# Thread: Find the length of the curve

1. ## Find the length of the curve

Find the length of the curve $r=\frac{e^\theta}{\sqrt2}$ on the interval $0\leq\theta\leq\pi$. I've plugged everything into the equation, but I can't seem to get the same answer that is on the answer sheet, which is $e^\pi-1$.

2. ## Re: Find the length of the curve

the answer is $e^{\pi-1}$ not $(e^\pi - 1)$ as you've written.

does that help any?

3. ## Re: Find the length of the curve

No, could you go over it step-by-step?

4. ## Re: Find the length of the curve

Do you have a rule that you can follow to find the arclength of a polar curve?

5. ## Re: Find the length of the curve

Yes, the equation is $L=\int_{a}^{b}\sqrt{r^2+(\frac{dr}{d\theta})^2$.

6. ## Re: Find the length of the curve

Well then, what is your \displaystyle \begin{align*} r^2 \end{align*} and \displaystyle \begin{align*} \left( \frac{dr}{d\theta} \right) ^2 \end{align*}? What are your a and b?

Also, the equation is actually \displaystyle \begin{align*} L = \int_a^b{\sqrt{r^2 + \left( \frac{dr}{d\theta} \right) ^2 }\,d\theta} \end{align*}...

7. ## Re: Find the length of the curve

$r^2=\frac{dr}{d\theta}=\frac{e^{\theta}}{\sqrt{2}}$; $a=0$; $b=\pi$

8. ## Re: Find the length of the curve

Originally Posted by accountholder
$r^2=\frac{dr}{d\theta}=\frac{e^{\theta}}{\sqrt{2}}$; $a=0$; $b=\pi$
no.

you are given $r=\frac{e^\theta}{\sqrt{2}}$

so $r^2=\frac{e^{2\theta}}{2}$

$\frac{d^2r}{d\theta^2}=\frac{dr}{d\theta}=\frac{e^ \theta}{\sqrt{2}}$

so $\left(\frac{d^2r}{d\theta^2}\right)^2=\frac{e^{2 \theta}}{2}$

now take your arc length formula and do the integration. Your limits are correct.