🔬 Grade 12 · Lesson 4 of 13
The Related-Rates Ladder: When One Rate Drives Another
As a ladder's base slides out, its top races down — and calculus tells you exactly how their speeds are linked.
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The big idea: When two or more quantities are tied together by an equation and both change over time, their rates of change are tied together too. Differentiate the relating equation with respect to time, and a known rate lets you solve for an unknown one. The critical discipline: differentiate first while everything is still a variable, and only plug in the specific numbers at the very end.
Two speeds, one connection
A ladder leans against a wall. Someone tugs the base away from the wall at a steady speed, and the top slides down. The base’s speed and the top’s speed are not independent — the rigid ladder links them. Knowing how fast the base moves, we can find exactly how fast the top drops at any instant.
Problems like this are called related rates: rates of change connected through a shared equation.
The ladder set-up
Let x be the distance from the wall to the foot of the ladder and y the height of the top up the wall. The ladder’s length L is fixed, so x, y and L form a right triangle. As time passes, x grows and y shrinks, but they must always satisfy the Pythagorean relation.
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Differentiate the relation with respect to time
Write an equation linking the quantities, then differentiate both sides with respect to time t. Every changing quantity carries a rate (like dx/dt or dy/dt) via the chain rule. This turns a relation between quantities into a relation between their rates.
\[ x^2 + y^2 = L^2 \]
The ladder relation: base distance and height are the legs of a right triangle with fixed hypotenuse L.
🎮 Related-Rates Ladder LIVE
Slide a ladder down a wall and watch how the top's speed relates to the base's.
A four-step recipe
(1) Identify the quantities and their rates, and note which are given and which are wanted. (2) Find an equation relating the quantities. (3) Differentiate both sides with respect to t. (4) Substitute the instantaneous values and solve. The order matters: differentiate before you substitute.
\[ 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0 \;\Longrightarrow\; \frac{dy}{dt} = -\frac{x}{y}\,\frac{dx}{dt} \]
Differentiating x squared plus y squared equals L squared (L constant) links the two speeds.
📝 Worked example: A 10 ft ladder leans on a wall. Its base slides away from the wall at 1 ft/s. How fast is the top sliding down when the base is 6 ft from the wall?
- Relation: \( x^2 + y^2 = 10^2 \). When \( x = 6 \), \( y = \sqrt{100 - 36} = 8 \) ft.
- Differentiate with respect to t: \( 2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0 \).
- Given \( \dfrac{dx}{dt} = 1 \). Substitute x = 6, y = 8: \( 2(6)(1) + 2(8)\dfrac{dy}{dt} = 0 \).
- Solve: \( 12 + 16\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{12}{16} = -0.75 \) ft/s.
✓ The top slides down at <strong>0.75 ft/s</strong> (the minus sign shows y is decreasing).
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Substitute numbers only AFTER differentiating
It is tempting to plug x = 6 into the relation before differentiating — but then x is frozen as a constant and its rate vanishes, giving nonsense. The quantities are varying; treat them as variables through the differentiation step, and insert the instantaneous values (6, 8, and dx/dt = 1) only at the final substitution.
📝 Worked example: A stone dropped in a pond makes a circular ripple whose radius grows at 2 cm/s. How fast is the enclosed area growing when the radius is 5 cm?
- Relation between area and radius: \( A = \pi r^2 \).
- Differentiate with respect to t: \( \dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt} \).
- Substitute \( r = 5 \) and \( \dfrac{dr}{dt} = 2 \): \( \dfrac{dA}{dt} = 2\pi(5)(2) \).
✓ The area grows at <strong>20π cm²/s</strong> (about 62.8 cm²/s).
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The rate keeps changing
The top of the ladder does not descend at a constant speed. As x grows and y shrinks, the factor −x/y grows in magnitude, so the top accelerates — racing toward the ground as the ladder nears flat. A related-rates answer is an instantaneous snapshot valid at that one configuration.
Check your understanding
1. Differentiating x² + y² = L² (with L constant) with respect to time gives:
Each variable term gets a time-rate by the chain rule; the constant L² differentiates to 0, giving 2x·(dx/dt) + 2y·(dy/dt) = 0.
2. For the 10 ft ladder with base 6 ft out (top at 8 ft) and dx/dt = 1 ft/s, dy/dt is:
From 2(6)(1) + 2(8)(dy/dt) = 0, dy/dt = −12/16 = −0.75 ft/s. The negative sign means the top descends.
3. For a growing circle, A = πr². If dr/dt = 2 and r = 3, then dA/dt equals:
dA/dt = 2πr·(dr/dt) = 2π(3)(2) = 12π.
4. Why must you substitute the specific numbers only after differentiating?
The values are instantaneous. Substituting before differentiating treats a varying quantity as constant, so its rate wrongly disappears.
5. As the ladder's base keeps sliding out, the top's downward speed:
Since dy/dt = −(x/y)·(dx/dt) and x grows while y shrinks, the factor x/y grows, so the top descends faster and faster.
✅ Key takeaways
- Related-rates problems link two or more time-varying quantities through a single equation.
- Write the relating equation, then differentiate both sides with respect to time t.
- Each varying quantity contributes a rate (dx/dt, dy/dt, dA/dt…) by the chain rule.
- Substitute the instantaneous values only AFTER differentiating — never before.
- Ladder example: dy/dt = −(x/y)(dx/dt) gives −0.75 ft/s when x = 6, y = 8, dx/dt = 1.
- The answer is an instantaneous rate; a negative sign signals a decreasing quantity, and the rate itself can change over time.