Using the information above, answer the following question: If A is a positive n

Question

Using the information above, answer the following question:

If A is a positive number, the parametric curve looks like Figure 1 to the right. For one value of A, the "self-intersection" of the curve is perpendicular. Find this value of A. A picture of this sort of situation is shown in Figure 2 below. Figure 2: A special choice of A makes the self-intersection perpendicular.

Explanation / Answer

To find the point of self intersection, we need two values of t that yield the same values of x and y. Call these values of t: S and T. We know A is not 0, other wise y would always be 0 and you would have a straight line instead of a curve. So A/(1+S^2) = A/(1+T^2) => S^2 = T^2 => S = +/- T. Now look at x: T^3-T = (-T)^3-(-T) => 2T^3 -2T = 0 => T^3-T = 0 => T(T-1)(T+1) = 0, so T = 0 or T=1 or T =-1. But we know that S = +/- T, so the only possibility is T = 1 and S = -1. Note that dx/dt = 3t^2-1 and for t = 1 and t = -1 we get dx/dt = 2. Note that dy/dt = -2tA/(1+t^2)^2 so for t = 1 dy/dt = -2A/4 = -A/2 and for t = -1, dy/dt = A/2. For t = 1, dy/dx = (dy/dt)/(dx/dt) = (-A/2)/2 = -A/4 For t = -1, dy/dx = (dy/dt)/(dx/dt) = (A/2)/2 = A/4 For the tangents to be perpendicular we need the product of these slopes to be -1, so (-A/4)(A/4) = -1 => A^2 = 16 => A = +/- 4. Only A = 4 will give the shape shown in the picture (check by graphing).

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