How to show a curve sweeps out a unit circle?

Question

If we have two unit and perpendicular vectors $u$ and $v$ show that the curve $r(t) = u cos(t) + v sin(t)$ sweeps out an unit circle?
My attempt: my biggest problem is understanding the techniques they mean to apply in proving it "sweeps out". Any ideas how I should approach this problem?

Answer 1

Take scalar product of both sides with themselves: $$r(t)=u\cos(t)+v\sin(t)$$ $$r(t)\cdot r(t)=(u\cos(t)+v\sin(t))\cdot (u\cos(t)+v\sin(t))$$ $$|r(t)|^2=|u|^2 \cos^2(t) + |v|^2 \sin^2(t) + 2 u\cdot v \sin(t) \cos(t)$$ now $|u|=|v|=1$, and $u\cdot v=0$, so $$|r(t)|^2= \cos^2(t) + \sin^2(t) = 1$$

This shows that all points $r(t)$ lie on the unit circle. Technically, it does not show that all points on the unit circle are given by $r(t)$ for some $t$.

Answer 2

Square (or, doc product) both sides of the curve equation

$$r(t) = u cos(t) + v sin(t)$$

to obtain

$$r^2 = (ucos(t) + v sin(t))\cdot (ucos(t) + v sin(t))$$

$$=\cos^2(t) + \sin (2t)|u||v| \cos(90^\circ) +\sin^2(t)$$

As a result,

$$r^2=1$$

which indicates a unit radius, hence, a unit circle.