1. ## Parametric Equations

Find the arc length of the givenn curve on the given innterval
x=e^-tcost, y=e^-tsint, 0<t<pi/2

so I figured out dx/dt and dy/dt to be -e^-tsint-e^-tcost and e^-tcost-e^-tsint respectively.

And then ... integeral(0to pi/2) of square root of dy/dt^2 + dx/dt^2

However, it gets really mess and when I take the squares of dx/dt and dy/dt it becomes really messy. Is there a better way to work through this problem?

2. Originally Posted by andrewsx
Find the arc length of the givenn curve on the given innterval
x=e^-tcost, y=e^-tsint, 0<t<pi/2

so I figured out dx/dt and dy/dt to be -e^-tsint-e^-tcost and e^-tcost-e^-tsint respectively.

And then ... integeral(0to pi/2) of square root of dy/dt^2 + dx/dt^2

However, it gets really mess and when I take the squares of dx/dt and dy/dt it becomes really messy. Is there a better way to work through this problem?
There's no mess. You just simplify in a straight forward way.

$\frac{dx}{dt} = -e^{-t} (\cos t + \sin t)$

$\Rightarrow \left(\frac{dx}{dt} \right)^2 = e^{-2t} (\cos t + \sin t)^2 = e^{-2t} (\cos^2 t + \sin^2 t + 2 \cos t \sin t) = e^{-2t} (1 + 2 \cos t \sin t)$.

$\frac{dy}{dt} = e^{-t} (\cos t - \sin t)$

$\Rightarrow \left(\frac{dy}{dt} \right)^2 = e^{-2t} (\cos t - \sin t)^2 = e^{-2t} (\cos^2 t + \sin^2 t - 2 \cos t \sin t) = e^{-2t} (1 - 2 \cos t \sin t)$.

So $\left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2$ is a pretty simple expression.