Question 3. Find the point on the parabola y = x2 that is the closest to the poi

Question

Question 3. Find the point on the parabola y = x2 that is the closest to the point (1, 2) in the cartesian plane. To do so, follow the steps below. (a) Find the function f = f(x) whose minimizer x gives the point on the curve y = x2 closest to (1, 2). [Hint: write down the square of the distance of a point (x, y) to the point (1, 2), and substitute the curve y = x2 into the formula.] (b) Find the critical points of the function f. (c) For each critical point of f, determine whether it is a minimum or a maximum. (d) Which of the critical points gives the global minimum? Give the coordinates (x, y) of the point on the parabola that is closest to (1, 2).

Explanation / Answer

The distance between a point (x, y) on the parabola and the point (3, 0) is given by sqr((x - 3)^2 + (y - 0)^2) (where sqr means square root). We want to minimize this. We can ignore the square root and just minimize (x - 3)^2 + y^2 (the square of the distance). And since y = x^2, we can substitute and minimize (x - 3)^2 + x^4. So we want to find a minimum of f(x) = x^4 + x^2 - 6x + 9. The first derivative is given by f'(x) = 4x^3 + 2x - 6. Setting this equal to zero, we have to solve: 4x^3 + 2x - 6 = 0. By inspection we can see that one solution is x = 1. Technically we need to verify that this is indeed a minimum, but it looks like the point (1, 1) is going to turn out to be the one we want. Checking the second derivative: f"(x) = 12x^2 + 2 This is positive for x = 1, so we've got a local minimum at x = 1. In fact it will turn out to be an absolute minimum as well, because the other two solutions of f'(x) = 0 will turn out to be imaginary. But I won't show that here. (Factor out x - 1 and look at the quadratic expression that's left.)

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