Question

What is the closest point to (0,3) on the curve y=2x^2 (use derivatives ? )

Answer 1

So we need a formula that relates to all of theinformation. Considering we want to know the closest point, this implies distance, so we can use the distance formula. What we do is plug in 0 and 3 into the distance formula, but we also plug in y = 2x^2. So it would look like this:

\[d = \sqrt{(x-x_{1})^{2} + (y-y_{1})^{2}}\]
\[d = \sqrt{(x-0)^{2}+(2x^{2}-3)^{2}} = \sqrt{x^{2} + 4x^{4}- 12x^{2} + 9}\implies \sqrt{4x^{4}- 11x^{2} + 9}\]
Now once we have the formula and all theappropriate values plugged in, we take the derivative. Now, we need not take the derivative of the sqrt, only whats inside. The reason is even if we take the derivative of the square root, we would still only be technically solving for the derivative of the inside anyway, so no point. So now we take the derivative of the inside and set it equal to 0. Setting the derivative to 0 is just like finding critical points, which we use to find minimums and maximums. so the derivative of the inside:
4x^4 -11x^2 + 9 becomes
16x^3 - 22x = 0 So factoring out a 2x we get
2x(8x^2- 11), meaning the 3 critical points we get are 0 and +/- sqrt(11/8). We can either use the first derivative test to see which of these points gives a minimum or just plug in the critical points into the equation we got from the distance formula. Not the original function, the distance formula one. Plugging in 0 into the 4x^4 function, we get 9. Plugging in sqrt(11/8) or -sqrt(11/8), we get 23/16, meaning these two x-values provide minimum distance. As for the full-coordinate point, now we go back to the original equation and plug in these minimum x-values. Plugging in each value into 2x^2 gives us the coordinate points:
\[(\sqrt{\frac{11}{8}}, \frac{11}{4}), (-\sqrt{\frac{11}{8}}, \frac{11}{4})\]