Related Rates Problems Solved Step by Step

Related rates problems ask how fast one quantity is changing when you know how fast another, connected quantity is changing. They look intimidating because they are wordy, but almost every one follows the same five-step recipe. Once you learn it, ladders, balloons and water tanks become routine marks.

The Big Idea

If two quantities are linked by an equation and both change with time \( t \), then differentiating that equation with respect to \( t \) links their rates of change. The tool that makes this work is implicit differentiation with the chain rule. For example, if \( x \) depends on time, then

$$ \frac{d}{dt}\left(x^2\right) = 2x \frac{dx}{dt}. $$

The Five-Step Method

1. Draw and label a diagram. Use letters for quantities that change, and numbers only for quantities that stay fixed.
2. Write down what is given and what is wanted, as derivatives. For instance, "the radius grows at 2 cm/s" means \( \frac{dr}{dt} = 2 \).
3. Find an equation connecting the variables (Pythagoras, area, volume, similar triangles, trigonometry).
4. Differentiate both sides with respect to \( t \).
5. Substitute the values at the given instant and solve for the unknown rate. Include units and interpret the sign.

The golden rule: substitute instantaneous values only after differentiating.

Worked Example 1: The Sliding Ladder

A 10 m ladder leans against a vertical wall. The bottom slides away from the wall at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall?

Step 1: Diagram. Let \( x \) be the distance from the wall to the bottom of the ladder and \( y \) the height of the top. The ladder length, 10 m, is fixed.

Step 2: Given and wanted.

$$ \frac{dx}{dt} = 1 \text{ m/s}, \qquad \text{find } \frac{dy}{dt} \text{ when } x = 6. $$

Step 3: Equation. By Pythagoras,

$$ x^2 + y^2 = 100. $$

Step 4: Differentiate with respect to \( t \).

$$ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}. $$

Step 5: Substitute. When \( x = 6 \), \( y = \sqrt{100 - 36} = 8 \). So

$$ \frac{dy}{dt} = -\frac{6}{8}(1) = -0.75 \text{ m/s}. $$

The negative sign means the height is decreasing: the top slides down at 0.75 m/s.

Worked Example 2: The Inflating Balloon

Air is pumped into a spherical balloon so that its volume increases at \( 100 \text{ cm}^3/\text{s} \). How fast is the radius increasing when the radius is 5 cm?

Step 1–2: Variables and rates. Let \( r \) be the radius and \( V \) the volume.

$$ \frac{dV}{dt} = 100 \text{ cm}^3/\text{s}, \qquad \text{find } \frac{dr}{dt} \text{ when } r = 5. $$

Step 3: Equation.

$$ V = \frac{4}{3}\pi r^3. $$

Step 4: Differentiate.

$$ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. $$

Step 5: Substitute.

$$ 100 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt} \quad\Longrightarrow\quad \frac{dr}{dt} = \frac{1}{\pi} \approx 0.318 \text{ cm/s}. $$

Notice that \( 4\pi r^2 \) is the surface area. This makes sense: the extra volume spreads over the surface, so a bigger balloon grows in radius more slowly.

Worked Example 3: Filling a Conical Tank

A conical tank stands with its vertex pointing down. It is 10 m tall with a top radius of 4 m. Water flows in at \( 2 \text{ m}^3/\text{min} \). How fast is the water level rising when the water is 5 m deep?

Step 1: Diagram. Let \( h \) be the depth of water and \( r \) the radius of the water surface.

Step 2: Given and wanted.

$$ \frac{dV}{dt} = 2, \qquad \text{find } \frac{dh}{dt} \text{ when } h = 5. $$

Step 3: Equation. The volume of a cone is \( V = \frac{1}{3}\pi r^2 h \). This has two changing variables, so eliminate \( r \) using similar triangles:

$$ \frac{r}{h} = \frac{4}{10} \quad\Longrightarrow\quad r = 0.4h. $$

Then

$$ V = \frac{1}{3}\pi (0.4h)^2 h = \frac{0.16\pi}{3} h^3. $$

Step 4: Differentiate.

$$ \frac{dV}{dt} = 0.16\pi h^2 \frac{dh}{dt}. $$

Step 5: Substitute \( h = 5 \):

$$ 2 = 0.16\pi (25)\frac{dh}{dt} = 4\pi \frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159 \text{ m/min}. $$

The level rises at about 0.159 m/min at that moment, and it slows down as the tank widens.

Common Mistakes

• Substituting too early. If you put \( x = 6 \) into the ladder equation before differentiating, \( x \) becomes a constant and its rate disappears.
• Wrong signs. A decreasing quantity has a negative rate. Write \( \frac{dV}{dt} = -3 \) for a leaking tank.
• Forgetting the chain rule. \( \frac{d}{dt}(y^2) \) is \( 2y\frac{dy}{dt} \), not \( 2y \).
• Keeping too many variables. In cone problems, use similar triangles to remove the extra variable before differentiating.
• Missing units. Rates always carry units such as m/s or \( \text{cm}^3/\text{s} \).

Tips

• Read the question twice and underline every rate and every "when" value.
• Use the "when" value with the original equation to find any other missing quantity, as we found \( y = 8 \) in the ladder problem.
• Ask whether your answer is sensible. A ladder top moving at 500 m/s means something went wrong.

Key Takeaways

• Related rates link the rates of change of connected quantities through an equation.
• Follow the routine: diagram, given and wanted, equation, differentiate with respect to \( t \), substitute.
• Substitute numerical values only after differentiating.
• Use similar triangles or geometry to reduce the number of variables.
• Always state units and interpret the sign of your answer.

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