Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 29

How fast is the area of the triangle increasing when the angle between the sides of fixed length is $\displaystyle \frac{\pi}{s}$
Given: Length of the sides of the triangle

$
\begin{equation}
\begin{aligned}
L_1 &= 4m\\
L_2 &= 5m
\end{aligned}
\end{equation}
$

The rate of change of the angle between them, $\displaystyle\theta = 0.06 \frac{\text{rad}}{x}$
Required: rate of change of the area of the triangle when the angle between the length is $\displaystyle \frac{\pi}{3}$




We use the formula $\displaystyle A = \frac{1}{2} L_1 L_2 \sin \theta$, for area given two sides and an included angle.

$
\begin{equation}
\begin{aligned}
A &= \frac{1}{2} L_1 L_2 \sin \theta\\
\\
\frac{dA}{dt} &= \frac{1}{2} L_1 L_2 \cos \theta \left( \frac{d \theta}{dt}\right) && \Longleftarrow \text{ derivative with respect to time}\\
\\
\frac{dA}{dt} &= \frac{1}{2} (4)(5) \cos \left( \frac{\pi}{3}\right) (0.06) && \Longleftarrow \text{ we use radian mode to be consistent with measurements}\\
\\

\end{aligned}
\end{equation}\\
\boxed{\displaystyle \frac{dA}{dt} = 0.3 \frac{m^2}{s}}
$