Question

A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?

Short answer

The rate of people moving apart at \(8.99\;{\rm{ft/s}}\).

Step-by-step solution

Step 1:Find the rate of change of distance in two directions

Let the distance traveled by the man bex and the distance traveled by the woman is y. Therefore;

\(\frac{{{\rm{d}}x}}{{{\rm{d}}t}} = 4\)and \(\frac{{{\rm{d}}y}}{{{\rm{d}}t}} = 5\)

The figure below represents the positioning of man and woman.

By the property of triangle:

\({z^2} = {\left( {x + y} \right)^2} + {500^2}\)

Find the expression for the rate at which people are moving apart

Differentiate the equation \({z^2} = {\left( {x + y} \right)^2} + {500^2}\).

\(\begin{aligned}2z\left( {\frac{{{\rm{d}}z}}{{{\rm{d}}t}}} \right) &= 2\left( {x + y} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + \frac{{{\rm{d}}y}}{{{\rm{d}}t}}} \right)\\\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \left( {\frac{{x + y}}{z}} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + \frac{{{\rm{d}}y}}{{{\rm{d}}t}}} \right)\\ &= \left( {\frac{{x + y}}{{\sqrt {{{\left( {x + y} \right)}^2} + {{500}^2}} }}} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + \frac{{{\rm{d}}y}}{{{\rm{d}}t}}} \right)\end{aligned}\)

The distance traveled by the woman is:

\(\begin{aligned}x &= \left( 4 \right)\left( {20} \right)\left( {60} \right)\\ &= 4800\end{aligned}\)

The distance traveled by the man is:

\(\begin{aligned}y &= \left( 5 \right)\left( {15} \right)\left( {60} \right)\\ &= 4500\end{aligned}\)

Find the rate at which people are moving apart

Substitute 4800 for x and 4500 for y, 4 for \(\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\) and 5 for \(\frac{{{\rm{d}}y}}{{{\rm{d}}t}}\) in the equation \(\frac{{{\rm{d}}z}}{{{\rm{d}}t}} = \left( {\frac{{x + y}}{{\sqrt {{{\left( {x + y} \right)}^2} + {{500}^2}} }}} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + \frac{{{\rm{d}}y}}{{{\rm{d}}t}}} \right)\).

\(\begin{aligned}\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \left( {\frac{{4800 + 4500}}{{\sqrt {{{\left( {4800 + 4500} \right)}^2} + {{500}^2}} }}} \right)\left( {4 + 5} \right)\\ &= \frac{{837}}{{\sqrt {8674} }}\\ \approx 8.99\;{\rm{ft/s}}\end{aligned}\)

Thus, the rate of people moving apart at \(8.99\;{\rm{ft/s}}\).