Source: AB Calculus Unit 3 Notes, Texas A&M University
Tags: related rates, implicit differentiation with respect to time, dA/dt, dV/dt, dr/dt, changing quantities, AP Calculus AB, Pythagorean theorem rates, similar triangles rates, cone volume rate
Related rates problems ask how fast one quantity is changing given information about how fast another quantity is changing. You set up a geometric or algebraic equation connecting the quantities, differentiate everything with respect to time using implicit differentiation, plug in the known values at the instant in question, and solve for the unknown rate.
Related rates
A class of problems where two or more quantities change with time and are connected by an equation. Differentiating that equation with respect to t links their rates of change.
Rate of change (with respect to time)
A derivative with respect to t. Written as dx/dt, dr/dt, dA/dt, dV/dt, etc. A positive rate means the quantity is increasing; a negative rate means it is decreasing.
Implicit differentiation with respect to t
The same technique as implicit differentiation with respect to x, but now every variable (x, y, r, A, V, etc.) is treated as a function of time t. Each one picks up its own d(·)/dt when differentiated.
This is a reliable step-by-step process. Following it methodically prevents the most common errors.
Draw and label a picture. Assign variables to every quantity that changes over time. If a quantity is constant throughout the problem, label it with its fixed numerical value instead.
Write down what you know as rates. Translate English into calculus notation. "The radius is growing at 2 ft/sec" becomes dr/dt = 2. "The volume is decreasing at 3 in³/min" becomes dV/dt = −3.
Write down what you want to find. State it as a rate at a specific instant, e.g. "Find dA/dt when r = 60."
Find an equation that relates the variables (not the rates). This is the geometry or physics of the situation: area of a circle, Pythagorean theorem, volume of a cone, similar triangles, trig relationship, etc.
Eliminate extra variables if needed. If your equation has three variables but you only have rate information for two, use a secondary equation (often similar triangles or a constraint) to reduce to two variables before differentiating.
Differentiate both sides with respect to t. Every variable gets a d(·)/dt via the chain rule.
Substitute known values at the instant in question and solve for the unknown rate.
Answer in a sentence with units. The AP exam expects this on free-response questions.
"Growing at 3 inches per year" → dh/dt = 3 in/yr
"Losing 5 cents per day" → dP/dt = −5 cents/day (negative because it is losing)
"Radius gets larger by 4 feet each hour" → dr/dt = 4 ft/hr
"Temperature drops by 5°F per minute" → dT/dt = −5 °F/min
Pythagorean theorem: a² + b² = c²
Differentiating: 2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
Area of a circle: A = πr²
Differentiating: dA/dt = 2πr(dr/dt)
Volume of a sphere: V = (4/3)πr³
Differentiating: dV/dt = 4πr²(dr/dt)
Surface area of a sphere: S = 4πr²
Differentiating: dS/dt = 8πr(dr/dt)
Volume of a cone: V = (1/3)πr²h
Differentiating (when both r and h change): dV/dt = (1/3)π[2rh(dr/dt) + r²(dh/dt)]
If the cone's shape is fixed (like a leaking snow cone), the ratio r/h is constant. Use similar triangles to write r in terms of h (or vice versa), substitute to eliminate one variable, then differentiate.
Trigonometric relationship: tan θ = x/k (where k is constant, e.g. the distance from a camera to a launch pad)
Differentiating: sec²θ · (dθ/dt) = (1/k)(dx/dt)
Expanding/shrinking circle (oil spill, ripple): Given dr/dt, find dA/dt. Use A = πr², differentiate, plug in the given r and dr/dt.
Example: Oil spill, dr/dt = 2 ft/s, find dA/dt when r = 60. dA/dt = 2π(60)(2) = 240π ft²/s.
Rectangle with changing sides: Given dl/dt and dw/dt, find dA/dt. Use A = lw, apply the product rule: dA/dt = (dl/dt)w + l(dw/dt). Watch signs: decreasing length means dl/dt is negative.
Camera-to-rocket (Pythagorean): The camera is a fixed horizontal distance from the launch pad. The rocket rises vertically. The camera-to-rocket distance is the hypotenuse. Use a² + b² = c² where a is fixed horizontal distance, b is height, c is the slant distance. Differentiate, noting da/dt = 0 since a is constant.
Angle of elevation (trig): Same setup, but now you want dθ/dt. Use tan θ = (height)/(horizontal distance), differentiate with respect to t.
Boats/cars moving apart (Pythagorean): Two objects move at right angles. Their distances from a common point are legs of a right triangle; the distance between them is the hypotenuse. Track positions at the specific time asked about, then differentiate.
Shadow problems (similar triangles): A light at height H casts the shadow of a person of height h. Similar triangles relate the person's distance from the pole to the shadow length. Set up the proportion, cross-multiply, differentiate.
Balloon inflation: Given dV/dt, find dS/dt (surface area rate). You can chain through the radius: find dr/dt from dV/dt, then use that to get dS/dt. With V = (4/3)πr³ and S = 4πr², the shortcut is dS/dt = (2/r) · dV/dt.
Cone draining/filling: The key move is using similar triangles to eliminate one variable (usually r in terms of h or vice versa) before differentiating, so you end up with one variable and one rate on each side.
Example: Paper cone with top radius 3 in and height 5 in. The ratio r/h = 3/5, so r = 3h/5. Substitute into V = (1/3)πr²h to get V in terms of h alone, differentiate, and solve for dh/dt or dr/dt.
Frequently used geometric formulas in related rates
Circle area: A = πr²
Circle circumference: C = 2πr
Sphere volume: V = (4/3)πr³
Sphere surface area: S = 4πr²
Cone volume: V = (1/3)πr²h
Pythagorean theorem: a² + b² = c²
Rectangle area: A = lw
Trigonometric: tan θ = opposite/adjacent
⚠️ The most common mistake is plugging in numerical values too early, before differentiating. Substitute numbers only after you have differentiated with respect to t.
⚠️ Watch the signs. A decreasing quantity has a negative rate. Forgetting the negative sign changes the entire answer.
⚠️ If the equation has three or more variables but you only have rates for two, use a constraint (similar triangles, a fixed ratio, the original equation at the specific instant) to eliminate a variable before differentiating.
⚠️ Always include units in your final answer. The AP free-response rubric awards a point for units and for a concluding sentence.
⚠️ "How fast is the distance changing" and "how fast is the angle changing" on the same setup (e.g. the rocket problem) use different equations. Distance uses Pythagorean theorem; angle uses a trig function. Know which one the question is asking for.
⚠️ Related rates are tested nearly every year on the AP AB free-response section. Expect one full problem or a significant part of a problem.
Q: Oil spreads in a circle with radius increasing at 2 ft/s. How fast is the area increasing when r = 60 ft?
A: A = πr², so dA/dt = 2πr · dr/dt = 2π(60)(2) = 240π ≈ 753.98 ft²/s.
Q: A rectangle's length decreases at 2 in/s and width increases at 3 in/s. When the length is 10 in and the width is 6 in, how fast is the area changing?
A: dA/dt = (dl/dt)(w) + (l)(dw/dt) = (−2)(6) + (10)(3) = −12 + 30 = 18 in²/s. The area is increasing at 18 in²/s.
Q: A spherical balloon is inflated at 5 in³/min. How fast is the surface area changing when r = 4 in?
A: From dV/dt = 4πr² · dr/dt: dr/dt = 5/(4π · 16) = 5/(64π). Then dS/dt = 8πr · dr/dt = 8π(4) · 5/(64π) = 160π/(64π) = 5/2 = 2.5 in²/min.
Q: A camera is 3000 ft from a launch pad. A rocket rises at 880 ft/s when it is 4000 ft high. How fast is the camera-to-rocket distance changing?
A: Let d be the slant distance. d² = 3000² + h². At h = 4000, d = √(9,000,000 + 16,000,000) = √25,000,000 = 5000. Differentiating: 2d(dd/dt) = 2h(dh/dt), so dd/dt = h(dh/dt)/d = 4000(880)/5000 = 704 ft/s.
Q: Same setup as above. How fast is the angle of elevation changing?
A: tan θ = h/3000. Differentiating: sec²θ · dθ/dt = (1/3000)(dh/dt). At h = 4000, tan θ = 4/3, so sec²θ = 1 + 16/9 = 25/9. Then dθ/dt = (880/3000) · (9/25) = 7920/75000 = 0.1056 rad/s.
Q: A paper cone has top radius 3 in and height 5 in. Liquid leaks out at 2 in³/min. How fast is the radius of the liquid surface changing when the radius is 2 in?
A: Similar triangles: r/h = 3/5, so h = 5r/3. V = (1/3)πr²(5r/3) = 5πr³/9. dV/dt = (15πr²/9)(dr/dt) = (5πr²/3)(dr/dt). With dV/dt = −2 and r = 2: −2 = (5π(4)/3)(dr/dt), so dr/dt = −6/(20π) = −3/(10π) ≈ −0.0955 in/min. The radius is decreasing at 3/(10π) in/min.
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