How to find the arc length of a curve?

Question

Given: $x=4e^tcos(t)$ and $y=4e^tsin(t)$ and a $t$ range from $0$ to $\pi$. I can find $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$ and setup the initial integral, but I am not sure where to go from there.

Answer 1

The arc length formula for parametric equations is: $\int^{a}_{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \; dt$

Put your $\frac{dx}{dt}$ and $\frac{dy}{dt}$ in the formula and there is a good chance that things will simplify nicely, perhaps to a square of sin or cos that you can then square root. Think about using identities such as $\sin^2(x) + \cos^2(x) = 1$ and such.

Answer 2

and we have $$\frac{dx}{dt}=4e^t\cos(t)-4e^t\sin(t)$$ $$\frac{dy}{dt}=4e^t\sin(t)+4e^t\cos(t)$$

Answer 3

I'm not sure what's blocking you because once you set it up the integral is easy. Along the path the position vector is $$\vec r=\langle4e^t\cos t,4e^t\sin t\rangle$$ So the vector element of arc length along the path is $$d\vec r=\langle4e^t\cos t-4e^t\sin t,4e^t\sin t+4e^t\cos t\rangle dt$$ Then the scalar element of arc length is $$\begin{align}ds&=\left|\left|d\vec r\right|\right|=\left|\left|\langle4e^t\cos t-4e^t\sin t,4e^t\sin t+4e^t\cos t\rangle dt\right|\right|\\ &=\sqrt{\left(4e^t\cos t-4e^t\sin t\right)^2+\left(4e^t\sin t+4e^t\cos t\right)^2}\,dt\\ &={\sqrt{16e^{2t}\cos^2t-32e^{2t}\sin t\cos t+16e^{2t}\sin^2t+16e^{2t}\sin^2t+32e^{2t}\sin t\cos t+16e^{2t}\cos^2t}\,dt}\\ &=\sqrt{32e^{2t}\left(\cos^2t+\sin^2t\right)}\,dt=4\sqrt2e^tdt\end{align}$$ So your integral should look like $$s=\int ds=\int_0^{\pi}4\sqrt2e^tdt=\left.4\sqrt2e^t\right|_0^{\pi}=4\sqrt2\left(e^{\pi}-1\right)$$