Answer to Question #271938 in Calculus for Angel Nodado

Question #271938

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)

1
Expert's answer
2021-12-06T04:34:38-0500
xt=e2t(sint+cost+2cost2sint)x'_t=e^{2t}(\sin t+\cos t+2\cos t-2\sin t)

=e2t(3costsint)=e^{2t}(3\cos t-\sin t)


yt=e2t(sintcost+2cost+2sint)y'_t=e^{2t}(\sin t-\cos t+2\cos t+2\sin t)

=e2t(cost+3sint)=e^{2t}(\cos t+3\sin t)

(xt)2+(yt)2=e4t(9cos2t6sintcost+sin2t(x'_t)^2+(y'_t)^2=e^{4t}(9\cos^2 t-6\sin t\cos t+\sin ^2 t

+cos2t+6sintcost+9sin2t)=10e4t+\cos^2 t+6\sin t\cos t+9\sin ^2 t)=10e^{4t}

L=11(xt)2+(yt)2dtL=\displaystyle\int_{-1}^1\sqrt{(x'_t)^2+(y'_t)^2}dt

=1110e4tdt=\displaystyle\int_{-1}^1\sqrt{10e^{4t}}dt

=102[e2t]11=10(e2e22)=\dfrac{\sqrt{10}}{2}[e^{2t}]\begin{matrix} 1 \\ -1 \end{matrix}=\sqrt{10}(\dfrac{e^2-e^{-2}}{2})

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


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