Question

How do you find the point on the curve `y=2x^2` closest to (2,1)?

Answer 1

See the explanation.

Explanation:

Every point on the curve has the form: `(x,2x^2)`

The closest point is the one whose distance is minimum.

The distance between `(x, 2x^2)` and `(2,1)` is `sqrt((x-2)^2+(2x^2-1)^2)`.

Because the square root is an increasing function, we can minimize the distance by minimizing:

`f(x) = (x-2)^2+(2x^2-1)^2`

`f(x) = 4x^4-3x^2-4x+5`

To minimize `f`,

`f'(x) = 16x^3-6x-4`

Now use whatever tools you have available to solve this cubic equation. (Formula, graphing technology, successive approximation, whatever you have.)

The critical number is approximately `0.824`.

Clearly `f'(0)` is negative and `f'(1)` is positive, so `f(0.824)` is a local minimum.

`f` is a polynomial with only one critical number, so "local" implies "global"

The minimum value of distance occurs at `x ~~ 0.824` and `f(0.824) ~~ 1.358`.

The closest point is about `(0.824, 1.358)`.