Question

Why is this one B? The picture of the problem is attached.

Answer 1

B

Explanation:

It comes from straightforward application of the integral for path length.

`\displaystyle S=\int{}\,ds=\int{√(dx^2+dy^2)}=\int_0^(2\pi){√((1-\sin(t))^2+(1-\cos(t))^2)}\,dt\\\\=\int_0^(2\pi){√((1-2\sin(t)+\sin^2(t))+(1-2\cos(t)+\cos(t)^2))}\,dt\\\\=\int_0^(2\pi){√((2+\sin^2(t)+\cos^2(t))-2\sin(t)-2\cos(t))}\,dt\\\\=\boxed{\int_0^(2\pi){√(3-2\sin(t)-2\cos(t))}\,dt}\qquad\text{matches B}`