Short on time? This page is the recap. For explanations, diagrams and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- A related rates problem finds one rate from others that are known, using an equation that links the quantities.
- Six steps: (1) draw and label; (2) list known rates (with signs), the instant and the rate wanted; (3) write an equation true at all times; (4) remove any extra variable; (5) differentiate with respect to t, then substitute; (6) answer with units and interpret.
- Letters for anything that changes, numbers only for quantities that never change.
- Find missing instant values (a hypotenuse, sec²θ) from the original equation or the triangle.
Key relationships
| Situation | Equation | After d/dt |
|---|---|---|
| Right triangle, one leg fixed at k | x² + k² = D² | 2x · dx/dt = 2D · dD/dt |
| Angle from a fixed point, adjacent side k | tan θ = x/k | sec²θ · dθ/dt = (1/k) · dx/dt |
| Cone, point down, rim radius R, depth H | r = (R/H)h, so V = (1/3)π(R/H)²h³ | dV/dt = π(R/H)²h² · dh/dt |
| Shadow from a lamp of height L, person of height p | s/p = (x + s)/L | solve for s first, then differentiate |
| Sphere | V = (4/3)πr³ | dV/dt = 4πr² · dr/dt |
Useful fact: for a container, dh/dt = (dV/dt) ÷ (area of the liquid surface).
Assumptions behind the method
- Every changing quantity is a differentiable function of time.
- The equation holds at all times near the instant, so it can be differentiated.
- Angles are in radians when you differentiate trig functions.
Mistakes to avoid
- Writing an instant value as a constant (x = 30 in the equation before differentiating).
- Fixing a radius that changes: the rim of a cone is fixed, the water’s radius is not.
- Keeping a variable whose rate is unknown instead of removing it.
- Missing sec²θ when differentiating tan θ.
- Answering with a number only. Say what changes, the direction, the units and the instant.
- Double negatives: write “decreasing at 0.5 cm/s”, not “decreasing at −0.5”.
Quick self-check
- A right triangle has one leg fixed at 12 and the other leg x increasing at 2 units/s. Find the rate of change of the hypotenuse when x = 9. (D = 15; dD/dt = (9/15)(2) = 1.2 units/s)
- A cone with its point down has rim radius 2 and depth 8. Write the volume of liquid in terms of the depth h. (r = h/4, so V = πh³/48)
- You find dh/dt = −0.5 m/min when h = 3 m. Interpret it. (At the moment the depth is 3 m, the depth is decreasing at 0.5 metres per minute.)
Next: practice questions.