Calculus I - Related Rates

Section 3.11 : Related Rates

8. Two people on bikes are at the same place. One of the bikers starts riding directly north at a rate of 8 m/sec. Five seconds after the first biker started riding north the second starts to ride directly east at a rate of 5 m/sec. At what rate is the distance between the two riders increasing 20 seconds after the second person started riding?

A sketch of the two bike riders. The initial position of both people is the point at the lower left. The “North” biker is the point at the upper left with an arrow showing them riding up and the “East” biker is the point at the lower right with an arrow showing them riding to the right. The line drawn between the two riders is labeled “Find Rate This Distance Changing”.

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Here is a sketch of this situation.

$A sketch of the two bike riders. The “North” biker is the point at the upper left with an arrow showing them riding up and the “East” biker is the point at the lower right with an arrow showing them riding to the right. The distance the north biker has gone is labeled $y$, the distance the east biker has gone is labeled $x$ and the line between the two riders is labeled $z$.$

We want to determine $z'$ after 20 seconds after the second biker starts riding east given that $x' = 5$, $y' = 8$ 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. Note that the biker riding east will be riding for 20 seconds and the biker riding north will be riding for 25 seconds (this biker started 5 seconds earlier…).

\[\begin{array}{c}x = \left( 5 \right)\left( {20} \right) = 100\hspace{0.5in}\hspace{0.25in}y = \left( 8 \right)\left( {25} \right) = 200\\ z = \sqrt {{{100}^2} + {{200}^2}} = \sqrt {50000} = 100\sqrt 5 = 223.6068\end{array}\]

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

\[z' = \frac{{\left( {100} \right)\left( 5 \right) + \left( {200} \right)\left( 8 \right)}}{{223.6068}} = \require{bbox} \bbox[2pt,border:1px solid black]{{9.3915}}\]