# Trigonometric Applications of Maximum and Minimum

Find the maximum value of $i=100\sin(50 \pi t +0.32)$, and the time when this maximum occurs.

\begin{align} \displaystyle \dfrac{di}{dt} &= 0 \\ 100\cos(50 \pi t+ 0.32) \times 50 \pi &= 0 \\ \cos(50 \pi t+ 0.32) &= 0 \\ 50 \pi t+ 0.32 &= \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \cdots \\ t &= \dfrac{1}{50 \pi}\Big(\dfrac{\pi}{2}-0.32\Big), \dfrac{1}{50 \pi}\Big(\dfrac{3\pi}{2}-0.32\Big) \cdots \\ t &= 0.0080, 0.028, \cdots \\ \dfrac{d^2i}{dt^2} &= -100\sin(50 \pi t + 0.32) \times 50^2 \pi^2 \\ \dfrac{d^2i}{dt^2} (t=0.0080) &= -100\sin(50 \pi \times 0.0080 + 0.32) \times 50^2 \pi^2 \\ &\lt 0 \text{ maximum}\\ \dfrac{d^2i}{dt^2} (t=0.028) &= -100\sin(50 \pi \times 0.028 + 0.32) \times 50^2 \pi^2 \\ &\gt 0 \text{ minimum}\\ i &= 100 \sin(50 \pi \times 0.0080+ 0.32) \\ &= 100 \end{align}
Therefore the maximum value is $i=100$ when $t=0.0080$.

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