Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 41

How fast is the distance from the plane to the radar station increasing after one minute?


Using cosine law,

$
\begin{equation}
\begin{aligned}
y^2 &= x^2 + 1^2 - 2 (x) (1) \cos (120^\circ)\\
\\
y^2 &= x^2 + 1 - 2x \left( \frac{-1}{2}\right)\\
\\
y^2 &= x^2 + 1 + x
\end{aligned}
\end{equation}
$


Taking the derivative with respect to time

$
\begin{equation}
\begin{aligned}
2y \frac{dy}{dt} &= 2x \frac{dx}{dt} + \frac{dx}{dt}\\
\\
\frac{dy}{dt} &= \frac{2x+1}{2y} \left( \frac{dx}{dt} \right) && \Longleftarrow \text{ Equation 1}
\end{aligned}
\end{equation}
$

The distance covered by the plane after one minute is
$\displaystyle x = \frac{300 \text{km}}{\cancel{\text{hr}}} \left( \frac{1 \cancel{\text{hr}}}{60 \cancel{\text{min}}} \right) ( 1 \cancel{\text{min}}) = 5 \text{km}$

So if $x = 5$,
$y = \sqrt{5^2 + 1 + 5} = \sqrt{31}$km

Now, plugging all the values we get in Equation 1

$
\begin{equation}
\begin{aligned}
\frac{dy}{dt} &= \frac{2(5) + 1}{2(\sqrt{31})} (300)\\
\\
\frac{dy}{dt} &= 296.3487 \frac{\text{km}}{\text{hr}}
\end{aligned}
\end{equation}
$