Mathematical Immaturity

14.19 Exercises

• A missile is designed to move directly toward its target. Due to mechanical failure, its direction in actual flight makes a fixed angle $\alpha \neq 0$ with the line from the missile to the target. Find the path if it is fired at a fixed target. Discuss how the path varies with $\alpha.$ Does the missile ever reach the target? (Assume the motion takes place in a plane.)
• If the direction of the missile makes a fixed angle $\alpha \neq 0$ with the target, then it implies that $\alpha$ is the angle between $\mathbf{v}(t)$ and $-\mathbf{r}(t).$ Use this along with the result of Exercise 14 to yield a separable differential equation in $r$ and $\theta.$
• Suppose the target is at the origin, and the missile is fired from starting position $(r_0, 0).$ Then, if the direction of the missile makes a fixed angle $\alpha \neq 0$ with the target, this implies that $\alpha$ is in fact the angle between $\mathbf{v}(t)$ and $-\mathbf{r}(t).$ Recall from Exercise 14 that if $\alpha$ is the angle between the position vector and velocity vector, where the curve is expressed in polar coordinates, then $r = v\sin\alpha$ and $\frac{dr}{d\theta} = -v\cos\alpha$ where $r = \|\mathbf{r}\|$ and $v = \|\mathbf{v}\|.$ This gives us the following separable differential equation: \begin{align*} \\ \frac{1}{r}\frac{dr}{d\theta} &= -\cot\alpha \end{align*} Separating terms by $r$ and $\theta$ gives us: \begin{align*} \\ \frac{dr}{r} &= -\cot\alpha\,d\theta \end{align*} Integrating from $x = 0$ to $x = \theta,$ applying the second fundamental theorem, we get: \begin{align*} \\ \int_{r_0}^{r}\frac{du}{u} &= -\cot\alpha\int_{0}^{\theta}dx \\ \\ \log\,r - \log\,r_0 &= -\theta\cot\alpha \end{align*} Rearranging terms and exponentiating both sides, we get \begin{align*} \ r &= r_0e^{-\theta\cot\alpha} \end{align*} The path of the missile can be described in three possible cases dependent on $\alpha:$ 1. If $0 < \alpha < \frac{\pi}{2},$ then $-\cot\alpha < 0$ and the path of the missile is a spiral whose radius approaches $0$ as $\theta$ grows indefinitely. 2. If $\alpha = \frac{\pi}{2},$ then $-\cot\alpha = 0$ and the path of the missile is a circle centered about the target with radius $r_0.$ 3. If $\frac{\pi}{2} < \alpha < \pi,$ then $-\cot\alpha > 0$ and the path of the missile is a spiral whose radius grows without bound as $\theta$ grows indefinitely. $\blacksquare$