AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/4/4-5)
Unit 4 · Topic 4.5
4.5 Solving Related Rates Problems
Solving a related rates problem is a process: draw a picture, name the variables, write an equation that links them, differentiate with respect to time, and only then substitute the values at the given instant. Geometry, such as the Pythagorean theorem or similar triangles, usually supplies the equation.
Key terms
- related rates
- Pythagorean theorem
- similar triangles
- volume formula
- substituting values at the instant
The process
Follow the same steps every time:
- Draw and label a picture. Use letters for quantities that change and numbers for ones that never change.
- Write down what you know (given rates, with signs) and what you want, in dy/dt form.
- Write an equation relating the variables. If it has an extra variable you don't have a rate for, eliminate it first (often with similar triangles).
- Differentiate both sides with respect to t.
- Find any missing values at the instant (often with the original equation), then substitute.
- Answer with units and a sentence.
Pythagorean problems
Ladders sliding down walls, two cars moving along perpendicular roads, a kite on a string: these all form right triangles, so x² + y² = z². Differentiating gives 2x·dx/dt + 2y·dy/dt = 2z·dz/dt, or after dividing by 2, x·dx/dt + y·dy/dt = z·dz/dt.
If one side is fixed (like a ladder's length), its rate is 0.
Similar triangles and cones
For a cone-shaped tank, the volume formula V = (1/3)πr²h has two changing variables, r and h. But the water forms a smaller cone with the same shape as the tank, so r/h equals the tank's radius over height. Use this ratio to replace r with an expression in h before differentiating. Then you only need dh/dt.
Problems with angles
When a question asks how fast an angle changes, link the angle to the sides with a trig ratio, then differentiate. Angles must be in radians, and dθ/dt comes out in radians per unit of time.
Example: a kite flies at a constant height of 60 ft and drifts horizontally away from you at 4 ft/s. With x the horizontal distance and θ the angle of the string above the ground, tan θ = 60/x. Differentiating gives sec²θ·dθ/dt = −(60/x²)·dx/dt. When x = 80, tan θ = 3/4, so sec²θ = 1 + 9/16 = 25/16. Then (25/16)·dθ/dt = −(60/6400)(4) = −0.0375, so dθ/dt = −0.024 radians per second. The angle is shrinking, which makes sense as the kite moves away.
Finishing the problem
Sometimes you need a value the problem didn't give directly. In the ladder problem, you're given the bottom's distance from the wall, but the equation also needs the top's height. Get it from the original equation at that instant: if the ladder is 13 ft and the bottom is 5 ft out, the top is √(169 − 25) = 12 ft high.
Finally, check that the sign makes sense. A ladder top sliding down should have a negative dy/dt.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
The sliding ladder
A 13-foot ladder leans against a wall. The bottom slides away from the wall at 2 ft/s. How fast is the top sliding down the wall when the bottom is 5 feet from the wall?
Show the solutionHide the solution
- Step 1: Let x be the distance from the wall to the bottom, and y the height of the top. Then x² + y² = 169, and dx/dt = 2.
- Step 2: Differentiate: 2x·dx/dt + 2y·dy/dt = 0, so x·dx/dt + y·dy/dt = 0.
- Step 3: At the instant, x = 5, so y = √(169 − 25) = 12.
- Step 4: Substitute: 5(2) + 12·dy/dt = 0, so dy/dt = −10/12 = −5/6.
Answer: dy/dt = −5/6 ft/s: the top is sliding down the wall at 5/6 foot per second.
- Example 2
Filling a cone
Water pours into a cone-shaped tank, point down, at 3 ft³/min. The tank is 10 ft tall with a top radius of 4 ft. How fast is the water level rising when the water is 5 ft deep?
Show the solutionHide the solution
- Step 1: Let h be the water's depth and r the radius of the water's surface. By similar triangles, r/h = 4/10, so r = 2h/5.
- Step 2: Volume of water: V = (1/3)πr²h = (1/3)π(4h²/25)h = (4π/75)h³.
- Step 3: Differentiate: dV/dt = (4π/25)h²·dh/dt.
- Step 4: Substitute dV/dt = 3 and h = 5: 3 = (4π/25)(25)·dh/dt = 4π·dh/dt.
- Step 5: Solve: dh/dt = 3/(4π).
Answer: dh/dt = 3/(4π) ≈ 0.239 ft/min
- Example 3
Trap: the sign of a rate toward a point
Car A is 6 miles north of an intersection, driving north (away from it) at 40 mph. Car B is 8 miles east of the intersection, driving west (toward it) at 20 mph. How fast is the distance between the cars changing?
Show the solutionHide the solution
- Step 1: Let y be A's distance north and x be B's distance east. Then y = 6, dy/dt = 40, x = 8, and dx/dt = −20. The negative sign matters: B's distance from the intersection is shrinking.
- Step 2: Distance between the cars: z² = x² + y², so z = √(64 + 36) = 10.
- Step 3: Differentiate: z·dz/dt = x·dx/dt + y·dy/dt.
- Step 4: Substitute: 10·dz/dt = 8(−20) + 6(40) = −160 + 240 = 80, so dz/dt = 8.
- Step 5: If you used dx/dt = +20, you'd get 10·dz/dt = 400 and dz/dt = 40, which is wrong.
Answer: The distance between the cars is increasing at 8 mph.
Common mistakes
- Using a positive rate for something that is decreasing, like a distance to a point you're approaching.
- Differentiating the cone volume with both r and h as variables when only dh/dt is known. Use similar triangles first.
- Forgetting to find a missing side length from the original equation before substituting.
On the exam
- Free-response related rates questions usually reward a correct differentiated equation and a correct answer with units, so write the derivative step clearly.
- Sometimes the equation is given (like a volume formula for a strange shape). Differentiate exactly what you're given, using the product rule if two variables are multiplied.
Connected topics
Videos
Check yourself
4 questions on 4.5 Solving Related Rates Problems. Pick an answer to see if you got it, and why.
A 20-foot ladder leans against a vertical wall. The bottom of the ladder slides away from the wall along level ground at a constant rate of 1.5 feet per second. At the instant the bottom of the ladder is 7 feet from the wall, what is the rate of change of the height of the top of the ladder, in feet per second?
Water is poured at a rate of 3 cubic feet per minute into a tank shaped like a cone with its vertex pointing down. The tank has a radius of 4 feet at the top and a height of 10 feet. How fast is the depth of the water rising at the instant the water is 5 feet deep? (The volume of a cone is V = (1/3)πr²h.)
A kite flies at a constant height of 60 feet and moves horizontally away from the person holding the string at 4 feet per second. Let θ be the angle between the string and the level ground, and assume the string stays straight. At the instant the kite is 80 feet horizontally from the person, how fast is θ changing?
A person 6 feet tall walks away from a 15-foot-tall lamppost at a rate of 5 feet per second. How fast is the length of the person's shadow increasing?
0 of 4 answered