Math · Calculus I · Concept
Related rates problems, step by step
To solve a related rates problem, write an equation that links the changing quantities, differentiate both sides with respect to time t using the chain rule, then substitute the values at the instant you care about and solve for the unknown rate.
Linked quantities have linked rates
When two quantities are tied together by an equation and both change over time, their rates of change are tied together too. As a ladder slides down a wall, its base and its top move at related speeds, because x² + y² stays equal to the square of the ladder’s length.
Differentiate with respect to time
Treat every changing quantity as a function of t, so each term needs the chain rule: the derivative of x² with respect to t is 2x·dx/dt, not just 2x. A constant, such as the ladder’s length, has derivative zero. This is implicit differentiation with t as the variable.
Substitute only after differentiating
Put in the values at the instant you care about only after differentiating. If you substitute x = 6 first, the base looks like a constant, its rate becomes zero, and the information you need is lost.
A method that works every time
Most related rates problems follow the same six steps.
| Step | What to do |
|---|---|
| 1 | Draw a picture and name each changing quantity with a variable. |
| 2 | Write the given rate and the wanted rate as derivatives, with units. |
| 3 | Write an equation that links the quantities. |
| 4 | Differentiate both sides with respect to t. |
| 5 | Substitute the values at that instant and solve. |
| 6 | Check the sign and the units. |
Read the sign and the units
A positive rate means the quantity is growing; a negative rate means it is shrinking. The units follow the derivative, such as feet per second or cubic centimeters per second. A rate with the wrong sign or units usually signals a slip in the equation.
Common mistakes
- Substituting the instant’s values before differentiating, which turns a changing length into a constant.
- Leaving out the chain-rule factor: the derivative of y² with respect to t is 2y·dy/dt.
- Forgetting that a constant, such as a ladder’s length, differentiates to zero.
- Reporting a rate without its sign or units: dy/dt = −0.75 ft/s means the top falls at 0.75 feet per second.
Key terms
- Related rates
- A problem in which quantities linked by an equation change over time. Differentiating the equation with respect to time links their rates, so a known rate gives an unknown one.
- Chain rule
- The rule for differentiating a function inside another function: differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside.
- Implicit differentiation
- Finding dy/dx from an equation in x and y without solving for y: differentiate both sides with respect to x, multiply the derivative of each y term by dy/dx, then solve for dy/dx.
- Instantaneous rate of change
- How fast a quantity is changing at one instant: the limit of average rates of change over smaller and smaller intervals. It equals the derivative at that point.
- Derivative
- The instantaneous rate of change of a function: the limit of the average rate of change as the step shrinks to zero, when that limit exists. On a graph it is the slope of the tangent line.
Work through an example
A 10 ft ladder leans against a vertical wall. Its base slides away from the wall at 1 ft/s. How fast is the top sliding down when the base is 6 ft from the wall?
Solve a sliding ladder problem →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
See the ladder’s top and its tangent in Graph Open worked example on a board Related-rates relationship in Math ReferenceYour existing work stays on this device. Examples open as editable copies.