# How do you graph the polar equation r=1.5theta?

Jul 20, 2017

Polar equations vary $r$ as $\theta$ cycles counterclockwise through from $0$ to $2 \pi$, $2 \pi - 4 \pi$, etc.

$r = 1.5 \theta$

This equation has $r$ increase by $1.5$ for every $1$ radian that we pass through (${\left(\frac{180}{\pi}\right)}^{\circ}$). If $r = \theta$, we would have gotten a spiral.

Here, with $r = 1.5 \theta$ in $\left[0 , 8 \pi\right]$, we just have a bigger spiral: For example, if you look at $\theta = \frac{\pi}{2}$, you should see

$r = 1.5 \times \frac{\pi}{2} \approx \underline{2.35}$ on the vertical axis.

If we pass through $\frac{\pi}{2}$ radians, we move to

$r = 1.5 \times \left(\frac{\pi}{2} + \frac{\pi}{2}\right) \implies \underline{- 4.71}$ on the horizontal axis

If we pass through $\frac{3 \pi}{2}$ radians, we move to

$r = 1.5 \times \left(\frac{\pi}{2} + \pi\right) \implies \underline{- 7.07}$ on the vertical axis.

If we pass through $2 \pi$ radians, we move up to

$r = 1.5 \times \left(\frac{\pi}{2} + \frac{3 \pi}{2}\right) \approx \underline{9.42}$ on the horizontal axis.

Then, if we pass through $\frac{5 \pi}{2}$ radians, we move up to

$r = 1.5 \times \left(\frac{\pi}{2} + 2 \pi\right) \approx \underline{11.78}$ on the vertical axis.

Once you have those major points, connect them in a spiral.