In this section, we shall look at ‘related rates.’ In this class of problems, two or more related quantities are changing. The rate of change of one quantity is given, and we seek to determine the rate of change of the other related quantities. To start, we need to define the average and instantaneous rates of change; these are not new concepts but new names for old, familiar ones.

**Definition 1.**If \(y=f(x)\), then the**average rate of change**of \(y\) with respect to \(x\) over the interval \(\left[x_{0},x_{1}\right]\) is the slope of the secant line, \(m_{\text{secant}}\), passing through \((x_{0},f(x_{0}))\) and \(\left(x_{1},f\left(x_{1}\right)\right)\) (which are on the graph of \(f\)) (Figure 1(a)).*If \(y=f(x)\), then the instantaneous rate of change of \(y\) with respect to \(x\) at the point \(x_{0}\) is the slope of the tangent line, \(m_{\text{tangent}}\), to the graph of \(f\) at the point \(x_{0}\) (Figure 1(b)).*

*If a rate is positive, it is often called a rate of increase.**If a rate is negative, its absolute value is often called a rate of decrease.**If we simply say “rate,” we mean instantaneous rate.*

(a) The slope of the secant, \(\dfrac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\), is the average rate of change of \(y\) with respect to \(x\) over the interval \([x_{0},x_{1}]\). | (b) The slope of the tangent line at the point \(x_{0}\), \(m_{\text{tangent}}=f^\prime(x_{0})\), is the instantaneous rate of change of \(y\) at \(x_{0}\). |

*Figure 1*

As René Descartes says in Discourse on Method, “Each problem that I solved became a rule, which served afterward to solve other problems.” Similarly, we can explain the rule to solve problems of this kind as follows.

Strategy for Solving Related Rates Problems.

- Assign appropriate letters to the variables and constants that remind you their actual meaning. For example, represent the volume by \(V\), the length by \(L,x\), or \(y\), temperature by \(T\), and time by \(t\).
- Write down the given information and what you are asked to find.
- Find an equation that relates the quantity whose rate you do not know to those other quantities whose rates are already known. You may need to combine two or more equations to get a single equation. Drawing a simple picture sometimes facilitates this step.
- Differentiate both sides of the equation with respect to the independent variable (which is usually time \(t\)) thereby you can express the rate you want in terms of the known rates and variables.
- Substitute the given values into the resulting equation and solve for the unknown rate using the known values.

WARNING: Always substitute the given information for the varying quantities AFTER differentiation. For example, in the above example, if we had substituted \(r=50\) before differentiation, we would have gotten \(A=50^{2}\pi\Rightarrow dA/dt=0\) (because \(50\pi^{2}\) is a constant and the derivative of a number is zero).

We are filming a rocket launch with a camera installed 900 m away from the launching pad (Figure 3). When the rocket lifts vertically, obviously the elevation angle increases. The next two examples deal with this problem, assuming that the camera’s focal point changes automatically in such a way that we can get a clear view of the rocket at all times.