I'm struggling with the following exercise:
Let $C$ be the curve parametrized by
$$\alpha:[0,2\pi]\to\Bbb R^2\\
\alpha(t):=(5\cos t-\cos5t,5\sin t-\sin5t)$$
Find the arc length of $C$ using the following identities:
$$\cos4t=\cos t\cos5t+\sin t\sin5t\\
\sin^2t=\frac{1-\cos2t}2$$
So far, I've tried the following:
$$L(C)=\int_0^{2\pi}\Vert\alpha'(t)\Vert\ dt$$
$$=\int_0^{2\pi}\sqrt{(5\cos t-\cos5t)^2+(5\sin t-\sin5t)^2}\ dt$$
$$=\int_0^{2\pi}\sqrt{26-10\cos4t}\ dt$$
$$=\int_0^{2\pi}\sqrt{16+10-10\cos4t}\ dt$$
$$=\int_0^{2\pi}\sqrt{16+10(1-\cos4t)}\ dt$$
$$=\int_0^{2\pi}\sqrt{16+20\sin^22t}\ dt$$
And at this point I got stuck. I've unsuccessfully tried some hyperbolic sine substitutions. Any ideas?
Best Answer
You have failed to take the derivative of the $x$- and $y$-components when computing the arc length, instead just plonking the expressions into the formula without further consideration. $$x'(t)=-5\sin t+5\sin5t$$ $$y'(t)=5\cos t-5\cos5t$$ The integral for arc length should have been $$\int_0^{2\pi}\sqrt{x'(t)^2+y'(t)^2}\ dt$$ $$=\int_0^{2\pi}\sqrt{(-5\sin t+5\sin5t)^2+(5\cos t-5\cos5t)^2}\ dt$$ $$=\int_0^{2\pi}\sqrt{ \color{red}{25\sin^2t} \color{green}{-50\sin t\sin5t} \color{blue}{+25\sin^25t} \color{red}{+25\cos^2t} \color{green}{-50\cos t\cos5t} \color{blue}{+25\cos^25t}}\ dt$$ $$=\int_0^{2\pi}\sqrt{ \color{red}{25}\color{blue}{+25} \color{green}{-50(\sin t\sin5t+\cos t\cos5t)}}\ dt$$ $$=\int_0^{2\pi}\sqrt{50-50\cos4t}\ dt$$ $$=\int_0^{2\pi}\sqrt{50(1-\cos4t)}\ dt$$ $$=\int_0^{2\pi}\sqrt{50(2\sin^22t)}\ dt$$ $$=\int_0^{2\pi}|10\sin2t|\ dt$$ $$=10\cdot4\int_0^{\pi/2}\sin2t\ dt=40$$