Related Rates in Calculus

A related rates problem describes two or more quantities that change over time and are linked by an equation. You are usually told how fast one quantity is changing and asked to find how fast another quantity is changing at a specific instant. Because both quantities depend on time, you differentiate the connecting equation implicitly with respect to \(t\), which brings in \(\frac{dx}{dt}\), \(\frac{dy}{dt}\), or similar rate terms through the chain rule.

This topic assumes you are already comfortable differentiating implicitly. If you need a refresher on taking derivatives of both sides of an equation with respect to a variable, review implicit differentiation before working through the examples below.

Almost every related rates problem can be solved by following the same sequence of steps.

1. Draw a picture of the situation and label every quantity that changes with a variable.
2. Write down the rate you are given and the rate you need to find, using derivative notation such as \(\frac{dx}{dt}\).
3. Find an equation that connects the variables at any moment in time, often from geometry such as the Pythagorean theorem, similar triangles, or a volume formula.
4. Differentiate both sides of that equation with respect to \(t\), treating each variable as a function of time.
5. Substitute the known numerical values, including the given rate, only after differentiating.
6. Solve the resulting equation for the unknown rate and check that the sign and units make sense.

A 10 foot ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at \(\frac{dx}{dt} = 2\) ft/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?

The wall, ground, and ladder form a right triangle, so at every moment \(x^2 + y^2 = L^2\). Since the ladder length is fixed at 10, this becomes \(x^2 + y^2 = 100\). Differentiating both sides with respect to \(t\) gives:

\(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)

When \(x = 6\), the Pythagorean theorem gives \(y = 8\) (a 6 to 8 to 10 triangle). Substituting \(x = 6\), \(y = 8\), and \(\frac{dx}{dt} = 2\):

\(2(6)(2) + 2(8)\frac{dy}{dt} = 0\)

\(24 + 16\frac{dy}{dt} = 0\)

\(\frac{dy}{dt} = -1.5\) ft/s

The negative sign tells you the height on the wall is decreasing, so the top of the ladder is sliding down at 1.5 feet per second at that instant.

Water drains from a conical tank with a top radius of 5 meters and a height of 10 meters, vertex pointing down. Water leaves the tank at \(\frac{dV}{dt} = -3\) cubic meters per minute. How fast is the water level dropping when the water is 4 meters deep?

The volume of a cone is \(V = \frac{1}{3}\pi r^2 h\), but \(r\) and \(h\) are not independent here: the water always forms a smaller cone similar to the tank, so \(\frac{r}{h} = \frac{5}{10} = \frac{1}{2}\), giving \(r = \frac{h}{2}\). Substituting this into the volume formula eliminates \(r\):

\(V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi}{12}h^3\)

Differentiating with respect to \(t\):

\(\frac{dV}{dt} = \frac{\pi}{4}h^2\frac{dh}{dt}\)

Now substitute \(h = 4\) and \(\frac{dV}{dt} = -3\):

\(-3 = \frac{\pi}{4}(16)\frac{dh}{dt}\)

\(\frac{dh}{dt} = \frac{-3}{4\pi} \approx -0.24\) meters per minute

The water level is falling at about 0.24 meters per minute when the depth is 4 meters. This same similar triangle trick, replacing one variable so the formula involves only a single changing quantity, appears in many related rates cone and shadow problems.

A few errors show up again and again in related rates work.

• Substituting the given numbers before differentiating. Once you plug in a specific value for a variable, that variable stops behaving like a variable, so any derivative taken afterward is meaningless.
• Forgetting to apply the chain rule when a variable is not \(t\) itself, which is exactly why \(\frac{d}{dt}(x^2) = 2x\frac{dx}{dt}\) rather than just \(2x\).
• Losing track of which rate is given and which rate is unknown, especially when a problem gives several rates at once.
• Ignoring the sign of an answer. A negative rate usually means a quantity is decreasing, which is important information, not a mistake to fix.

Ladder, cone, and shadow problems are the three setups you will see most often, but the same six step strategy applies to any scenario, including expanding circles, moving cars, and inflating balloons. Draw a fresh diagram for every problem, write out what is given and what is unknown before touching any algebra, and always differentiate the general equation first.

Once you are confident with related rates, you are ready for other applications of the derivative in optimization. Many of those problems ask you to locate the critical numbers of a function to find a maximum or minimum value. Rates of change also show up outside geometry, for instance in economics when working with the price elasticity of demand formula, which measures how sensitive demand is to a changing price.

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