Calculus I - Related Rates

Section 3.11 : Related Rates

7. Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later?

A sketch of the person and the elevator. The initial position of both people is the point at the lower right. The person walking away is the point at the lower left with an arrow showing them moving left, and the elevator is the point at the upper right with an arrow showing it moving up. The line drawn between the person and the elevator is labeled “Find Rate This Distance Changing”.

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Here is a sketch for this part.

$A sketch of the person and the elevator. The person on the ground is the point at the lower left with an arrow showing them walking away and the elevator is the point at the upper right with an arrow showing it going up. The horizontal distance is labeled $x$, the height of the elevator is labeled $y$ and the line between the two people is labeled $z$.$

We want to determine $z'$ after 15 seconds given that $x' = 2$, $y' = 7$ and assuming that they start at the same point.

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Hopefully it’s clear that we’ll need the Pythagorean Theorem to solve this problem so here is that.

\[{z^2} = {x^2} + {y^2}\]

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Finally, let’s differentiate this with respect to $t$ and we can even solve it for $z'$ so the actual solution will be quick and simple to find.

\[2z\,z' = 2x\,x' + 2y\,y'\hspace{0.5in} \Rightarrow \hspace{0.5in}z' = \frac{{x\,x' + y\,y'}}{z}\]

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To finish off this problem all we need to do is determine all three lengths of the triangle in the sketch above. We can find $x$ and $y$ using their speeds and time while we can find $z$ by reusing the Pythagorean Theorem.

\[\begin{array}{c}x = \left( 2 \right)\left( {15} \right) = 30\hspace{1.0in}y = \left( 7 \right)\left( {15} \right) = 105\\ z = \sqrt {{{30}^2} + {{105}^2}} = \sqrt {11925} = 15\sqrt {53} = 109.2016\end{array}\]

The rate of change of the distance between the two people is then,

\[z' = \frac{{\left( {30} \right)\left( 2 \right) + \left( {105} \right)\left( 7 \right)}}{{109.2016}} = \require{bbox} \bbox[2pt,border:1px solid black]{{7.2801}}\]