Section 4.1 Related Rates
References.
- Calculus, Early Transcendentals by Stewart, Section 3.9.
In many physical situations, several related quantities change with time, such as the pressure \(P\) and volume \(V\) of a fixed amount of a gas at a fixed temperature (\(PV=c\text{.}\)) When the changing quantities are related by a known formula, their rates of change are also related, so that a measurement of one rate of change can be used to determine the other. In the above example, if the applied pressure is increased at a given rate, one can predict how fast the volume will be decreasing.
Since the rates of change are with respect to the variable time, this leads to implicit differentiation of the formula relating the two quantities. For example, the above equation could be spelt out as \(P(t) \cdot V(t) = c\text{,}\) with \(c\) a known constant.
Strategy for Related Rates Problems.
Firstly, what identifies a related rates problem is that you are asked to find one rate of change (a derivative, such as a velocity) using information about one or more other rates of change (other derivatives), and you often have that rate of change information only at one time, not as functions of time. Most often the independent variable is time, \(t\text{,}\) so I assume that here. In the following strategy, always keep track of your main goal: to compute the derivative of a certain variable with respect to \(t\text{.}\) Keep your eye on that “key” variable!Strategy for Related Rates Problems.
- Read all the given information carefully, and identify the main goal: What is the quantity whose rate of change you wish to know?
- Give names to all relevant variables; in particular name variable quantities whose rate of change is either wanted or is known. (Constants usually do not need names.)
- Draw a diagram relating all the variables and other known values, if appropriate.
- Describe the rate of change sought as the derivative of a certain variable.
- Seek an equation (or several equations) relating just the relevant variables. That is, variables whose derivatives are either known or wanted. Make sure that these equations are true at all times, not just at one moment.
- Differentiate the equation[s] with respect to time. This is implicit differentiation, with “hidden” compositions.
- Substitute known values into the equation[s] got by differentiating. At this stage only, the values might only be valid at the one time of interest.
- Solve the resulting equation[s] for the desired rate of change (derivative). It might be necessary to also substitute all known values into the original equations (from step 5), and use this larger collection of equations to solve for the desired rate of change.
- Answer the original question! That is, relate your mathematical results back to the question asked, preferably as a verbal statement of the result. Put back in physical units, and interpret the sign of the derivative as saying whether the quantity is increasing or decreasing.
Air is being pumped into a spherical balloon at a rate of 100cm\({}^3\)/s. How fast is the radius increasing when the diameter is 50cm?
Solution.Writing \(V=V(t)\) for the volume, a function of time, and \(r=r(t)\) for the radius:
- What we want to know is \(dr/dt\) at a certain moment.
- what we know is that \(dV/dt=100\text{cm}^3\)/s and \(V=\frac{4}{3}\pi r^3\) at all times, and at that moment, \(r=50/2\) cm (half the diameter.)
\begin{equation*}
V(t)=\frac{4}{3}\pi [r(t)]^3,
\quad \text{ or just }\quad
V= \frac{4}{3}\pi r^3.
\end{equation*}
This can be differentiated with respect to time, involving the Chain Rule, or implicit differentiation of \(r\) with respect to t:
\begin{equation*}
\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}.
\end{equation*}
At the moment of interest, \(r=25\) and \(dV/dt=100\text{,}\) and substituting in these numbers: \(100 = 4\pi (25)^2 \displaystyle\frac{dr}{dt}\text{,}\) so
\begin{equation*}
\frac{dr}{dt} = \frac{100}{4 \pi (25)^2}= \frac{1}{25 \pi} \approx 0.0127\text{cm/s.}
\end{equation*}
We could also solve for the unknown derivative at any moment in terms of known quantities:
\begin{equation*}
\frac{dr}{dt} = \frac{dV/dt}{4 \pi r^2}\text{.}
\end{equation*}
Exercises Exercises
Study Calculus Volume 1, Section 4.1 2 ; in particular the Problem Solving Strategy, all Examples and Checkpoints, and Exercises 1, 3, 5, 7, 9, 17, and 25.
openstax.org/books/calculus-volume-1/pages/4-1-related-ratesopenstax.org/books/calculus-volume-1/pages/4-1-related-rates
