Related rates: the sliding ladder

TL;DR:

• Related-rates problems connect the rates of two quantities that are tied by an equation.

• Differentiate with respect to time first, and substitute the numbers after.

• For a 10 ft ladder, if the foot slides out at 2 ft/s, the top falls at 1.5 ft/s when the foot is 6 ft away.

A 10 ft ladder leans against a wall. Someone pulls the foot of the ladder away from the wall at a steady 2 ft/s. How fast is the top sliding down at the moment the foot is 6 ft from the wall? The two ends move in a linked way, and the link is a right triangle.

The method in four moves

1. Draw the situation and label every quantity that changes with a letter.

2. Write an equation that is true at every moment and ties those letters together.

3. Differentiate both sides with respect to time t, using the chain rule.

4. Only then substitute the numbers for the moment you are asked about, and solve for the unknown rate.

The order matters. Numbers that change cannot be plugged in before you differentiate, because that would freeze them.

Worked example: a 10 ft ladder sliding down a wall

Let x be the distance of the foot from the wall and y the height of the top. Both depend on t. The ladder is the hypotenuse, so x² + y² = 100 at all times. We are told dx/dt = 2 ft/s, and want dy/dt when x = 6.

1. Differentiate: 2x · dx/dt + 2y · dy/dt = 0. The right side is 0 because 100 never changes.

2. Solve for the unknown: dy/dt = −(x/y) · dx/dt.

3. Find y at the moment: y = √(100 − 36) = √64 = 8 ft.

4. Substitute: dy/dt = −(6/8) × 2 = −1.5 ft/s.

The top is moving down at 1.5 ft/s. The negative sign means y is decreasing.

Check the answer with a small time step

After 0.01 s, the foot has moved 0.02 ft, so x = 6.02. Then y = √(100 − 36.2404) = √63.7596 ≈ 7.98497. The drop is 8 − 7.98497 = 0.01503 ft in 0.01 s, a speed of about 1.503 ft/s. That is very close to 1.5, as it should be for a tiny time step.

What the sign and size tell you

The top moves slower than the foot (1.5 against 2), because the ladder is still fairly upright. As the foot moves further out, the top speeds up. Try x = 8, then y = 6 and dy/dt = −(8/6) × 2 ≈ −2.67 ft/s. Near the ground, as x approaches 10 ft, y goes to 0 and the formula says the speed grows without limit. A real ladder would leave the wall before that, so the model stops being true. Always ask whether the answer is physically sensible.

Another quick related-rates idea: a growing circle

For a circle of radius r, the area is A = πr². Differentiate in time: dA/dt = 2πr · dr/dt. If the radius grows at 0.5 cm/s when r = 4 cm, the area grows at 2π × 4 × 0.5 = 4π ≈ 12.6 cm² per second. The pattern is the same: an equation, a time derivative, then numbers.

Try one yourself: an inflating balloon

A spherical balloon has volume V = (4/3)πr³. Differentiate in time: dV/dt = 4πr² · dr/dt. If the radius is 3 cm and grows at 0.2 cm/s, then dV/dt = 4π × 9 × 0.2 = 7.2π ≈ 22.6 cm³/s. Check the units: cm² times cm/s gives cm³/s. Check the size: the surface area at that moment is 4π × 9 ≈ 113 cm², and a shell 0.2 cm thick has volume about 113 × 0.2 ≈ 22.6 cm³. Same answer, found a different way.

How to check your answer

• Units. Feet divided by seconds gives ft/s.

• Direction. Things that shrink get a negative rate.

• Relative sizes. Compare with the other rate. The ladder example shows the top slower than the foot when x is less than y.

• Time step. Move forward a small time and compute the change numerically.

Common mistakes

• Plugging in x = 6 before differentiating. Then x is a constant, and dx/dt disappears from the equation.

• Forgetting the factor dx/dt that the chain rule gives.

• Using the wrong sign for a quantity that is shrinking.

• Mixing up which rate is given and which is wanted.

• Writing the equation for one instant only, instead of one that holds at all times.

Where you will use it next

Related rates is the chain rule at work, as explained in the chain rule with one picture. The next application of derivatives is finding best values, in optimization: the biggest open box from a square sheet. The triangle behind the ladder is the same one in the Pythagorean theorem.

Safety note: A real ladder can slip. Set it at a safe angle, on firm ground, with someone holding the base, and never rely on a maths model to judge safety.

Frequently asked questions

Why differentiate both sides of the equation?

Both sides are equal at every moment, so they change at the same rate as time passes. Differentiating each side with respect to time keeps the equation true.

Do I always need a right triangle?

No. The equation can come from any relationship, such as a volume formula or a similar-triangle ratio. The method is the same.