Question

Find the arc length of the curve on the given interval. $$ x=e^{-t} \cos t, \quad y=e^{-t} \sin t \quad 0 \leq t \leq \frac{\pi}{2} $$

Short answer

The arc length of the curve over the interval from 0 to \( \frac{\pi}{2} \) is \( -\sqrt{2}(e^{-\frac{\pi}{2}} - 1) \)

Step-by-step solution

Calculate Derivatives

Calculate the derivatives of x and y with respect to t: \n\n \(dx/dt = e^{-t}(-\sin(t) - \cos(t))\), \n \(dy/dt = e^{-t}(\cos(t) - \sin(t)).\)

Square the Derivatives

Square the derivatives and add them together: \n\n \((dx/dt)² + (dy/dt)² = e^{-2t}(\sin²(t) + 2\sin(t)\cos(t) + \cos²(t) + \cos²(t) + 2\sin(t)\cos(t) + \sin²(t)) = 2e^{-2t}.\)

Integrate Over the Interval

Integrate the square root of the sum of the squares of the derivatives over the interval [0, pi/2]: \n\n \(L = \int_{0}^{\frac{\pi}{2}} \sqrt{2 e^{-2t}} dt = \sqrt{2} \int_{0}^{\frac{\pi}{2}} e^{-t} dt = -\sqrt{2}e^{-t}|_{0}^{\frac{\pi}{2}}. \) This gives \( L= -\sqrt{2}(e^{-\frac{\pi}{2}} - 1)\).