Related rates problems ask: "If one quantity is changing at a known rate, how fast is another quantity changing?" The key is to find an equation that relates the quantities, then differentiate both sides with respect to time.
🎯 In this section you will learn
- A step-by-step strategy for solving related rates problems
- How to differentiate equations implicitly with respect to time
- Classic examples: expanding balloon, sliding ladder, moving shadow, and more
📌 Step-by-Step Strategy
①
Read the problem carefully
Identify what quantities are changing and what rates are given/asked.
②
Draw a diagram
Label all relevant variables. Assign variable names to changing quantities.
③
Write an equation
Find a relationship between the variables (geometry, similar triangles, Pythagorean theorem, etc.).
④
Differentiate implicitly
Differentiate both sides with respect to time $t$. Every variable is a function of $t$.
⑤
Substitute known values
Plug in the given rates and values at the instant of interest.
⑥
Solve for the unknown rate
Interpret the result with correct units.
💡 Key Insight
When you differentiate, don't forget to multiply by $\frac{dx}{dt}$ for each variable $x$ (since $x$ is a function of time).
A spherical balloon is being inflated at a rate of $3$ cm³/s. How fast is the radius increasing when the radius is $5$ cm?
①
Known and unknown
Given: $\frac{dV}{dt} = 3$ cm³/s. Find: $\frac{dr}{dt}$ when $r = 5$ cm.
②
Volume of a sphere
$V = \frac{4}{3}\pi r^3$
③
Differentiate with respect to $t$
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
④
Plug in known values
$3 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}$
⑤
Solve
$\frac{dr}{dt} = \frac{3}{100\pi} \approx 0.00955$ cm/s
A ladder 10 ft long rests against a vertical wall. The bottom is sliding away from the wall at 1 ft/s. How fast is the top sliding down when the bottom is 6 ft from the wall?
①
Diagram variables
Let $x$ = distance from wall to bottom, $y$ = height of top. Ladder length constant: $x^2 + y^2 = 100$.
②
Given and find
$\frac{dx}{dt} = 1$ ft/s. Find $\frac{dy}{dt}$ when $x = 6$ ft.
③
Find $y$ at that instant
$6^2 + y^2 = 100$ ⇒ $y^2 = 64$ ⇒ $y = 8$ ft
④
Differentiate $x^2 + y^2 = 100$
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ ⇒ $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$
⑤
Solve for $\frac{dy}{dt}$
$(6)(1) + (8)\frac{dy}{dt} = 0$ ⇒ $8\frac{dy}{dt} = -6$ ⇒ $\frac{dy}{dt} = -\frac{3}{4} = -0.75$ ft/s
⑥
Interpret
The top is sliding down at $0.75$ ft/s.
Water is being poured into a conical tank (vertex down) at 2 m³/min. The tank has height 10 m and radius 4 m. How fast is the water level rising when the water is 5 m deep?
①
Similar triangles
For a cone: $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$ ⇒ $r = \frac{2}{5}h$
②
Volume of cone
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{2}{5}h\right)^2 h = \frac{1}{3}\pi \cdot \frac{4}{25}h^3 = \frac{4\pi}{75}h^3$
③
Differentiate
$\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \frac{dh}{dt} = \frac{4\pi}{25} h^2 \frac{dh}{dt}$
④
Plug in values
Given $\frac{dV}{dt} = 2$ m³/min, $h = 5$ m:
$2 = \frac{4\pi}{25} (25) \frac{dh}{dt} = 4\pi \frac{dh}{dt}$
⑤
Solve
$\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} \approx 0.159$ m/min
A man 6 ft tall walks away from a 15 ft tall streetlight at 4 ft/s. How fast is the tip of his shadow moving?
①
Diagram variables
Let $x$ = distance from man to light, $y$ = distance from light to shadow tip. By similar triangles: $\frac{15}{y} = \frac{6}{y-x}$
②
Simplify equation
$15(y-x) = 6y$ ⇒ $15y - 15x = 6y$ ⇒ $9y = 15x$ ⇒ $y = \frac{5}{3}x$
③
Differentiate
$\frac{dy}{dt} = \frac{5}{3} \frac{dx}{dt}$
④
Given $\frac{dx}{dt}=4$ ft/s
$\frac{dy}{dt} = \frac{5}{3}(4) = \frac{20}{3} \approx 6.67$ ft/s
⑤
Shadow tip speed relative to ground = $\frac{dy}{dt}$
The shadow tip is moving at $\frac{20}{3}$ ft/s.
⚠️ Common Mistakes
- Forgetting to differentiate all variables — every variable is a function of time.
- Plugging in numbers before differentiating — you must differentiate first, then substitute.
- Mixing up rates — keep track of what each derivative represents (positive = increasing, negative = decreasing).
- Using the wrong formula — make sure the equation correctly models the situation.
🔍 Key Takeaways
- Related rates problems link multiple changing quantities through an equation.
- Follow the 6-step strategy: read → diagram → equation → differentiate → substitute → solve.
- Do not substitute known values until after differentiating.
- Common setups: Pythagorean theorem, similar triangles, volumes/areas.
- Interpret negative rates as "decreasing."
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