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Suppose a target moves with constant speed $v > 0$ on a circular path of radius $r$, and a missile, also having constant speed $v$, pursues the target by starting from the center of the circle, always remaining between the center and the target. When does the missile hit the target?

Hey! I'm trying to solve this problem, but the answer I found does not really work. I called the position of the target $T(t)$ and the position of the missile $m(t)$. Then I set $T(t) = (r \sin(t), r \cos(t))$ and $m'(t) = \frac{v}{r}T(t)$. Then I found that $m(t) = (-r \cos(t) + r, r \sin(t))$. However, when I set $m(t) = T(t)$, this does not seem to work, can anybody help?

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  • $\begingroup$ Please use MathJax to type the equations. $\endgroup$ – Matti P. May 16 '18 at 5:38
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Basic approach. I think the initial insight is that the target moves at constant linear speed at a constant radial distance from the center, so its angular speed is the same, and equal to $v/r$. That is, if we denote the angular position of the target at time $t$ by $\theta(t)$, then

$$ \theta(t) = \frac{v}{r} t $$

But because the missile is always between the center and the target, that must be the angular position of the missile as well, and the only thing you must figure out is the radial position at time $t$, which we will denote by $x = x(t)$.

Since the missile also has constant linear speed $v$, we can determine the rate of change in $x(t)$ by drawing a little right triangle. The radial leg has infinitesimal length $dx$, the transverse leg has infinitesimal length $x\,d\theta$, and the hypotenuse—the path actually travelled by the missile—is $v\,dt$. That is to say,

$$ v^2 \,dt^2 = dx^2 + x^2\,d\theta^2 $$

Dividing both sides by $dt^2$, we get

$$ v^2 = \left(\frac{dx}{dt}\right)^2 + x^2\,\left(\frac{d\theta}{dt}\right)^2 $$

Use the fact that $d\theta/dt = v/r$, and you should obtain an expression for $dx/dt$ that you can solve using a trigonometric substitution.

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