri; panose-1:2 15 5 2 2 2 4 3 2 4; mso-font-charset:0; mso-generic-font-family:auto; mso-font-pitch:variable; mso-font-signature:-536870145 1073786111 1 0 415 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-unhide:no; mso-style-qformat:yes; mso-style-parent:""; margin:0mm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} .MsoChpDefault {mso-style-type:export-only; mso-default-props:yes; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} @page WordSection1 {size:612.0pt 792.0pt; margin:72.0pt 72.0pt 72.0pt 72.0pt; mso-header-margin:36.0pt; mso-footer-margin:36.0pt; mso-paper-source:0;} div.WordSection1 {page:WordSection1;} --></style> <br /><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="color: blue;"><span style="color: #274e13;">MONDAY:</span></span> This is the ‘root’ task for the week. </span></div><div class="MsoNormal"><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-T6Y4F5DH5Lk/WneLOWnZKII/AAAAAAAAAEs/UKVWw5JHBMottXPO-yM2dJcrsH1j_LLQwCLcBGAs/s1600/Algrabya150-1A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-T6Y4F5DH5Lk/WneLOWnZKII/AAAAAAAAAEs/UKVWw5JHBMottXPO-yM2dJcrsH1j_LLQwCLcBGAs/s1600/Algrabya150-1A.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The t<span style="font-size: small;">ask i</span>s designed to help students focus on the structure of the equation, rather than to reach for a familiar procedure for solving it (be it an informal strategy like trial and improvement or the use of formal transformation rules). We are not really interested in the solution per se - although it is fairly easy to find, if one uses the given information and observes the equation's structure.</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">- </span></div><span style="font-family: "times new roman"; font-size: 12.0pt;"><span style="color: blue;"><span style="color: #274e13;">TUESDAY:</span></span> Tweaking a term in an equation can help us see what role the term plays in the equation and how it interacts with other terms. Here's another tweak of the original equation, one whose effect is relatively easy to see:</span><br /><div class="separator" style="clear: both; text-align: center;"><span style="font-family: "times new roman"; font-size: 12.0pt;"><a href="https://2.bp.blogspot.com/-LfwyQbUHagY/WnjizxIIrJI/AAAAAAAAAE4/JPaPulOSoLwhu54fgP3PMHruW7amWre2ACLcBGAs/s1600/Algrabya150-1B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-LfwyQbUHagY/WnjizxIIrJI/AAAAAAAAAE4/JPaPulOSoLwhu54fgP3PMHruW7amWre2ACLcBGAs/s1600/Algrabya150-1B.jpg" /></a></span></div><span style="font-family: "times new roman"; font-size: 12.0pt;">You or students might like to think of some other tweaks - which ones seem easy to construe, and which are more opaque?</span><br /><span style="font-family: "times new roman"; font-size: 12.0pt;">-</span><br /><span style="font-family: "times new roman"; font-size: 12.0pt;">WEDNESDAY: The tweak lined up for the middle of the week is nice and simple. Actually, I've decided on an extra tweak. Both still quite simple, but they seem to work in subtly different ways. What's going on?!</span><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-Ha_JDljz6mY/WnpEzjXcB6I/AAAAAAAAAFQ/I48klr4_maEOaNR9Tbq1kRujhtzsghUEwCLcBGAs/s1600/ALG-1Cab.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="371" data-original-width="1112" height="212" src="https://2.bp.blogspot.com/-Ha_JDljz6mY/WnpEzjXcB6I/AAAAAAAAAFQ/I48klr4_maEOaNR9Tbq1kRujhtzsghUEwCLcBGAs/s640/ALG-1Cab.png" width="640" /></a></div>It might strike us as odd (at least momentarily), that doing the same thing <i>additively</i> to the terms 77 and 203 has no effect, while <i>multiplicatively</i> it does ....<br />- <br />THURSDAY: This is Thursday's variant. If it seems too simple, you could change 42 to 84, say, or wouldn't that make enough difference? I'd be interested to know.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-CnU8FJW01lw/Wntz5UHbiJI/AAAAAAAAAFg/Mi6OHT62ecwR9bk6RsAiyXERuFOYq2XRwCLcBGAs/s1600/Algrabya150-1D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-CnU8FJW01lw/Wntz5UHbiJI/AAAAAAAAAFg/Mi6OHT62ecwR9bk6RsAiyXERuFOYq2XRwCLcBGAs/s1600/Algrabya150-1D.jpg" /></a></div>-<br /><div class="separator" style="clear: both; text-align: center;"></div><span style="font-family: "times new roman";">FRIDAY: Here's the last ALG 1 variant (for now?). With some classes, you might decide to start with an open task such as this, rather than the earlier, more closed variants.</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-gRsFNEbRcc0/WnzeThh3lQI/AAAAAAAAAFw/_2D8RMyKVmg-0VvNgh3oMpUXMOdycCEUwCLcBGAs/s1600/Algrabya150-1E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-gRsFNEbRcc0/WnzeThh3lQI/AAAAAAAAAFw/_2D8RMyKVmg-0VvNgh3oMpUXMOdycCEUwCLcBGAs/s1600/Algrabya150-1E.jpg" /></a></div><span style="font-family: "times new roman";">And you might (at some stage) want something more open still, eg along the lines of "Change one of the numbers in the equation - what does this do to the solution?". You might also want to investigate equations with a different form from <i>ax</i> + <i>b</i> = <i>c</i>.</span> ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com9tag:blogger.com,1999:blog-9058930865720822726.post-34619046119754540812018-02-01T11:53:00.000+00:002018-02-01T11:59:58.653+00:00About this site<div class="entry-content"><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-hJuGPbpkClA/WnL_WkMrVQI/AAAAAAAAAEY/Y_hEw-0AvlATG-ka7uJ-jU3He6i8UQQWQCLcBGAs/s1600/alge2.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="239" data-original-width="547" height="137" src="https://3.bp.blogspot.com/-hJuGPbpkClA/WnL_WkMrVQI/AAAAAAAAAEY/Y_hEw-0AvlATG-ka7uJ-jU3He6i8UQQWQCLcBGAs/s320/alge2.png" width="320" /></a></div><span style="font-size: small;">The aim of this site is to publish tasks designed to help school students get a better feel for algebra.</span><br /><span style="font-size: small;">The plan is to post a new task at the beginning of a week, and then to post variations of the task during the week, together with comments and guidance. Let us know how the tasks work with your students.</span></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0