Find the value of $\theta$ on $\pi/2 \le \theta \le \pi$ at which the curve $r=\theta - \sin (3\theta)$ is closest to the pole.

Question

Find the value of $\theta$ on $\pi/2 \le \theta \le \pi$ at which the curve $r=\theta - \sin (3\theta)$ is closest to the pole.

How can I approach this problem? I thought to find the values of theta where $r=0$, but apparently that's not right. Calculators are allowed.

Answer 1

The problem is that the curve does not cross zero in this interval. The closer that it gets to it is at $\theta = \frac{1}{3}\arccos\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)\pi\approx 2.5047$. The plot looks like this:

Answer 2

You are right that the first step is to find where $r=0$. As you have found, the only solution is $\theta=0$ which is outside your allowed values of $\theta$, so this step fails. The given curve does not go through the origin.

Your next step is to find where $r$ is a minimum. The function for $r$ is continuous so we can use the usual calculus methods. We find the derivative and find where it equals zero.

$$\begin{align} 0 & = \frac d{d\theta}\left( r \right) \\[2ex] & = \frac d{d\theta}\left( \theta-\sin(3\theta) \right) \\[2ex] & = 1 - 3\cos(3\theta) \\[2ex] \cos(3\theta) & = \frac 13 \\[2ex] 3\theta & = 2k\pi\pm\cos^{-1}\left(\frac 13\right) \\[2ex] \theta & = \frac{2k\pi}3\pm\frac 13\cos^{-1}\left(\frac 13\right) \\[2ex] \end{align}$$

The only values of $\theta$ that fit in your required interval have $k=1$:

$$\begin{align} \theta & = \frac{2\pi}3\pm\frac 13\cos^{-1}\left(\frac 13\right) \\[2ex] & \approx 1.6840752966129, \quad 2.5047149081735 \end{align}$$

I'll let you finish form here. Note that there is no value of $\theta$ that makes $r$ undefined, so we have found all the critical points. Find which of those two values of $\theta$ has the minimum value of $r$ (the other has a maximum value). Compare that value of $r$ with those at the endpoints of the given interval and find the absolute minimum of $r$ with its corresponding value of $\theta$. Ask if you need more help.

Here is a polar graph of your problem, done on the TI-Nspire CX Graphing Calculator emulator. This confirms that the correct answer is the larger of the two values of $\theta$ above, $2.5047149081735$. This graph also shows the corresponding value of $r$ and the cartesian coordinates.