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Related Rates Master Reference cheat sheet - grade 11-12

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Related rates problems use derivatives to connect quantities that change over time. This cheat sheet helps students organize word problems, choose equations, and differentiate correctly with respect to time. It is especially useful for motion, geometry, water level, ladder, and shadow problems where several variables change at once. The main idea is to write an equation relating the variables, then apply implicit differentiation with respect to tt. Known values are substituted after differentiating, not before, so changing quantities keep their derivatives. Common formulas include the Pythagorean theorem, area and volume formulas, trigonometric relationships, and similar triangle ratios.

Key Facts

  • In related rates, every changing variable is treated as a function of time, so ddt[x2]=2xdxdt\frac{d}{dt}[x^2] = 2x\frac{dx}{dt}.
  • The standard process is identify variables, write a relation, differentiate with respect to tt, substitute known values, and solve for the unknown rate.
  • For a right triangle, x2+y2=z2x^2 + y^2 = z^2 gives 2xdxdt+2ydydt=2zdzdt2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}.
  • For a circle, A=πr2A = \pi r^2 gives dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r\frac{dr}{dt}.
  • For a sphere, V=43πr3V = \frac{4}{3}\pi r^3 gives dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}.
  • For a cone, V=13πr2hV = \frac{1}{3}\pi r^2h, and similar triangles often create a relation such as r=khr = kh before differentiating.
  • A positive rate means the quantity is increasing, and a negative rate means the quantity is decreasing, such as dxdt<0\frac{dx}{dt} < 0 for distance getting smaller.
  • In shadow problems, similar triangles give proportional equations such as HL=hs\frac{H}{L} = \frac{h}{s} before differentiating.

Vocabulary

Related rates
A calculus method for finding how fast one quantity changes by using its relationship to another changing quantity.
Implicit differentiation
Differentiating an equation without first solving for one variable, while using the chain rule for variables that depend on tt.
Rate of change
A derivative such as dxdt\frac{dx}{dt} that measures how a quantity changes with respect to time.
Constraint equation
An equation that connects the variables in a problem, such as x2+y2=z2x^2 + y^2 = z^2 or V=πr2hV = \pi r^2h.
Similar triangles
Triangles with equal angle measures and proportional side lengths, often used to relate heights, distances, and shadows.
Instantaneous rate
The rate of change at one exact moment, found after differentiating and then substituting the given values.

Common Mistakes to Avoid

  • Substituting values before differentiating is wrong because changing variables lose their time dependence. Differentiate first, then plug in the instant given in the problem.
  • Forgetting the chain rule is wrong because variables such as xx, rr, and hh depend on time. For example, ddt[r2]\frac{d}{dt}[r^2] must be 2rdrdt2r\frac{dr}{dt}, not 2r2r.
  • Using the wrong sign for a rate is wrong because direction matters. If a distance is decreasing, write its rate as negative, such as dxdt=3\frac{dx}{dt} = -3.
  • Mixing units is wrong because all quantities in one equation must use consistent units. Convert before differentiating, such as inches to feet or minutes to seconds.
  • Treating a constant dimension as changing is wrong because constants have derivative 00. In a ladder problem, the ladder length LL stays fixed, so dLdt=0\frac{dL}{dt} = 0.

Practice Questions

  1. 1 A ladder 1010 ft long leans against a wall. The bottom slides away at dxdt=2\frac{dx}{dt} = 2 ft/s. How fast is the top sliding down when the bottom is 66 ft from the wall?
  2. 2 A spherical balloon is inflated so that dVdt=48π\frac{dV}{dt} = 48\pi cm3^3/s. How fast is the radius increasing when r=4r = 4 cm?
  3. 3 A 66 ft person walks away from a 1515 ft lamp at 44 ft/s. How fast is the tip of the person's shadow moving away from the lamp?
  4. 4 Why should the known values for xx, yy, or rr usually be substituted after differentiating, not before differentiating?