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Problem 50:  More curve sketching ( ) 2001 Paper II

The curve C1 passes through the origin in the xy plane and its gradient is given by dy dx = x(1 x2)ex2 .

Show that C1 has a minimum point at the origin and a maximum point at 1, 1 2e1 . Find the coordinates of the other stationary point. Give a rough sketch of C1.

The curve C2 passes through the origin and its gradient is given by dy dx = x(1 x2)ex3 .

Show that C2 has a minimum point at the origin and a maximum point at (1,k), where k > 1 2e1. (You need not find k.)


No work is required to find the x coordinate of the stationary points, but you have to integrate the differential equation to find the y coordinate. For the second part, you cannot integrate the equation — other than numerically, or in terms of rather obscure special functions that you almost certainly haven’t come across, such as the incomplete gamma function defined by

Γ(x,a) =0atx1etdt.

However, you can obtain an estimate, which is all that is required, by comparing the gradients of C1 with C2 and thinking of the graphs for 1 x 1. This is perhaps a bit tricky; an idea that you may well not alight on under examination conditions.

Solution to problem 50

C1 has stationary points when dy dx = 0, i.e. when x = 0, x = +1 or x = 1. To find the y coordinates of the stationary points, we integrate the differential equation, using integration by parts:

y =(1 x2)xex2 dx = (1 x2)(1 2ex2 ) xex2 dx = 1 2(1 x2)ex2 + 1 2ex2 + const = 1 2x2ex2 ,

where we have used the fact that C1 passes through the origin to evaluate the constant of integration. The coordinates of the stationary points are therefore (1, 1 2e1), (0,0) and (1, 1 2e1).

One way of classifying the stationary points is to look at the second derivative:

dy dx = (x x3)ex2 d2y dx2 = (1 3x2)ex2 2x2(1 x2)ex2 = (1 5x2 + 2x4)ex2 ,

which is positive when x = 0 (indicating a minimum) and negative when x = 1 and x = 1 (indicating maxima).

Since y 0 as x ±, the sketch is:


For C2, the stationary points are again at x = 0 and x = ±1. To classify the stationary points, we calculate the second derivative:

d2y dx2 = (1 3x2 3x3 + 3x5)ex3 ,

which is positive at x = 0 (a minimum) and negative at x = 1 (a maximum).

Since we cannot integrate this differential equation explicitly, we must compare C2 with C1 to see whether the value of y at the maxima is indeed greater for C2 than for C1.

For 0 < x < 1, x3 < x2, so ex3 < ex2 and ex3 > ex2 . Thus the gradient of C2 is greater than the gradient of C1. Since both curves pass through the origin, we deduce that C2 lies above C1 for 0 < x 1 and therefore the maximum point (1,k) on C2 has k > 1 2e1.


You were probably surprised that the examiners didn’t ask you to sketch C2; they usually do. It looks as if it could be done, because you know that C2 passes through the origin and you can relate its slope, roughly at least, to that of C1, which you did sketch. The difficulty lies in determining its behaviour as x ; does y 0 as for C1? In fact, it doesn’t; no reason why it should. Instead, y 0.153... . We shouldn’t have expected C1 to asymptote to the x-axis either; the fact that it does so is due to a rather delicate balancing of the two terms that determine its gradient.