Weekly Challenge 7: Gradient match

7060 /www/nrich/html/content/id/7060/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/7092&part=">Try this next</a> </li> <li> <a href="http://en.wikipedia.org/wiki/Mean_value_theorem">Read all about it</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/149887.html?1287990619">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/6603&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/7097&part=solution">Last week's solution</a> </li> </ul> <div> On 1cm paper neatly draw a pair of coordinate axes about 10cm by 10 cm and accurately mark the points  $(0, 0)$, $(8, 8)$  and $(4, 6)$. Smoothly draw any curve between $(0, 0)$ and $(8, 8)$. Locate a point with gradient $1$ Smoothly draw any curve between $(0, 0)$ and $(4, 6)$. Locate a point with gradient $1.5$ Smoothly draw any curve between $(4, 6)$ and $(8, 8)$. Locate a point with gradient $0.5$   How did I know that the curves you drew would necessarily have such points? Create a geometric proof. </div> <div class="framework"> Did you know ... ? Geometric proofs are useful for gaining intuition concerning calculus, but concepts concerning 'smoothness' need to be made clear and fully mathematical before analytical proofs of such statements can be created. These ideas form the basis of first year undergraduate courses in calculus and analysis which include proof of the 'Mean Value Theorem'.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">There are various intuitive ways to think about such results but creating really clear arguments is somewhat more difficult. The descriptions given here do not constitute proofs as such, but do introduce advanced analytical ways of thinking which are refined at university. Consider the first case; the others are similar. Clearly the special case of the curve joining the two points $(0, 0)$ and $(8, 8)$ with a straight line has gradient $1$ everywhere. Any other curve between these two points can be considered as a deformed version of this line. Wherever the curve is deformed outwards point bulges will necessarily occur. Imagine dragging the line $y=x$ up or down. It will cut through each bulge but eventually pass out of each bulge. As it passes out of each bulge it will touch each bulge at a single point. These are the points with gradient 1. <mdo:image height="355" width="449" src="88image.png" alt=""></mdo:image> Analytical proofs proceed along these sorts of lines: Sketch: Imagine the curve being sketched starting from the origin. Imagine that the gradient of your curve is always less than or equal to some number $M$ which satisfies $M< 1$. Then in $8$ units of $x$ the $y$ value of the curve can increase by at most $8M$, which is less than $8$. Thus, it could not pass through the point $(8, 8)$; therefore the maximum achieved gradient cannot be less than $1$. Similarly the minimum achieved gradient cannot be greater than $1$. It is thus intuitively clear that a gradient of $1$ is achieved somewhere. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">There are various intuitive ways to think about such results but creating really clear arguments is somewhat more difficult. The descriptions given here do not constitute proofs as such, but do introduce advanced analytical ways of thinking which are refined at university. Consider the first case; the others are similar. Clearly the special case of the curve joining the two points $(0, 0)$ and $(8, 8)$ with a straight line has gradient $1$ everywhere. Any other curve between these two points can be considered as a deformed version of this line. Wherever the curve is deformed outwards point bulges will necessarily occur. Imagine dragging the line $y=x$ up or down. It will cut through each bulge but eventually pass out of each bulge. As it passes out of each bulge it will touch each bulge at a single point. These are the points with gradient 1. <mdo:image height="355" width="449" src="88image.png" alt=""></mdo:image> Analytical proofs proceed along these sorts of lines: Sketch: Imagine the curve being sketched starting from the origin. Imagine that the gradient of your curve is always less than or equal to some number $M$ which satisfies $M< 1$. Then in $8$ units of $x$ the $y$ value of the curve can increase by at most $8M$, which is less than $8$. Thus, it could not pass through the point $(8, 8)$; therefore the maximum achieved gradient cannot be less than $1$. Similarly the minimum achieved gradient cannot be greater than $1$. It is thus intuitively clear that a gradient of $1$ is achieved somewhere. </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 7: Gradient match A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Differentiation and Integration AS Secondary Mapping Document DisplayCabinet