Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 17

How fast are the people moving apart 15min after the woman starts walking?

Illustration










$
\begin{equation}
\begin{aligned}

\cancel{2} \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)

\end{aligned}
\end{equation}
$


The distance covered by the man is $\displaystyle x = (4ft/ \cancel{s}) \left( 20 \cancel{min} \cdot \frac{60 \cancel{s}}{1 \cancel{min} }\right) = 4800 ft$. We use $20 min$ since the man starts walking $5 mins$ ahead to the woman. On the other hand, the distance covered by the woman is

$\displaystyle y = (5 ft/s) \left( 15 \cancel{min} \cdot \frac{60s}{1 \cancel{min}} \right) = 4500 ft.$

We can use equation 1 to solve for $z$. Then,


$
\begin{equation}
\begin{aligned}

z^2 =& (4800 + 4500)^2 + 500^2
\\
\\
z =& 9313.4312 ft

\end{aligned}
\end{equation}
$


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


$
\begin{equation}
\begin{aligned}

\frac{dz}{dt} =& \left( \frac{4800 + 4500}{9313.4312} \right) (4 + 5)
\\
\\
\frac{dz}{dt} =& 8.987 ft/s

\end{aligned}
\end{equation}
$


This means that the distance between the man and the woman changes at a rate of $8.987 ft/s$ after 4 hours.