Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 21

At what rate is the distance between the ships changing 4 hours later?

Illustration:







By using Pythagorean Theorem we have,

$z^2 =(x + y)^2 + 100^2 \qquad$ Equation 1

Taking the derivative with respect to time


$
\begin{equation}
\begin{aligned}

\cancel{2} z \frac{dz}{dt} =& \cancel{2} (x + y) \left( \frac{dx}{dt} + \frac{dy}{dt} \right)
\\
\\
\frac{dz}{dt} =& \frac{x + y}{z} \left( \frac{dx}{dt} + \frac{dy}{dt} \right) \qquad \text{Equation 2}

\end{aligned}
\end{equation}
$


The distance covered by boat A after 4 hours is $x = 35km /\cancel{hr} (4 \cancel{hr}) = 140$ while boat B is $\displaystyle y = \frac{25 km}{\cancel{hr}} (4\cancel{hr}) = 100 km $. We can use equation 1 to solve for $z$. Then,


$
\begin{equation}
\begin{aligned}

z^2 =& (140 + 100)^2 + 100^2
\\
\\
z^2 =& 260 km

\end{aligned}
\end{equation}
$


Now, using equation 2 to solve for the unknown, we have


$
\begin{equation}
\begin{aligned}

\frac{dz}{dt} =& \frac{(140 + 100)}{260} (35 + 25)
\\
\\
\frac{dz}{dt} =& 55.3846 km/hr

\end{aligned}
\end{equation}
$


This means that the distance between the ships, is changing at a rate of $55.3846 km/hr$ after 4 hours.