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%TCIDATA{<META NAME="Title" CONTENT="Reference/Mathematics/Related Rates">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
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\section{Related Rates\label{Related rates}}

In related rates problems, you compute the rate of change of one quantity in
terms of the rate of change of another quantity. The procedure is to find an
equation that relates the two quantities and then to use the chain rule to
differentiate both sides of the equation with respect to time.

\begin{example}
Air is being pumped into a spherical balloon so that its volume increases at
a rate of $100\unit{cm}/\unit{s}$. How fast is the radius of the balloon
increasing when the diameter is $50\unit{cm}$? The equation that relates the
radius of the balloon to its volume is 
\[
V=\frac{4}{3}\pi r^{3}
\]
Differentiating both sides of the equation with respect to time gives 
\[
\frac{dV}{dt}=\frac{4}{3}\pi \left( 3r^{2}\right) \frac{dr}{dt}
\]
Solving for $\dfrac{dr}{dt}$ and then substituting the value $\frac{dV}{dt}%
=100\frac{\unit{cm}}{\unit{s}}$ gives the equations 
\[
\frac{dr}{dt}=\frac{1}{4\pi r^{2}}\frac{dV}{dt}=\frac{1}{4\pi r^{2}}100\frac{%
\unit{cm}}{\unit{s}}=\frac{25}{\pi r^{2}}\frac{\unit{cm}}{\unit{s}}
\]
Now you can find the rate at which the radius of the balloon is increasing
when the diameter is $50\unit{cm}$, namely, 
\[
\left[ \frac{dr}{dt}\right] _{r=50}=\frac{25}{\pi 50^{2}}\frac{\unit{cm}}{%
\unit{s}}=\frac{1}{100\pi }\frac{\unit{cm}}{\unit{s}}
\]
\end{example}

\subsection{Related Reference Topics}

{\small \hyperref{Principles of Mathematical Writing}{}{}{%
Mathematical_writing.tex}}

{\small \hyperref{Principles of Problem Solving}{}{}{cal00_04.tex}}

\subsection{Related Scientific Notebook Topics}

{\small \hyperref{Computing implicit derivatives}{}{}{%
../../Help/DM7-3.tex#Implicit differentiation}}

{\small \hyperref{Differentiation}{}{}{../../Help/DM7-3.tex#Derivative}}

{\small \hyperref{Numbers, Functions, and Units}{}{}{../../Help/DM2.tex}}

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