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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,

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