Calculus 1: 10 ft Ladder Slides Away at 1 ft/s, How Fast Is the Top Sliding Down When the Bottom Is 6 ft From the Wall?

The Problem

A ladder 10 ft long leans against a vertical wall. The bottom of the ladder is pulled away from the wall at a constant rate of 1 ft/s. How fast is the top of the ladder sliding down the wall at the instant the bottom is 6 ft from the wall?

This is one of the most common related rates problems in Calculus 1 (it appears in nearly every textbook section on related rates), so expect it, or a version with different numbers, on a midterm.

Step 1: Name the Variables

• $x(t)$ = distance from the wall to the bottom of the ladder (ft)
• $y(t)$ = height of the top of the ladder on the wall (ft)
• Given: $\frac{dx}{dt} = 1$ ft/s (positive, because $x$ is growing)
• Find: $\frac{dy}{dt}$ when $x = 6$

The ladder length stays 10 ft the whole time. That constant is the key relation.

Step 2: Write the Relation That Holds at Every Moment

The wall, the ground, and the ladder form a right triangle:

$$ x^2 + y^2 = 10^2 = 100 $$

Step 3: Differentiate With Respect to Time

Both $x$ and $y$ depend on $t$, so use the chain rule on each term:

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

Solve for the unknown rate:

$$ \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} $$

Step 4: Find y at the Instant

When $x = 6$: $36 + y^2 = 100$, so $y^2 = 64$ and $y = 8$ ft (a 6-8-10 right triangle).

Step 5: Substitute

$$ \frac{dy}{dt} = -\frac{6}{8}\cdot 1 = -\frac{3}{4}\ \text{ft/s} $$

Answer: the top of the ladder slides down the wall at $\mathbf{3/4\ \text{ft/s} = 0.75\ \text{ft/s}}$.

Sanity Check

• The sign is negative: the top moves down while the bottom moves out. That matches the picture.
• The top moves slower than the bottom right now (0.75 vs 1 ft/s) because the ladder is still fairly steep ($y > x$). Later, when the bottom is 8 ft out, the same formula gives $\frac{dy}{dt} = -\frac{8}{6}\cdot 1 = -\frac{4}{3}$ ft/s, so the top speeds up as it nears the ground.

Common Follow-Up: How Fast Is the Angle Changing?

Let $\theta$ be the angle between the ladder and the ground, so $\cos\theta = \frac{x}{10}$. Differentiating gives $-\sin\theta\,\frac{d\theta}{dt} = \frac{1}{10}\frac{dx}{dt}$. At $x = 6$, $\sin\theta = \frac{8}{10}$, so

$$ \frac{d\theta}{dt} = -\frac{1}{10}\cdot\frac{10}{8} = -\frac{1}{8}\ \text{rad/s} $$

The angle with the ground is shrinking at 1/8 radian per second.

Traps That Cost Points

• Substituting too early. Plugging $x = 6$ into $x^2 + y^2 = 100$ before differentiating turns $x$ into a constant and kills $\frac{dx}{dt}$. Always differentiate the general relation first.
• Dropping the sign. If the question asks "how fast is the top sliding down", report 3/4 ft/s and say "down". If it asks for $\frac{dy}{dt}$, the answer is $-3/4$ ft/s.
• Forgetting the chain rule. $\frac{d}{dt}(y^2) = 2y\frac{dy}{dt}$, not $2y$.
• Treating the ladder length as a variable. The 10 ft is constant, so its derivative is 0.

The 5-Step Template for Any Related Rates Problem

1. Draw the picture and name every changing quantity.
2. List the given rate(s) and the rate you need, with signs.
3. Write one equation linking the quantities that is true at every moment.
4. Differentiate with respect to $t$ (chain rule on every variable).
5. Only now substitute the instant values and solve.

For a related implicit differentiation problem, see the tangent line to x³ + y³ = 6xy at (3, 3).