20. Related Rates

Adapted from: Classical single-variable calculus (time rates linked by a geometric or algebraic equation). Rewritten for teaching — not a verbatim textbook dump.

If a quantity $y$ depends on time $t$, its rate of change is $dy/dt$. When several time-dependent quantities are tied together by an equation, differentiate that equation with respect to $t$ to relate their rates.

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The method

1. Draw a figure; introduce variables for every quantity that changes with time.
2. Write an equation that relates those variables (geometry, similar triangles, volume formulas, …).
3. Identify which rates are known and which rate is asked for.
4. Differentiate both sides with respect to $t$ (Chain Rule). Constants stay constant.
5. Then plug in the values at the specific instant of interest; solve for the unknown rate.
6. Interpret the sign (e.g. negative $dy/dt$ may mean “sliding down”).

Timing

Substitute numerical values after differentiating, unless a quantity is truly constant for all $t$ (like a fixed ladder length).

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Example 20.1 — Sliding ladder

A $25$-foot ladder leans against a vertical wall. The bottom slides away from the wall at $3$ ft/s. How fast is the top sliding down when the bottom is $7$ ft from the wall?

Fig. 20-1 — x² + y² = 625

Let $x$ = distance of the bottom from the wall, $y$ = height of the top. Given $dx/dt=3$; find $dy/dt$ when $x=7$.

Pythagoras:

$$x^{2}+y^{2}=25^{2}=625. \tag{20.1}$$

Differentiate in $t$:

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \implies x\frac{dx}{dt}+y\frac{dy}{dt}=0. \tag{20.2}$$

At $x=7$: $49+y^{2}=625$ ⇒ $y=24$ (positive height). With $dx/dt=3$:

$$7\cdot 3+24\cdot\frac{dy}{dt}=0 \implies \frac{dy}{dt}=-\frac{7}{8}.$$

So the top is sliding down at $\dfrac{7}{8}$ ft/s.

Check: ladder

Quiz

Sliding ladder

1 / 2

From x² + y² = 625, differentiating in t gives…

Pick an answer to see if you are right.

Hands On: Compute dy/dt

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Key takeaways

• Related rates: one equation among time-varying quantities → differentiate in $t$ → solve for the unknown rate
• Plug the specific instant after differentiating
• Sign of the rate answers “up or down / in or out”
• Ladder model: $x^{2}+y^{2}=L^{2}$ ⇒ $dy/dt=-(x/y)\\,dx/dt$

Further Reading

• 19. Rectilinear & Circular Motion — $ds/dt$ as a rate
• 19b. Motion — Practice — prior drills
• 17. Differentiation of Trig Functions — trig related rates (angles)
• 16. Review of Trigonometry — right triangles
• 20b. Related Rates — Practice — balloons, funnels, ships, shadows
• 21. Differentials · Newton's Method — linear approx and Newton