Having Trouble Finding The Length Of A Parametric Curve

Hey everyone. I've been trying to find the length of this parametric curve, but haven't had much luck.

$\displaystyle x=cos{t}$

$\displaystyle y=t+sin{t}$

$\displaystyle 0\leq t \leq \pi$

After finding $\displaystyle \frac{dy}{dt}$ and $\displaystyle \frac{dx}{dt}$ and substituting them into the formula for curve length, I got:

$\displaystyle \int^\pi_0 \sqrt{2+2cost}\ dt$

After this, I tried u-substitution, but couldn't figure out how to eliminate the resultant $\displaystyle t$:

$\displaystyle u = 2+2cost $

$\displaystyle du = -2sint\ dt$

$\displaystyle dt = \frac{du}{-2sint} $

I consulted Wolfram Alpha as a last result, but couldn't figure out how they got the $\displaystyle \int \frac{1}{\sqrt{4-u}}\ du$

Help would be greatly appreciated!

Re: Having Trouble Finding The Length Of A Parametric Curve

Multiply top and bottom by the conjugate.

Re: Having Trouble Finding The Length Of A Parametric Curve

Hello, Algebrah!

Quote:

I've been trying to find the length of this parametric curve.

. . $\displaystyle \begin{Bmatrix}x \:=\:\cos t \\ y\:=\:t+\sin t \end{Bmatrix} \quad 0\:\leq\:t\:\leq\:\pi $

After finding $\displaystyle \tfrac{dy}{dt}$ and $\displaystyle \tfrac{dx}{dt}$ and substituting, I got:

. . $\displaystyle L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt \;=\; \int^\pi_0 \sqrt{2+2\cos t}\ dt$ . . Good!

We have: .$\displaystyle \sqrt{2(1+\cos t)} \;=\; \sqrt{4\cdot \tfrac{1\:+\:\cos t}{2}} \;=\;\sqrt{4\cos^2\tfrac{t}{2}} \;=\;2\cos\tfrac{t}{2} $

Re: Having Trouble Finding The Length Of A Parametric Curve

Quote:

Originally Posted by

**Soroban** Hello, Algebrah!

We have: .$\displaystyle \sqrt{2(1+\cos t)} \;=\; \sqrt{4\cdot \tfrac{1\:+\:\cos t}{2}} \;=\;\sqrt{4\cos^2\tfrac{t}{2}} \;=\;2\cos\tfrac{t}{2} $

This is beautiful, but the OP wanted to know how Wolfram got the particular substitution they did. I expect it was by multiplying top and bottom by a particular conjugate, but when I do I get this...

$\displaystyle \displaystyle \begin{align*} \int_0^{\pi}{\sqrt{2 + 2\cos{t}}\,dt} &= \int_0^{\pi}{\frac{\sqrt{2 + 2\cos{t}}\,\sqrt{2 - 2\cos{t}}}{\sqrt{2 - 2\cos{t}}}\,dt} \\ &= \int_0^{\pi}{\frac{\sqrt{4 - 4\cos^2{t}}}{\sqrt{2 - 2\cos{t}}}\,dt} \\ &= \int_0^{\pi}{\frac{2\sin{t}}{\sqrt{2 - 2\cos{t}}}\,dt} \end{align*}$

And now the substitution $\displaystyle \displaystyle \begin{align*} u = 2 - 2\cos{t} \implies du = 2\sin{t}\,dt \end{align*}$ is appropriate.

Re: Having Trouble Finding The Length Of A Parametric Curve

Sorry, I turned off my computer last night and then proceeded to solve the problem using the conjugate method.

I also really appreciate Soroban's method, which is a nice use of the power-reducing formula.

Thanks a lot for the help!