Question

calculate the length of polar curve
r=e^2x where 0<=x<=2pi

Answer 1

So the arc length in polar coordinates is simply derived from the one in cartesian coordinates. I'll walk you through it.

The pythagorean theorem says for a little bit of arc length (hypotenuse) it's just:
\[ds^2=dx^2+dy^2\]
Now we solve for ds, the arc length bit \[ ds=\sqrt{ dx^2+dy^2}\]
That's super awkward. Let's multiply by a fancy version of 1.\[ds= \frac{ d \theta }{ d \theta } \sqrt{ dx^2+dy^2}\] Now let's bring the bottom part inside the integral by squaring it. \[ds= \sqrt{ (\frac{ dx }{ d \theta })^2+(\frac{dy }{ d \theta })^2} d \theta\] now remembering tha:
\[x=r*\cos \theta\]\[y=r*\sin \theta\]

We can differentiate these with respect to theta, keeping in mind that r is a function of theta to get:

\[dS=\sqrt{ r^2+(\frac{dr }{ d \theta })^2} d \theta \]

Now we just add up all the little bits from 0 to 2pi\[S=\int\limits_{0}^{2 \pi}\sqrt{ r^2+(\frac{dr }{ d \theta })^2} d \theta \]

So now we have the general formula fairly simply, we can just plug and chug our r=e^(2 theta) into our formula and integrate.