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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?

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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.

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

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