Answer to Question #271938 in Calculus for Angel Nodado

Question

Find the exact arc length of the curve

x = e ^ 2t* (sin t + cos t) , y=e^2t(sin t - cos t) ; (−1≤t≤1)

Answer 1

$$x'_t=e^{2t}(\sin t+\cos t+2\cos t-2\sin t)$$

$$=e^{2t}(3\cos t-\sin t)$$

$$y'_t=e^{2t}(\sin t-\cos t+2\cos t+2\sin t)$$

$$=e^{2t}(\cos t+3\sin t)$$

$$(x'_t)^2+(y'_t)^2=e^{4t}(9\cos^2 t-6\sin t\cos t+\sin ^2 t$$

$$+\cos^2 t+6\sin t\cos t+9\sin ^2 t)=10e^{4t}$$

$$L=\displaystyle\int_{-1}^1\sqrt{(x'_t)^2+(y'_t)^2}dt$$

$$=\displaystyle\int_{-1}^1\sqrt{10e^{4t}}dt$$

$$=\dfrac{\sqrt{10}}{2}[e^{2t}]\begin{matrix} 1 \\ -1 \end{matrix}=\sqrt{10}(\dfrac{e^2-e^{-2}}{2})$$

$$=\sqrt{10}\sinh(2)\ (units)$$