Given curve $r(t)=\langle\underline a\cos t,\underline b\sin t\rangle$ is the length of the curve $1$?

Question

Given curve $r(t)=\langle\underline a\cos t,\underline b\sin t\rangle$,$0\le t\le 2\pi$, $\quad\underline a,\underline b$ are constant unit vectors, orthogonal to each other find if the length of the curve is $1$ and whether the vector $\underline a\times\underline b$ is orthogonal to the curve.

First I think that $\underline a\times\underline b$ is orthogonal to the curve because it's orthogonal to both $\underline a$ and $\underline b$ while they are orthogonal to each other.

Regarding whether the length is $1$ I thought to represent the curve like this: $$ \underline a=\langle a_1,a_2\rangle\\ \underline b=\langle b_1,b_2\rangle\\ r(t)=\langle a_1\cos t+b_1\sin t,a_2\cos t+b_2\sin t\rangle $$ In order to find the length of the curve we can use integration: $$ \int_0^{2\pi} ||r'(t)|| \,dt;\\ r'(t)=\langle -a_1\sin t+b_1\sin t,-a_2\sin t+b_2\cos t\rangle\\ \int_0^{2\pi} ||r'(t)|| \,dt=\int_0^{2\pi} \sqrt{a_1^2\sin^2t-2a_1b_1\cos t\sin t+b_1^2\cos t+a_2^2\sin^2t-2a_2b_2\cos t\sin t+b^2_2\cos^2t}\,dt=\\ \int_0^{2\pi}\sqrt{a_1^2+a_2^2+b_1^2+b_2^2-\sin 2t(a_1b_1-a_2b_2)}=\\ \int_0^{2\pi}\sqrt{1+1-\sin 2t(a_1b_1-a_2b_2)} $$ Doesn't look like I can find a solution from here?

Answer 1

Instead of doing the integral, you could note that $\mathbf{r}(t)$ attains $\pm \mathbf{a}$, and hence has length at least $4|\mathbf{a}|$.

Answer 2

First of all, the correct expression is $$r'(t)=\langle -a_1\sin t+b_1\cos t,-a_2\sin t+b_2\cos t\rangle,$$ but that's only a typo, you use this, for a short moment. Also, you're right that $a_1^2+a_2^2=1$ and $b_1^2+b_2^2=1$, orthogonality adds $a_1b_1+a_2b_2=0$. But then, the correct expression under the root sign becomes $$(a_1^2+a_2^2)\sin^2t+(b_1^2+b_2^2)\cos^2t-\sin 2t(a_1b_1+a_2b_2)=\sin^2t+\cos^2t=1,$$ and your integral is $2\pi.$ Of course, one could arrive at that result without tedious calculations, since your curve is simply a circle with radius $1$.