Geometric Related Rates
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.1: “Related Rates” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: Watch the Ripple Widen
A stone is dropped into a still pond. The ripple forms a circle that expands outward at 2 m/s.
Before any formula:
- When the radius is 3 m, how fast is the area of the circle growing?
- Will the area grow faster or slower when the radius reaches 10 m?
Reason: the area is $A = \pi r^2$. When $r$ is larger, even a small increase in $r$ adds a larger ring of area. The area should grow faster at larger radii.
Computing these rates exactly takes the chain rule applied to the geometric relationship.
Quantity-First Framing
In related-rates problems, two or more quantities are changing simultaneously. Each depends on time $t$ (even if time is not explicit in the problem). The key equation is a geometric relationship between the quantities, and differentiation with respect to $t$ (the chain rule) connects their rates.
The derivative $\dfrac{dA}{dt}$ answers: “at this moment, how fast is the area growing?” The derivative $\dfrac{dr}{dt}$ answers: “at this moment, how fast is the radius growing?” The chain rule connects them: $\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt}$.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
The related-rates procedure for geometry:
- Draw a picture. Label the quantities that are changing.
- Write a geometric equation relating those quantities (Pythagorean theorem, area, volume formula, similar triangles).
- Differentiate both sides with respect to $t$.
- Substitute the known values of the quantities and rates at the specific moment of interest.
- Solve for the unknown rate.
Key Concepts
1. Expanding Circle
Example 1. The radius of a circle increases at $2$ m/s. How fast is the area increasing when $r = 5$ m?
Step 1: Draw a circle with radius $r(t)$.
Step 2: Geometric equation: $A = \pi r^2$.
Step 3: Differentiate with respect to $t$: \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt}. \]
Step 4: At the moment of interest: $r = 5$ m, $\frac{dr}{dt} = 2$ m/s.
Step 5: \[ \frac{dA}{dt} = 2\pi(5)(2) = 20\pi \approx 62.8 \text{ m}^2/\text{s}. \]
Predict-then-check: At $r = 10$ m: $\frac{dA}{dt} = 2\pi(10)(2) = 40\pi$, twice as large. Your intuition was right: the area grows faster at larger radius.
Two representations:
Formula: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Verbal: The ring of area added per second is a narrow band of circumference $2\pi r$ and width $\frac{dr}{dt}$, so the area rate is (circumference) $\times$ (rate of radius change).
2. The Ladder Problem
A 10 m ladder leans against a wall. The base slides away from the wall at $0.5$ m/s. How fast is the top sliding down when the base is 6 m from the wall?
Step 1: Let $x$ = distance of base from wall, $y$ = height of top on wall, both in meters. Both change with time.
Step 2: Pythagorean theorem: $x^2 + y^2 = 100$.
Step 3: Differentiate with respect to $t$: \[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. \]
Step 4: When $x = 6$: $y = \sqrt{100 - 36} = 8$ m. Given $\frac{dx}{dt} = 0.5$ m/s.
Step 5: \[ 2(6)(0.5) + 2(8)\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{6}{16} = -\frac{3}{8} = -0.375 \text{ m/s}. \]
The top is sliding down at $0.375$ m/s. The negative sign confirms the height is decreasing.
Predict-then-check: As the base moves farther out, the top slides faster (near the floor, a small base movement causes a large drop). Does the formula confirm this? At $x = 9$: $y = \sqrt{19} \approx 4.36$, $\frac{dy}{dt} = -9(0.5)/(4.36) \approx -1.03$ m/s. Yes, faster.
3. Filling a Cone
Water fills a conical tank (vertex down) at $2$ m$^3$/min. The cone has height $10$ m and radius $4$ m at the top. How fast is the water level rising when the water is $5$ m deep?
Step 1: Let $h$ = depth of water (m), $r$ = radius of water surface (m), $V$ = volume (m$^3$).
Step 2: By similar triangles: $r/h = 4/10$, so $r = 2h/5$. Volume of cone: \[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{2h}{5}\right)^2 h = \frac{4\pi h^3}{75}. \]
Step 3: Differentiate: \[ \frac{dV}{dt} = \frac{4\pi \cdot 3h^2}{75} \frac{dh}{dt} = \frac{4\pi h^2}{25}\frac{dh}{dt}. \]
Step 4: At $h = 5$: $\frac{dV}{dt} = 2$ m$^3$/min.
Step 5: \[ 2 = \frac{4\pi(25)}{25}\frac{dh}{dt} = 4\pi\frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159 \text{ m/min}. \]
4. Multiple Representations and Translation
For the expanding circle:
| $r$ (m) | $dA/dt$ (m$^2$/s) |
|---|---|
| 1 | $4\pi \approx 12.6$ |
| 5 | $20\pi \approx 62.8$ |
| 10 | $40\pi \approx 125.7$ |
(All at $dr/dt = 2$ m/s.)
Translation prompt: The table shows $dA/dt$ doubling when $r$ doubles. Why? The formula $dA/dt = 2\pi r \cdot (dr/dt)$ is linear in $r$, so doubling $r$ doubles the rate.
5. Ask Why: Why Substitute AFTER Differentiating?
A common approach is to substitute the specific $x$ or $y$ value before differentiating. This produces a wrong answer.
Example: In the ladder problem, substituting $x = 6$ into $x^2 + y^2 = 100$ before differentiating gives $36 + y^2 = 100$, a constant equation whose derivative is 0 -- losing all rate information. The equation relating $x$ and $y$ must be differentiated first as a relationship between functions of $t$; specific values are substituted only at Step 4.
Named Misconception: height-vs-slope
In the ladder problem, when $x = 6$ and $y = 8$, the statement “$y$ is 8” tells you the height (the function value). The statement “$dy/dt = -0.375$” tells you the rate of change of height (the slope of $y$ as a function of $t$). These are completely different things.
Confusing the value of a quantity with its rate of change is the height-vs-slope misconception. Always ask: “Is this number a quantity ($y = 8$) or a rate ($dy/dt$)?”
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Substituting specific values before differentiating | Plugging $x = 6$ into $x^2 + y^2 = 100$ then differentiating | Differentiate the equation in general, then substitute values |
| Wrong sign on decreasing rate | Writing $dy/dt = +0.375$ for a falling top | The top’s height decreases, so $dy/dt < 0$ |
| Forgetting the chain rule factor | $\frac{d}{dt}[r^2] = 2r$ (missing $dr/dt$) | $\frac{d}{dt}[r^2] = 2r\frac{dr}{dt}$; $r$ depends on $t$ |
Common Misconceptions
a rate of change in a related-rates problem is a single fixed number that applies throughout the problem.
This is the rate-as-fixed-number error. In the expanding-circle problem, $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ shows that the area rate depends on the current value of $r$; at $r = 5$ m the area grows at $20\pi$ m$^2$/s, but at $r = 10$ m it grows at $40\pi$ m$^2$/s even though $\frac{dr}{dt}$ has not changed. Treating the rate as constant overlooks the geometric amplification that a larger radius produces.
in the ladder problem, the value of $y$ and the rate $dy/dt$ convey the same information.
This is the height-vs-slope error. When the base is 6 m from the wall, the fact that $y = 8$ m tells where the top is at that instant; the fact that $dy/dt = -0.375$ m/s tells how fast the top is moving at that instant. A quantity and its rate of change are entirely different pieces of information, related by differentiation, not by arithmetic.
Leveled Practice
Level 1 -- Expanding Circle and Sphere
Problem 1. The radius of a sphere increases at $3$ cm/s. How fast is the volume increasing when $r = 4$ cm? (Volume of sphere: $V = \frac{4}{3}\pi r^3$.)
Show answer
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. At $r = 4$: $\frac{dV}{dt} = 4\pi(16)(3) = 192\pi \approx 603$ cm$^3$/s.
Problem 2. The circumference of a circle grows at $6$ cm/s. How fast is the area growing when the circumference is $10\pi$ cm?
Show answer
$C = 2\pi r$, so $r = C/(2\pi)$. When $C = 10\pi$: $r = 5$ cm.
$dC/dt = 2\pi\,dr/dt$, so $dr/dt = (6)/(2\pi) = 3/\pi$ cm/s.
$dA/dt = 2\pi r\,dr/dt = 2\pi(5)(3/\pi) = 30$ cm$^2$/s.
Level 2 -- Ladder and Related Geometry
Problem 3. A ladder 13 m long rests against a wall. Its base slides away at $0.4$ m/s. How fast is the top sliding down when the base is 5 m from the wall?
Show answer
$x^2 + y^2 = 169$. $2x\,dx/dt + 2y\,dy/dt = 0$.
At $x = 5$: $y = 12$ m. $2(5)(0.4) + 2(12)\,dy/dt = 0 \Rightarrow dy/dt = -1/(6) \approx -0.167$ m/s.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) A square’s side length increases at $2$ cm/s. How fast is the area growing when the side is 6 cm?
(b) (Mid) For the expanding circle problem, find $d^2A/dt^2$ (the acceleration of area growth) when $r = 5$ m and $dr/dt = 2$ m/s. Assume $dr/dt$ is constant.
(c) (Ceiling) A cylindrical tank (radius 3 m) is being filled at $5$ m$^3$/min. Water is simultaneously draining at a rate proportional to the water depth: $Q_{out} = 0.1h$ m$^3$/min. Set up and solve for $dh/dt$ when $h = 4$ m. At what depth is the water level steady?
Show answer
(a) $A = s^2$, $dA/dt = 2s\,ds/dt = 2(6)(2) = 24$ cm$^2$/s.
(b) $A = \pi r^2$, $dA/dt = 2\pi r\,dr/dt$. Differentiate again: $d^2A/dt^2 = 2\pi(dr/dt)^2 + 2\pi r\,d^2r/dt^2$.
If $dr/dt = 2$ = constant: $d^2r/dt^2 = 0$. So $d^2A/dt^2 = 2\pi(4) = 8\pi \approx 25.1$ m$^2$/s$^2$.
(c) Volume: $V = \pi r^2 h = 9\pi h$. Net rate: $dV/dt = 5 - 0.1h$.
$dV/dt = 9\pi\,dh/dt$, so $dh/dt = (5 - 0.1h)/(9\pi)$.
At $h = 4$: $dh/dt = (5 - 0.4)/(9\pi) = 4.6/(9\pi) \approx 0.163$ m/min (rising).
Steady: $5 - 0.1h = 0 \Rightarrow h = 50$ m. If the tank is tall enough, steady state at $h = 50$ m.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Related-rates problems are a movie, not a photograph. The geometric equation is always true (the ladder always satisfies $x^2 + y^2 = 100$ as long as it touches the wall). Differentiating that equation with respect to time converts a static geometric fact into a dynamic rate relationship. The equation “at every moment, the rates must balance” is what you extract.
Substitute the specific snapshot values (specific $x$, specific $y$, specific $dx/dt$) only at the very end, to answer the question about “right now.”
Connections
Within Calculus I (MATH161)
- Chain rule: Every related-rates computation uses the chain rule, differentiating with respect to $t$ while the variables ($x$, $y$, $r$) are themselves functions of $t$.
- Implicit differentiation: Related rates generalize implicit differentiation: instead of differentiating with respect to $x$, you differentiate with respect to $t$.
- Optimization: Both related rates and optimization use derivatives to connect changing quantities. In optimization, you find when a rate is zero; in related rates, you find what a rate equals.
Back to Calculus I Skills | Next: Solving Related Rates Problems