A procedure for related rates

The calculus in a related-rates problem is usually straightforward. The harder part is translating the situation into an equation that connects the changing quantities, and once that equation is correct, the rest follows a stable procedure.

A five-step routine

1. Draw the situation and name the changing quantities.
2. Write the rates you know and the rate you want.
3. Find an equation relating the variables.
4. Differentiate the equation with respect to time.
5. Substitute the values that describe the particular instant and solve.

The order matters. A value that is true only at one instant should usually be substituted after the differentiation, and a relationship that is true for all time may be used before differentiating.

The most common mistake

Suppose the area of a circle is \(A=\pi r^2\). If a problem asks about the instant when \(r=5\), substituting 5 before differentiating gives \(A=25\pi\). Now \(r\) is gone, and the expression no longer records that the radius is changing.

Instead, differentiate first:

\[\frac{dA}{dt} = 2\pi r\frac{dr}{dt}\]

Then substitute \(r=5\) and the relevant value of \(dr/dt\). The snapshot value belongs after the derivative because it is true only at that instant.

Example 1: a sliding ladder

A 13-foot ladder leans against a wall. Its base moves away from the wall at 2 feet per second. How fast is the top moving when the base is 5 feet from the wall?

Let \(x\) be the distance from the wall to the base and \(y\) the height of the top. We know \(\tfrac{dx}{dt}=2\), and we want \(dy/dt\). The ladder length is constant, so \(x^2+y^2=169\), and differentiating with respect to \(t\) gives

\[2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\]

At the instant when \(x=5\), the Pythagorean theorem gives \(y=12\). Now substitute to get \(2(5)(2) + 2(12)\tfrac{dy}{dt} = 0\), and therefore \(\tfrac{dy}{dt} = -\tfrac56\). The top is moving downward at \(\tfrac56\) foot per second, and the negative sign records the direction.

The base moves at a constant rate, but the top does not. As the ladder becomes flatter, the magnitude of \(dy/dt\) increases, and the relationship between the two rates depends on the geometry at that instant. This is why the position values cannot be inserted before differentiating.

Example 2: a draining cone

An inverted conical tank has radius 6 meters and height 12 meters. Water drains at 3 cubic meters per minute. How fast is the water level falling when the water is 4 meters deep?

Let \(h\) be the water depth, \(r\) the radius of the water surface, and \(V\) the volume. Both \(r\) and \(h\) change, but the shape of the tank gives a relationship that is true at every instant, and similar triangles give \(\tfrac{r}{h} = \tfrac{6}{12}\), so \(r=\tfrac{h}{2}\). Because this relationship is always true, it can be substituted before differentiation, and the volume is

\[V = \frac13\pi r^2h = \frac13\pi\left(\frac h2\right)^2h = \frac{\pi h^3}{12}\]

Differentiating gives \(\tfrac{dV}{dt} = \tfrac{\pi h^2}{4}\tfrac{dh}{dt}\), and at the instant in question, \(\tfrac{dV}{dt}=-3\) and \(h=4\). So \(-3 = 4\pi\tfrac{dh}{dt}\), which gives \(\tfrac{dh}{dt} = -\tfrac{3}{4\pi}\), and the water level is falling at \(\tfrac{3}{4\pi} \approx0.239\) meter per minute.

Notice the two kinds of substitution. The relationship \(r=h/2\) is structural and holds for all time, while the value \(h=4\) is a snapshot and waits until after differentiation.

Example 3: a camera tracking a rocket

A camera is 100 meters from a launch pad. A rocket rises vertically at 50 meters per second. How fast is the camera’s angle of elevation increasing when the rocket is 100 meters high?

Let \(y\) be the rocket’s height and \(\theta\) the angle of elevation. The horizontal distance is constant, so \(\tan\theta = \tfrac{y}{100}\), and differentiating with respect to time gives

\[\sec^2\theta\frac{d\theta}{dt} = \frac{1}{100}\frac{dy}{dt}\]

When \(y=100\), \(\theta=\tfrac{\pi}{4}\), so \(\sec^2\theta=2\). Using \(\tfrac{dy}{dt}=50\), we get \(2\tfrac{d\theta}{dt} = \tfrac{50}{100}\), and therefore \(\tfrac{d\theta}{dt} = \tfrac14\). The angle is increasing at one quarter radian per second, and radians matter here because the standard trigonometric derivative formulas assume radian measure.

A final check

Before finishing a related-rates problem, check four things.

• Does the sign agree with the story?
• Do the units work?
• Did every changing variable receive its own rate during differentiation?
• Does the final sentence state the value, units, and direction?

If the tank is draining, \(dV/dt\) should be negative. If a term such as \(y^2\) was differentiated with respect to time, the result should contain \(dy/dt\), and if the units do not balance across the differentiated equation, something in the setup is wrong.