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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?

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Start Solution

Here is a sketch of this situation.

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.

Show Step 2

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

\[{z^2} = {x^2} + {y^2}\] Show Step 3

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.

\[2z\,z' = 2x\,x' + 2y\,y'\hspace{0.5in} \Rightarrow \hspace{0.5in}z' = \frac{{x\,x' + y\,y'}}{z}\] Show Step 4

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,

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