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

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Answer 1 · 2015-09-02T17:59:14

See the 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 $(0.824, 1.358)$ .

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