Minimising the length of the vector $r(t) = \sqrt{2}\sin{t}\mathbf{i}+\cos{2t}\mathbf{j}$ for $t \in (0, \pi/2)$ I want to minimise the length of the vector $r(t) = \sqrt{2}\sin{t}\mathbf{i}+\cos{2t}\mathbf{j}$ for $t \in (0, \pi/2)$
I have found that $|r(t)| = \sqrt{\sin^2(t)+1}$ but I can't minimise this for $t \in (0, \pi/2)$ because $\sin(t)$ is minimum for $t = 0 \not \in (0, \pi/2)$and the next time it assumes a minimum value it's $t=\pi \not \in (0, \pi/2)$. 
EDIT: It should be $|r(t)| = \sqrt{2\sin^2(t)+\cos^2{2t}}.$
 A: $$\begin{align}
|r(t)|^2 &= (\sqrt 2\sin t)^2+(\cos 2t)^2 \\
 &= 2\sin^2 t+(1-2\sin^2 t)^2 \\
 &= 2\sin^2 t + 1 - 4\sin^2 t + 4\sin^4 t \\
 &= 4\sin^4 t-2\sin^2 t+1 \\
 &= 4u^2-2u+1
\end{align}$$
where $u=\sin^2 t$ for $0<u<1$. Now we minimize that for $u$ then find the corresponding value of $t$.
The derivative of that function of $u$ is $8u-2$ which has the root $u=\frac 14$. The second derivative of that function is $8$ which is positive, so that is a minimum. Checking the "endpoints" of $0$ and $1$ show that $u=\frac 14$ is indeed the minimum. (If you know how to analyze a quadratic function, there are other ways to get the same results.)
So $\sin^2t=\frac 14$, $\sin t=\frac 12$, thus

$$t=\frac{\pi}6$$

is your answer. The minimum length of the vector is then $|r(\frac{\pi}6)|=\frac{\sqrt 3}2$.
A quick check of this on my TI-Nspire CX calculator confirms that is the correct answer.

A: You can minimize the square of the length, because the square root is an increasing function. The square of the length is
$$
f(t)=2\sin^2t+\cos^22t
$$
and
$$
f'(t)=4\sin t\cos t-4\sin2t\cos2t=2\sin2t-4\sin2t\cos2t=
2\sin2t(1-2\cos2t)
$$
which vanishes, in the interval $(0,\pi/2)$, only where $\cos2t=1/2$, that is, $2t=\pi/3$ and so $t=\pi/6$.
Since $f(0)=1$, $f(\pi/2)=2$ and
$$
f(\pi/6)=2\frac{1}{4}+\frac{1}{4}=\frac{3}{4}
$$
we found the minimum.
