Question

Find the point on the parabola y=x^2/(2) that is closest to the point (4, 1).

Answer 1

Right, distances between points
\[r = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\]
To avoid the hassle of square roots, just calculate the smallest (distance)^2
\[r^2 = f(x) = (x_1-x_2)^2 + (y_1-y_2)^2 = \Big(4-x\Big)^2 + \Big(1-\frac{x^2}2\Big)^2\]
Now you want to differentiate f(x) and set it to zero
\[\frac{\text{d}f(x)}{\text{d}x} = 0\]
Then solve for x.

Answer 2

The derivative of x^2/(x) = x

Answer 3

So I find the derivative of the function and then I set it to 0 and then what do I do?

Answer 4

Am I taking the derivative of (4−x)^2 +(1=x^2/2) or x^2/2

Answer 5

The whole thing. It's a bit easier to expand it out first.

Answer 6

d/dx of (4−x)^2 +(1-x^2/2)^2 = x^3-8

Answer 7

yeah, so you can set that to zero then get x, then get y

Answer 8

x=2

Answer 9

Are the x and ys that I find the answer?