Related Rates Problems Step by Step (Worked Examples)

In a related rates problem, every changing quantity is a function of time $t$. If you have an equation connecting the quantities, you can differentiate both sides with respect to $t$ to get an equation connecting their rates.

For example, if $x$ and $y$ both change over time and $x^2 + y^2 = 100$, then implicit differentiation with respect to $t$ gives:

Every variable picks up its own rate, like $\frac{dx}{dt}$ or $\frac{dy}{dt}$, because of the chain rule.

1. Draw and label a picture. Give every changing quantity a letter. Only use numbers for things that never change.
2. List what you know and what you want. Write each rate with units, like $\frac{dx}{dt} = 2$ ft/s. Decreasing quantities get a negative rate.
3. Write an equation that connects the variables (Pythagorean theorem, volume, area, similar triangles).
4. Differentiate both sides with respect to $t$.
5. Plug in the values at the moment you care about. Do this only after differentiating. Use the original equation to find any missing values.
6. Solve for the unknown rate and state it with units and in words.

Problem: A 10 ft ladder leans against a wall. The bottom slides away from the wall at 2 ft/s. How fast is the top sliding down the wall when the bottom is 6 ft from the wall?

1. Picture: Let $x$ be the distance from the wall to the bottom of the ladder, and $y$ the height of the top. The ladder length stays 10.
2. Know and want: $\frac{dx}{dt} = 2$ ft/s. Find $\frac{dy}{dt}$ when $x = 6$.
3. Equation: The wall, ground, and ladder form a right triangle: $x^2 + y^2 = 10^2 = 100$.
4. Differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.
5. Plug in: First find $y$ when $x = 6$: $36 + y^2 = 100$, so $y = 8$. Now substitute: $2(6)(2) + 2(8)\frac{dy}{dt} = 0$, which is $24 + 16\frac{dy}{dt} = 0$.
6. Solve: $\frac{dy}{dt} = -\frac{24}{16} = -1.5$ ft/s.

The negative sign means $y$ is decreasing. The top of the ladder is sliding down at 1.5 ft/s.

Problem: Air is pumped into a spherical balloon at $100 \text{ cm}^3/\text{s}$. How fast is the radius increasing when the radius is 5 cm?

1. Picture: A sphere with radius $r$ and volume $V$, both changing.
2. Know and want: $\frac{dV}{dt} = 100 \text{ cm}^3/\text{s}$. Find $\frac{dr}{dt}$ when $r = 5$.
3. Equation: Volume of a sphere: $V = \frac{4}{3}\pi r^3$.
4. Differentiate: $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$.
5. Plug in: $100 = 4\pi(5)^2\frac{dr}{dt} = 100\pi\frac{dr}{dt}$.
6. Solve: $\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318$ cm/s.

Follow-up: How fast is the surface area growing at that moment? Surface area is $S = 4\pi r^2$, so $\frac{dS}{dt} = 8\pi r\frac{dr}{dt} = 8\pi(5)\left(\frac{1}{\pi}\right) = 40 \text{ cm}^2/\text{s}$.

Notice that the radius grows more slowly as the balloon gets bigger, because the same amount of air is spread over a larger surface.

Problem: Water flows into an upside-down cone tank at $2 \text{ ft}^3/\text{min}$. The tank is 12 ft tall with a top radius of 4 ft. How fast is the water level rising when the water is 6 ft deep?

1. Picture: The water forms a smaller cone with height $h$ and radius $r$.
2. Know and want: $\frac{dV}{dt} = 2$. Find $\frac{dh}{dt}$ when $h = 6$.
3. Equation: $V = \frac{1}{3}\pi r^2 h$ has two changing variables, so remove one. By similar triangles, $\frac{r}{h} = \frac{4}{12}$, so $r = \frac{h}{3}$. Substitute:

4. Differentiate: $\frac{dV}{dt} = \frac{3\pi h^2}{27}\frac{dh}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt}$.
5. Plug in: $2 = \frac{\pi (36)}{9}\frac{dh}{dt} = 4\pi\frac{dh}{dt}$.
6. Solve: $\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} \approx 0.159$ ft/min.

• Signs: If a quantity is shrinking (water draining, a ladder top falling, a distance closing), its rate is negative. If your answer has an unexpected sign, recheck which way things move.
• Units: A rate is always "units of the quantity per unit of time." Volume rates use cubic units, area rates use square units.
• Reasonableness: In the ladder problem, the top moved slower (1.5 ft/s) than the bottom (2 ft/s) at that moment. That makes sense because the ladder was still fairly upright.

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