Having Trouble Finding The Length Of A Parametric Curve

• August 26th 2012, 08:26 PM
Algebrah
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.

$x=cos{t}$
$y=t+sin{t}$
$0\leq t \leq \pi$

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

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

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

$u = 2+2cost$

$du = -2sint\ dt$

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

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

Help would be greatly appreciated!
• August 26th 2012, 08:37 PM
Prove It
Re: Having Trouble Finding The Length Of A Parametric Curve
Multiply top and bottom by the conjugate.
• August 27th 2012, 05:46 AM
Soroban
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.

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

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

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

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

• August 27th 2012, 06:02 AM
Prove It
Re: Having Trouble Finding The Length Of A Parametric Curve
Quote:

Originally Posted by Soroban
Hello, Algebrah!

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

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 \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 \begin{align*} u = 2 - 2\cos{t} \implies du = 2\sin{t}\,dt \end{align*} is appropriate.
• August 27th 2012, 08:09 AM
Algebrah
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!