18. Related Rates

a. For Variables Related by an Equation

When two quantities are related and one of them changes, the other changes in a prescribed way. If we know the rate of change of one quantity, we can calculate the related rate of change of the other quantity. If one variable is given explicitly as a function of the other variable, then the related rate can be found simply by using the chain rule. For example:

We differentiate the volume equation remembering that \(r\) and \(V\) are both functions of time and using the Chain Rule: \[ \dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt} \] (a) Since \(r=2\) and \(\dfrac{dr}{dt}=5\), we calculate: \[ \dfrac{dV}{dt}=4\pi (2)^2(5)=80\pi\,\dfrac{\text{cm}^3}{\text{sec}} \]
(b) Conversely, since \(r=2\) and \(\dfrac{dV}{dt}=6\), we calculate: \[ 6=4\pi (2)^2\dfrac{dr}{dt} \qquad \text{or} \qquad \dfrac{dr}{dt}=\dfrac{3}{8\pi}\,\dfrac{\text{cm}}{\text{sec}} \]

The two variables are not always related by an explicit function. They may be implicitly related. For example:

Suppose \(x\) and \(y\) satisfy the equation \[ x^3y=6-x^2y^2 \] We are currently at the point \((x,y)=(1,2)\).
(a) If the rate of change of \(x\) is currently \(\dfrac{dx}{dt}=3\), find the current rate of change of \(y\).
(b) If the rate of change of \(y\) is currently \(\dfrac{dy}{dt}=6\), find the current rate of change of \(x\).

We first check that the point \((x,y)=(1,2)\) satisfies the equation. Since \((1)^3(2)=2\) while \(6=(1)^2(2)^2=2\) it does satisfy the equation. If it didn't, you should complain to your instructor, that the problem makes no sense.
We now implicitly differentiate the relation equation using the Product and Chain Rules: \[ 3x^2\dfrac{dx}{dt}y+x^3\dfrac{dy}{dt}=-2x\dfrac{dx}{dt}y^2-x^22y\dfrac{dy}{dt} \] We can now plug in the known current values and solve for the unknown current values.
(a) Currently, \(x=1\), \(y=2\) and \(\dfrac{dx}{dt}=3\). So: \[ 3(1)^2(3)(2)+(1)^3\dfrac{dy}{dt}=-2(1)(3)(2)^2-(1)^22(2)\dfrac{dy}{dt} \] We solve for \(\dfrac{dy}{dt}\): \[\begin{aligned} 18+\dfrac{dy}{dt}&=-24-4\dfrac{dy}{dt} \\ 5\dfrac{dy}{dt}&=-42 \\ \dfrac{dy}{dt}&=-\,\dfrac{42}{5} \end{aligned}\]
(b) Currently, \(x=1\), \(y=2\) and \(\dfrac{dy}{dt}=6\). So: \[ 3(1)^2\dfrac{dx}{dt}(2)+(1)^3(6)=-2(1)\dfrac{dx}{dt}(2)^2-(1)^22(2)(6) \] We solve for \(\dfrac{dx}{dt}\): \[\begin{aligned} 6\dfrac{dx}{dt}+6&=-8\dfrac{dx}{dt}-24 \\ 14\dfrac{dx}{dt}&=-30 \\ \dfrac{dx}{dt}&=-\,\dfrac{15}{7} \end{aligned}\]