# Length of parametric curve in polar coordinates

I have to find the length of this parametric curve: $$R(\theta)= \theta^4$$ with $$0 < \theta < 1$$

So, We have the formula: $$ds = \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta$$.

My question: As I have understood. I have to convert the $$\theta$$ into trig functions. How do I do that and proceed with my calculations?

We have \begin{align} ds &= \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \\ &= \sqrt{\theta^8 + 16\theta^6} \ d\theta \\ &= \theta^3 \sqrt{\theta^2 + 16} \ d\theta \end{align}

Now, take $$\theta = 4\tan\alpha$$, and integrate.

• Using $x=\sqrt{\theta^2+16}\implies xdx=\theta d\theta\implies\theta^3\sqrt{\theta^2+16}d\theta=(x^2-16)xdx$ is a bit easier.
– J.G.
Jan 10 at 11:28
• @Ishan Deo Thank you. How did you get the $\theta = 4 \tan α$? Jan 10 at 11:30
• That was a substitution I was assuming in order to integrate. Jan 10 at 12:31
• Ok. So go integrate: $(4 tan α)^3 \sqrt{(4 tan α)^2 +16 }dα$? Jan 10 at 12:43
• Oh sorry, yes. There should be a factor of $4$. Jan 10 at 20:27