Ladder Length Related Rates- Calculus Problem Solutions
What the Ladder Problem Actually Is
You've seen it. A ladder leans against a wall. Someone pulls the bottom away. You need to find how fast the top is sliding down when the bottom is 6 feet away.
This is the classic related rates problem. Two quantities change simultaneously, and you need to find the rate of change of one given the rate of change of the other.
The math behind it isn't complicated. But students mess it up because they skip steps or plug numbers in too early.
The Setup: What You're Working With
Every related rates problem has three parts you need to identify:
- The primary relationship — an equation connecting your variables
- The given rate — what you know (like "ladder sliding at 2 ft/sec")
- The requested rate — what you need to find
For the ladder problem, the primary relationship comes from the Pythagorean theorem:
x² + y² = L²
Where:
- x = distance from wall to bottom of ladder
- y = distance from floor to top of ladder
- L = ladder length (stays constant)
Why the Ladder Length Stays Constant
Here's where students get confused. The ladder doesn't stretch or shrink. Its length L is fixed. When you take the derivative, L² becomes 0 because it's a constant.
This gives you the working equation:
2x(dx/dt) + 2y(dy/dt) = 0
Or simplified:
x(dx/dt) + y(dy/dt) = 0
Step-by-Step Solution
Let's solve the standard problem:
"A 25-foot ladder rests against a wall. The bottom is pulled away at 3 ft/sec. How fast is the top sliding down when the bottom is 7 feet from the wall?"
Step 1: Identify Your Knowns
L = 25 ft (constant)
dx/dt = +3 ft/sec (bottom moving away, positive direction)
x = 7 ft (current position of bottom)
Step 2: Find y
Use the Pythagorean theorem with your known values:
7² + y² = 25²
49 + y² = 625
y² = 576
y = 24 ft
Always find both variables before taking derivatives. You need y to finish the problem.
Step 3: Apply the Related Rates Equation
Plug into: x(dx/dt) + y(dy/dt) = 0
7(3) + 24(dy/dt) = 0
21 + 24(dy/dt) = 0
24(dy/dt) = -21
dy/dt = -21/24 = -7/8 ft/sec
The negative sign means the top is sliding down. The ladder goes down as the bottom goes out. This is physics, not just math.
Common Mistakes That Sink Students
Mistake 1: Forgetting Units
Your answer isn't just a number. It's -7/8 feet per second. Missing units costs you points and shows you don't understand what you're calculating.
Mistake 2: Plugging In Before Differentiating
Some students substitute x = 7 into the original equation before taking the derivative. This gives you the wrong derivative. Take the derivative first, then plug in numbers.
Mistake 3: Ignoring the Sign
A positive answer means increasing. A negative answer means decreasing. In this problem, the top of the ladder must be going down. If you get a positive number, something's wrong.
Mistake 4: Using the Wrong Equation
Some problems use similar triangles or trigonometry. The ladder problem always uses the Pythagorean theorem. If your equation doesn't look like x² + y² = L², double-check your setup.
Comparison: Ladder Problem vs. Similar Problems
| Problem Type | Primary Equation | What Changes |
|---|---|---|
| Ladder against wall | x² + y² = L² | x and y; L is constant |
| Sphere inflating | V = (4/3)πr³ | V and r |
| Cone draining | V = (1/3)πr²h | V, r, and h (related by shape) |
| Circle area expanding | A = πr² | A and r |
The ladder problem is simpler because only two variables change. Problems with spheres or cones add another layer because you need to relate the variables (like r and h in a cone that keeps its shape).
Practice Problems
Problem 1
A 13-foot ladder leans against a wall. The bottom slides away at 2 ft/sec. Find the speed of the top when the bottom is 5 feet from the wall.
Answer: dy/dt = -5/12 ft/sec
Check: 5² + 12² = 25 + 144 = 169 = 13². The triangle checks out.
Problem 2
A ladder 20 ft long slides down a wall. The bottom moves at 4 ft/sec. How fast is the top moving when the bottom is 12 ft from the wall?
Answer: dy/dt = -3 ft/sec
Same setup, just different numbers. When x = 12, y = 16 (from 12² + y² = 20²).
How to Get Started on Any Related Rates Problem
- Draw a diagram. It doesn't have to be pretty. A right triangle with labels x, y, and L works fine.
- Identify all given rates and values. Write dx/dt, dy/dt, x, y next to your diagram.
- Write the equation connecting your variables. Usually Pythagorean theorem or a volume formula.
- Take the derivative of both sides. Use the chain rule on every variable that has t in it.
- Plug in known values. Everything except your target rate should be substituted now.
- Solve for your unknown rate. Isolate it and calculate.
- Check your sign. Does a positive or negative answer make sense given the situation?
When the Ladder Problem Gets Trickier
Sometimes the problem asks for a rate at a different moment than given. Or it might ask how fast the area between the ladder and wall is changing.
The approach stays the same. Find your equation, differentiate, plug in the specific values for that moment in time.
For area problems, use A = (1/2)xy. Differentiate to get dA/dt = (1/2)[x(dy/dt) + y(dx/dt)]. Same process, different geometry.
The Bottom Line
Ladder problems are straightforward once you internalize the process. Draw, identify, equation, differentiate, plug, solve. The Pythagorean theorem is your foundation. Derivatives with respect to time are your tool.
Stop memorizing. Start understanding why each step exists. The math makes sense once you visualize what's actually happening to the ladder.