Problem (Algebra):
If $f(t)=\sin \left(\pi t-\frac{\pi}{2}\right)$, what is the smallest positive value of $t$ at which $f(t)$ attains its minimum value?

Solution:
Since $t>0, \pi t-\frac{\pi}{2}>-\frac{\pi}{2}$. So $\sin \left(\pi t-\frac{\pi}{2}\right)$ first attains its minimum value when

$$
\begin{aligned}
\pi t-\frac{\pi}{2} & =\frac{3 \pi}{2} \\
t & =2 .
\end{aligned}
$$

Rewriting $f(t)$ as, $f(t)=\sin \left[\pi\left(t-\frac{1}{2}\right)\right]$.

Thus $f(t)$ has a period $\frac{2 \pi}{\pi}=2$ and appears in the diagram at the right.

Thus $f(t)$ attains its minimum at $t=2$. Note that $f(t)$ attains a minimum value at $t=0$ but since $t>0$, the required answer is $t=2$.



Final answer: 2