1
$\begingroup$

As the title says, I'm not sure what I'm doing wrong. Any help would be greatly appreciated. Here's the problem with my solution.

Find the exact length of the parametric curve $(x,y)=(\theta+\sin \theta,−\cos θ)$, where $0\le θ \le \pi$.

Solution: $$\frac{dx}{d\theta}=1+\cos \theta$$ $$\frac{dy}{d\theta}=\sin \theta$$

Using the formula $$\int_0^\pi \sqrt{\left( \frac{dx}{d\theta} \right)^2+ \left( \frac{dy}{d\theta} \right)^2} \, d\theta$$

I have $$\int_0^\pi \sqrt{2+2\cos \theta} \, d\theta$$

and then $$\left[ \frac{(4+4\cos \theta)\sqrt{2+2\cos \theta}}{3} \right]_{0}^{\pi}$$

I end up with $\dfrac{16}{3}$!

However, Wolfram is saying it's wrong. Please help?

$\endgroup$
  • $\begingroup$ I'm not sure how you integrated, but try using the substitution $t=\tan\frac{\theta}{2}.$ Also, make what you're integrating with respect to more explicit please. $\endgroup$ – garserdt216 Mar 8 '17 at 4:35
  • $\begingroup$ What I did was (2+2cos(theta))^(3/2))/(3/2) as my integration technique. $\endgroup$ – Infremo Mar 8 '17 at 4:41
  • $\begingroup$ @Infremo Maybe check Wolfram alpha to see if your in-definite integral is right. (It is not. That's not a valid integration rule.) $\endgroup$ – spaceisdarkgreen Mar 8 '17 at 4:55
  • $\begingroup$ @ Infremo It is a cycloid, obtainable by $x$ shift and $y$ reflection.. standard references would be of help, $ \theta = 2 \phi$ substitution is also helpful $\endgroup$ – Narasimham Mar 8 '17 at 10:09
  • $\begingroup$ @Infremo Yeah, you can't integrate (crud)$^n$ as (crud)$^{n+1}/(n+1)$, unless (crud)$=x$. There's some chain-rule-y stuff you have to unwind. $\endgroup$ – B. Goddard Mar 8 '17 at 14:09
0
$\begingroup$

Hint

First recall that integration is harder than differentiation and that you can't just use the power rule when there's a cosine running around inside (remember the chain rule? It's not as easy in reverse.)

As for how to do the integral, there's a popular trig identity involving $\sqrt{1+\cos(x)}$

$\endgroup$
0
$\begingroup$

Hint: $$2+2\cos\theta=2(1+\cos\theta)=2(2\cos^2\dfrac{\theta}{2})=4\cos^2\dfrac{\theta}{2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.