Closest point to a curve

Question

In one of my homework questions I am tasked with finding the shortest distance between a point $P = (4, 18, 9)$ and a vector curve $r(t) = (t^2, 2t, 2t)$. I believe the correct way to solving the problem is to find the values of t for which the distance between the point in line is the lowest; that is, I need to find t such that $D'(t) = 0$. But whenever I try to do this I get a ridiculous quadratic equation, which makes me think I'm plugging values into $D(t)$ improperly.

$$D(t)^2 = (t^2 -4)^2 + (2t-18)^2 + (2t-9)^2$$

Is this the correct formula for the distance between P and an arbitrary point on $r(t)$? Am I misunderstanding the distance formula, or do I have to do something with r(t) before putting its components in the equation?

Answer 1

Yes we have that in general the distance between $P(x,y,z)$ and $P_0(x_0,y_0,z_0)$ is given by

$$D=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$$

and in this case we have

$$D(t)^2=(t^2-4)^2+(2t-18)^2+(2t-9)^2$$

and since

$$x>y> 0 \iff x^2>y^2> 0$$

we can minimize $D(t)^2$ to obtain

$$[D(t)^2]'=4t(t^2-4)+4(2t-18)+4(2t-9)=4t^3-108=0$$

Refer also to the related

• Shortest distance from point to curve

Answer 2

Another approach: for the shortest distance we must have that $P-r(t)$ is perpendicular to $\dot r(t)$, which gives in two lines $t = 3$.

Now compute the distance between $P$ and $r(3)= (9,6,6)$.