AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/4/4-4)
Unit 4 · Topic 4.4
4.4 Introduction to Related Rates
In a related rates problem, several quantities change over time and are tied together by an equation. Differentiating that equation with respect to time t connects their rates. If you know how fast one quantity changes, you can find how fast another one does.
Key terms
- related rates
- rate with respect to time
- implicit differentiation
- dV/dt and dr/dt
The big idea
Suppose a balloon's radius r and volume V both change with time. They're linked by V = (4/3)πr³. Differentiate both sides with respect to t. Because r is a function of t, the chain rule gives dV/dt = 4πr²·dr/dt.
That one equation relates the two rates. Know dr/dt and r at some instant, and you can compute dV/dt at that instant.
Every variable depends on time
In these problems, treat every changing quantity as a function of t. When you differentiate a term with a variable, you multiply by that variable's rate, just like dy/dx in implicit differentiation:
- d/dt (r²) = 2r·dr/dt
- d/dt (x² + y²) = 2x·dx/dt + 2y·dy/dt
- d/dt (xy) = x·dy/dt + y·dx/dt (product rule)
- d/dt (πr²h) = π(2rh·dr/dt + r²·dh/dt) if both r and h change
When the product rule is needed
If two changing quantities are multiplied in the equation, differentiate with the product rule. For a rectangle with A = lw where both sides change: dA/dt = l·dw/dt + w·dl/dt.
Example: when l = 10 cm and w = 4 cm, the length grows at 2 cm/s and the width shrinks at 1 cm/s. Then dA/dt = 10(−1) + 4(2) = −2 cm²/s, so the area is decreasing even though the length is growing. Quotients work the same way with the quotient rule.
Constants vs. variables
Only quantities that stay fixed for the whole problem can be treated as constants, like the length of a ladder or a cone's fixed shape. A quantity that equals 10 only at the instant you care about is not a constant. Plug its value in after differentiating, never before.
Signs of rates
A rate is positive if the quantity is growing and negative if it's shrinking. If water drains at 3 cubic feet per minute, dV/dt = −3. If your answer for a rate comes out negative, it means that quantity is decreasing; say so in words.
Common formulas to have ready
Area of a circle A = πr², circumference C = 2πr, sphere volume V = (4/3)πr³, cone volume V = (1/3)πr²h, cylinder volume V = πr²h, and the Pythagorean theorem a² + b² = c². The exam expects you to know these; it won't always give them.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Inflating a balloon
Air is pumped into a spherical balloon so that its radius grows at 0.5 cm/s. How fast is the volume increasing when the radius is 10 cm?
Show the solutionHide the solution
- Step 1: Equation: V = (4/3)πr³.
- Step 2: Differentiate with respect to t: dV/dt = 4πr²·dr/dt.
- Step 3: Now substitute the instant's values, r = 10 and dr/dt = 0.5: dV/dt = 4π(100)(0.5) = 200π.
- Step 4: Units: cm³ per second.
Answer: dV/dt = 200π ≈ 628.3 cm³/s
- Example 2
Trap: plugging in too early
A ripple spreads in a pond. Its radius grows at 3 ft/s. How fast is the area inside the ripple growing when r = 8 ft? A student starts by writing A = π(8)² = 64π and differentiates to get dA/dt = 0. Fix the work.
Show the solutionHide the solution
- Step 1: The student turned r into the constant 8 before differentiating, so the derivative was 0. But r changes over time.
- Step 2: Correct: A = πr², so dA/dt = 2πr·dr/dt.
- Step 3: Now plug in r = 8 and dr/dt = 3: dA/dt = 2π(8)(3) = 48π.
- Step 4: Units: square feet per second.
Answer: dA/dt = 48π ≈ 150.8 ft²/s
Common mistakes
- Substituting the instant's values before differentiating, which turns variables into constants.
- Forgetting the dr/dt (or other rate) factor from the chain rule.
- Using a positive rate for a shrinking quantity. Draining, falling and moving closer all mean negative rates.
On the exam
- Related rates appear in both multiple choice and free response. In free response, a correct differentiated equation often earns a point even if later arithmetic slips.
- Include units with your final rate, and state whether the quantity is increasing or decreasing if asked.
Connected topics
Videos
Check yourself
4 questions on 4.4 Introduction to Related Rates. Pick an answer to see if you got it, and why.
Air is pumped into a spherical balloon at a rate of 100 cubic centimeters per second. How fast is the radius of the balloon increasing at the instant the radius is 5 centimeters? (The volume of a sphere is V = (4/3)πr³.)
The edge length of a cube, in centimeters, is s(t) = 2 + ln(1 + t²), where t is measured in seconds. What is the rate of change of the volume of the cube, in cubic centimeters per second, at time t = 3?
Each edge of a cube is increasing at a rate of 0.2 centimeter per second. How fast is the volume of the cube increasing at the instant each edge is 5 centimeters long?
The variables x and y are differentiable functions of t and satisfy x² + 3xy = 28. At the instant when x = 2 and y = 4, dx/dt = 3. What is dy/dt at that instant?
0 of 4 answered