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

Sep 2, 2015

See the explanation.

Explanation:

Every point on the curve has the form: $\left(x , 2 {x}^{2}\right)$

The closest point is the one whose distance is minimum.

The distance between $\left(x , 2 {x}^{2}\right)$ and $\left(2 , 1\right)$ is $\sqrt{{\left(x - 2\right)}^{2} + {\left(2 {x}^{2} - 1\right)}^{2}}$.

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

$f \left(x\right) = {\left(x - 2\right)}^{2} + {\left(2 {x}^{2} - 1\right)}^{2}$

$f \left(x\right) = 4 {x}^{4} - 3 {x}^{2} - 4 x + 5$

To minimize $f$,

$f ' \left(x\right) = 16 {x}^{3} - 6 x - 4$

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 $f ' \left(0\right)$ is negative and $f ' \left(1\right)$ is positive, so $f \left(0.824\right)$ is a local minimum.

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

The minimum value of distance occurs at $x \approx 0.824$ and $f \left(0.824\right) \approx 1.358$.

The closest point is about $\left(0.824 , 1.358\right)$.