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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?

Understanding Related Rates Master Reference

A related rates problem describes a system with a built in constraint. The quantities are linked by the shape, the equipment, or a physical rule. For a ladder against a wall, the ladder length stays fixed while its two endpoints move.

For water in a tank, the container shape fixes the connection between water depth and surface radius. This means the variables cannot change freely. A useful first step is to draw the situation at one particular instant and label only the lengths that matter.

Separate fixed measurements from changing measurements. A ladder length, a cone angle, or a person’s height may be constant even though nearby distances change.

The derivative in these problems measures an instant, not a total change over a long interval. A balloon can have the same radius increase each second at two moments, yet its volume increase can be much larger at the larger radius. This happens because a larger sphere has more surface area available as it expands.

The chain rule captures that changing sensitivity. In words, the rate of area change for a circle equals two pi times the current radius times the radius rate.

The current radius must remain in the result because the size of the circle matters at that instant. This is why numerical measurements belong near the end of a solution.

Geometry details often decide whether a model is correct. In a cone problem, the radius and water height may both change, but they are connected by the cone’s fixed shape. Similar triangles provide that connection.

Use the water surface radius with the water depth, not the full cone radius with the water depth unless the full cone is actually filled. In ladder problems, distinguish the ladder length from its horizontal distance and vertical height.

In shadow problems, identify which distance is from the light source and which is the shadow length. A small labeling error can produce a perfectly differentiated equation that describes the wrong triangle.

Units give a strong check on every answer. A length rate might be in centimeters per minute. An area rate must be in square centimeters per minute.

A volume rate must be in cubic centimeters per minute. If a calculation for changing volume ends in centimeters per minute, a factor or formula is missing. Signs matter just as much.

A negative horizontal rate for a ladder can mean its base moves toward the wall, while the height rises. Before finishing, compare the sign with the physical motion and estimate whether the size of the answer makes sense. These habits are useful in engineering, fluid flow, camera tracking, construction, and any situation where one measurement changes because another one moves.