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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?

user
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Joa
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2 Answers2

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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 $r(3)= (9,6,6)$.

Michael Hoppe
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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

user
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  • Dang. This seems maddeningly simple in retrospect. Well, thanks for the help – Joa Sep 04 '21 at 21:43
  • @JoshAbel Note that $$\left(\sqrt{f(t)}\right)'=\frac{f'(t)}{2\sqrt{f(t)}}$$ which also leads to the same simple equation, being $f(t)>0$. You are welcome! Bye – user Sep 04 '21 at 21:47
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    For an algebraic verification, one may write $$D(t)^2=t^4-108t+421=(t-3)^2((t+3)^2+18)+178.$$ The first term is strictly nonnegative, which makes the minimum very easy to spot. (Of course, obtaining this decomposition is much easier once one has done the above calculus!) – Semiclassical Sep 04 '21 at 22:00
  • @Semiclassical Well I suppose that quadratic equation looks a little less ridiculous when you know what to do with it. But how did you turn the left side into the right side? – Joa Sep 04 '21 at 23:19
  • The key point is knowing that $D(t)^2$ has a local minimum at $D(3)^2=178$. That means $D(t)^2-D(0)^2=t^4-108t+243$ must have a double root at $t=3$, from which long division gives $$t^4-108t+178=(t-3)^2(t^2+6t+27)$$ The only extra step is completing the square on the quadratic factor. – Semiclassical Sep 05 '21 at 01:07