Luke Skyrunner Improper Integral Problem

Question

I came across a challenging calculus problem in Calculus II by Jerrold Marsden and Alan Weinstein. The question can be found in section 11.3 on page 533. The problem is titled "Example 11."

"Luke Skyrunner has just been knocked out in his spaceship by his archenemy, Captain Tralfamadore. The evil captain has set the controls to send the spaceship into the sun! His perverted mind insists on a slow death, so he sets the controls so that the ship makes a constant angle of 30˚ with the sun. What path will Luke's ship follow? How long does Luke have to wake up if he is 10 million miles from the sun and his ship travels at a constant velocity of a million miles per hour."

Figure 11.3.5 Luke Skyrunner's ill-fated ship

Here is the solution they give:

We use polar coordinates to describe a curve $(r(t),\theta (t))$ such that the radius makes a constant angle $\alpha$ with the tangent ($\alpha = 30^o$ in the problem). To find $\frac{dr}{d\theta}$, we observe, from Fig. 11.3.6(a), that $$\Delta r \approx \frac{r \Delta \theta}{\tan{\alpha}}$$ so $$ \frac{dr}{d \theta} = \frac{r}{\tan{\alpha}} \hspace{2cm}(1) $$

Figure 11.3.6. The geometry of Luke's path.

We can derive formula (1) rigorously, but also more laboriously, by calculating the slope of the tangent line in polar coordinates and setting it equal to $\tan{(θ+α)}$ as in Fig. 11.3.6(b). This approach gives

$$\begin{array}{rcl} \displaystyle{\frac{\tan{\theta} \left(\frac{dr}{d\theta}+r\right)}{\frac{dr}{d\theta} -r\tan{\theta}}} & = & \tan{(\theta + \alpha)} \\ & = & \displaystyle{\frac{\tan{\theta}+\tan{\alpha}}{1-\tan{\theta}\tan{\alpha}}}\end{array}$$

so that again $\frac{dr}{dθ} = \frac{r}{\tan{\alpha}}$.

The solution of equation (1) is

$$\mathbf{r(\theta) = r(0)e^{\frac{\theta}{\tan{\alpha}}}}$$

I don't understand how the bolded equation above is found. I also don't understand how the derivative $$ \frac{dr}{d \theta} = \frac{r}{\tan{\alpha}} \hspace{2cm}(1) $$ is found. I do understand how $$\mathbf{r(\theta) = r(0)e^{\frac{\theta}{\tan{\alpha}}}}$$ is used to find how long Luke has to wake up using improper integrals.

Answer 1

This is a simple differential equation. $r$ is a function of $\theta$ such that $$\frac{dr}{d\theta} = \frac{r}{\tan \alpha}.$$ Note $\alpha$ is a constant. So by separation of variables, we have $$\frac{1}{r} \, dr = \frac{1}{\tan \alpha} \, d\theta.$$ Integrating both sides gives $$\int \frac{1}{r} \, dr = \frac{1}{\tan \alpha} \int \, d\theta$$ or $$\log r = \frac{1}{\tan \alpha} \theta + C,$$ where $C$ is a constant of integration. Solving for $r$ gives $$r(\theta) = e^{\theta/\tan\alpha + C} = e^C e^{\theta/\tan \alpha} = r(0) e^{\theta/\tan \alpha},$$ where $r(0)$ is an initial condition that gives the distance of the ship from the sun at angle $\theta = 0$.

You should review the section of the text (apparently, Chapter 8) that discusses ordinary linear differential equations.

Answer 2

The book's derivation of the differential equation is pretty poor in my opinion. I'm going to use an analysis similar to Figure (a) but defining $\theta$ to be an increasing function of time. Basic trigonometry shows that, if $\theta$ increases by $\Delta\theta$, then $r$ changes by $$ \Delta r = -\frac{r\sin(\Delta\theta)}{\tan\alpha} \;\Longrightarrow\;\frac{\Delta r}{\Delta\theta} = -\frac{r}{\tan\alpha}\frac{\sin(\Delta\theta)}{\Delta\theta}. $$ Note the minus sign: $r$ decreases as $\theta$ increases. Now we take the limit as $\Delta\theta \rightarrow 0$. If we treat $r$ as a function of $\theta$, then $\Delta r = r(\theta + \Delta \theta) - r(\theta)$, and we get $$ \lim_{\Delta\theta\rightarrow0}\frac{\Delta r}{\Delta\theta} = \lim_{\Delta\theta\rightarrow0}\frac{r(\theta+\Delta\theta) - r(\theta)}{\Delta\theta} =\frac{dr}{d\theta} = \lim_{\Delta\theta\rightarrow0} -\frac{r}{\tan\alpha} \frac{\sin(\Delta\theta)}{\Delta\theta} = -\frac{r}{\tan\alpha} $$ where I've used the definition of the derivative and the fact that $\lim_{x\rightarrow 0}\sin(x)/x = 1$. So we have an ordinary differential equation defining $r$ as a function of $\theta$ given by $$ \frac{dr}{d\theta} = -\frac{r}{\tan\alpha}. $$ Separation of variables easily solves this equation as $r(\theta) = r(0)\exp(-\theta/\tan\alpha)$.

Finding the time it takes is much easier. The $r$-component of the velocity vector is constant ($v_r = -v\cos\alpha$), so $r(t) = r(0) - vt\cos\alpha$, and simple algebra gives $t_\mathrm{final} = [r(0) - r_\mathrm{final}]/(v\tan\alpha)$.