Question

Find a point on the curve y= x^2 that is closest to the point (18, 0).? Find a point on the curve y= x^2 that is closest to the point (18, 0). After taking the derivative of this one, what should I do? plug in different values into the derivative?

Answer 1

You're trying to minimize the distance between the point $(18,0)$ and an arbitrary point on the curve, $(x,y)=(x,x^2)$.

The distance between two such points is given by the function

$d(x)=\sqrt{(x-18)^2+(x^2-0)^2}=\sqrt{x^4+x^2-36x+324}$

so this is the function whose derivative you should check.

But before you do that, it's helpful to know that $d(x)$ is minimized at the same point as the modified distance function $d^*(x)=d^2(x)=x^4+x^2-36x+324$.

Differentiating, you have

$\frac{\mathrm d}{\mathrm dx}d^*(x)=4x^3+2x-36$

Set this to zero and solve for the critical points.

$4x^3+2x-36=2(2x^3+x-18)=0$

You can use the rational root theorem to find some potential candidates for roots to the cubic. The constant term has factors $\pm1,\pm2,\pm3,\pm6,\pm9,\pm18$, while the leading coefficient has factors $\pm1,\pm2$. The only candidates for rational roots are $\pm1,\pm\dfrac12,\pm2,\pm3,\pm\dfrac32,\pm6,\pm9,\pm\dfrac92,\pm18$. The only one of these that works is $2$, so $x=2$ is a root to the cubic above.

Polynomial division reveals that we can factor the cubic as

$2(x-2)(2x^2+4x+9)=0$

which has only one real root at $x=2$. Checking the value of the derivative of $d^*$ to the left and right of this point confirms that a minimum occurs here.

Therefore the closest point on the curve to $(18,0)$ is $(2,4)$.