Related rates — setting up and solving

WHAT is a related-rates problem?

The hidden engine is the chain rule: if $V = f(r)$ and $r = r(t)$, then $\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}.$

HOW to set one up (the 6-step recipe)

Worked Example 1 — Expanding circle (oil spill)

An oil spill is a circle whose radius grows at $\frac{dr}{dt}=2$ m/s. How fast is the area growing when $r=10$ m?

Step 3 — link: $A = \pi r^2$. Why? Area depends only on radius for a circle.

Step 4 — differentiate in $t$: $\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} = 2\pi r\,\frac{dr}{dt}.$ Why this step? $r$ is a function of $t$, so $\frac{d}{dt}(r^2)=2r\frac{dr}{dt}$ (chain rule).

Step 5 — substitute the instant: $r=10,\ \frac{dr}{dt}=2$: $\frac{dA}{dt} = 2\pi(10)(2) = 40\pi \approx 125.7\ \text{m}^2/\text{s}.$ Why now? Only after differentiating is it safe to fix $r=10$.

Worked Example 2 — Sliding ladder

A 5 m ladder leans on a wall. Its base slides away at $\frac{dx}{dt}=1$ m/s. How fast does the top slide down when the base is $x=3$ m from the wall?

Step 3 — link (Pythagoras): $x^2 + y^2 = 5^2$. Why? Wall, ground, ladder form a right triangle; ladder length (5) is constant.

Step 4 — differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \;\Rightarrow\; \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}.$ Why this step? RHS is constant $\Rightarrow$ its derivative is $0$. The minus sign will tell us $y$ decreases.

Find $y$ at the instant: $y=\sqrt{25-9}=4$ m.

Step 5 — substitute: $\frac{dy}{dt} = -\frac{3}{4}(1) = -0.75\ \text{m/s}.$ The negative sign means the top moves down — physically correct. Why? As the base moves out, the top must drop.

Worked Example 3 — Conical tank (where to be careful)

Water fills a cone (apex down), radius $R=4$ m at top, height $H=8$ m, at $\frac{dV}{dt}=2$ m³/min. How fast is the depth $h$ rising when $h=4$ m?

Step 3 — link: $V=\frac13\pi r^2 h$. But $r$ and $h$ both change — eliminate $r$ using similar triangles: $\frac{r}{h}=\frac{R}{H}=\frac{4}{8}=\frac12$, so $r=\frac h2$. Why eliminate? We want only $h$; fewer variables means fewer unknown rates.

$V=\frac13\pi\left(\frac h2\right)^2 h=\frac{\pi}{12}h^3.$

Step 4 — differentiate: $\frac{dV}{dt}=\frac{\pi}{12}\cdot 3h^2\frac{dh}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.$

Step 5 — substitute $h=4$, $\frac{dV}{dt}=2$: $2=\frac{\pi(16)}{4}\frac{dh}{dt}=4\pi\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{2\pi}\approx0.159\ \text{m/min}.$

Flashcards

Connections

• Chain rule — the core mechanism enabling related rates.
• Implicit differentiation — same technique, differentiating a relation w.r.t. $t$.
• Derivatives as rates of change — interpreting $\frac{dx}{dt}$ as velocity/speed.
• Similar triangles — used to eliminate variables in cone/shadow problems.
• Pythagorean theorem — links in ladder/distance problems.
• Optimization — next application of derivatives in modelling.

Concept Map

Hinglish (regional understanding)