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?

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$\begin{equation} \begin{aligned} \cancel{2} \frac{dz}{dt} =& \cancel{2}(x + y) \left( \frac{dx}{dt} + \frac{dy}{dt} \right) \\ \\ \frac{dz}{dt} =& \frac{x + y}{z} \left(\frac{dx}{dt} + \frac{dy}{dt} \right) \end{aligned} \end{equation}$


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.