# Length of Curve (Parametric)

• Mar 25th 2009, 06:50 PM
Hellreaver
Length of Curve (Parametric)
Find the length of the following parametric curve

$y= a(cos t + ln tan \frac {t}2$)
$y= a sint$

from the point (0,a) to the point (x,y).

I need to use the formula
$L= \int \sqrt{( \frac {dx}{dt}^2)+ (\frac {dy}{dt}^2)}dt$

but I'm not sure what the limits of integration are so I can evaluate the result.

As well, would $\frac {dx}{dt}$ be $a(-sin t+ \frac {1}{tan \frac {t}2})$? Or am I differentiating the ln incorrectly?
• Mar 25th 2009, 09:37 PM
NonCommAlg
Quote:

Originally Posted by Hellreaver

Find the length of the following parametric curve $\color{red}x$ $= a(\cos t + \ln \tan(t/2)), \ \ y= a \sin t,$ from the point (0,a) to the point (x,y).

I need to use the formula
$L= \int \sqrt{( \frac {dx}{dt}^2)+ (\frac {dy}{dt}^2)}dt$

but I'm not sure what the limits of integration are so I can evaluate the result.

As well, would $\frac {dx}{dt}$ be $a(-sin t+ \frac {1}{tan \frac {t}2})$? Or am I differentiating the ln incorrectly?

the curve is defined over $0 < t < \pi.$ the points (0,a) and (x,y) correspond to $t=\frac{\pi}{2}$ and, say $t_0,$ where $t_0$ is any number between $0$ and $\pi.$ we also have $\frac{dx}{dt}=a(-\sin t + \csc t)$ and $\frac{dy}{dt}=a \cos t.$

so the length of the curve between two given points is: $\ell = \int_{t_0}^{\frac{\pi}{2}} \sqrt{a^2(-\sin t + \csc t)^2 + a^2 \cos^2t} \ dt=-a \ln \sin t_0.$ (i'm assuming that $a>0$)
• Mar 25th 2009, 09:53 PM
Hellreaver
So there is no definite answer then? Thats a pain in the butt...
• Mar 26th 2009, 05:09 AM
HallsofIvy
There is no "definite answer" because there is no definite question! You asked for the arclength from the specific point (0, a) to the variable point (x, y). Obviously the answer will depend upon (x,y). Since you are integrating with respect to t, NonCommAlg converted that to the variable $t_0$, the value of t corresponding to (x,y).