Answer
$\frac{d \theta}{d t}=-\frac{9 \sqrt{3}}{20} \\
\frac{d \theta}{d t}=-0.779423 \frac{r a d}{s}$
Work Step by Step
We are given that
\[
\operatorname{Arca}(A)=\frac{a b \sin \theta}{2}, \quad \text { with } \frac{d a}{d t}=3, \quad \frac{d b}{d t}=3
\]
By theorem 13.5.2:
\[
\frac{d A}{d t}=b \sin \theta \frac{d a}{d t}+a \sin \theta \frac{d b}{d t}+a b \cos \theta \frac{d \theta}{d t}
\]
We know $a=5 \mathrm{cm}, \quad \mathrm{b}=4 \mathrm{cm}$
\[
\begin{array}{c}
0=5\left(\frac{1}{2}\right)(3)+4\left(\frac{1}{2}\right)(3)+20\left(\frac{\sqrt{3}}{2}\right) \frac{d \theta}{d t} \\
0=10 \sqrt{3} \frac{d \theta}{d t}+\frac{27}{2} \\
\frac{d \theta}{d t}=-\frac{9 \sqrt{3}}{20} \\
\end{array}
\]
