Related rates
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In differential calculus, related rates problems involve finding a rate that a quantity changes by relating that quantity to other quantities whose rates of change are known. The rate of change is usually with respect to time.
[edit] Procedure
The most common way to approach related rates problems is the following:
- Identify the known rates of change and the rate of change that is to be found.
- Construct an equation relating the quantities whose rates of change are known to the quantity whose rate of change is to be found.
- Differentiate both sides of the equation with respect to time (or other rate of change).
- Substitute the known rates of change and the known quantities into the equation.
- Solve for the wanted rate of change.
Errors in this procedure are often caused by plugging in the known values for the variables before (rather than after) finding the derivative with respect to time. Doing so will yield an incorrect result.
[edit] Example
A 10-meter ladder is leaning against the wall of a building, and the base of the ladder is sliding away from the building at a rate of 3 meters per second. How fast is the top of the ladder sliding down the wall when the base of the ladder is 6 meters from the wall?
The distance between the base of the ladder and the wall, x, and the height of the ladder on the wall, y, represent the sides of a right triangle with the ladder as the hypotenuse, h. The object is to find dy/dt, the rate of change of y with respect to time, t, when h, x and dx/dt, the rate of change of x, are known.
From the Pythagorean theorem, the equation
,
describes the relationship between x, y and h, for a right triangle. Differentiating both sides of this equation with respect to time, t, yields
which, when solved for the wanted rate of change, dy/dt, gives us
The given values are h = 10, x = 6, and dx/dt = 3. Solving the Pythagorean theorem for y when x = 6 and h = 10 gives us y = 8. Plugging these values into the equation for dy/dt gives us the answer:
The top of the ladder is sliding down the wall at a rate of 9⁄4 meters per second.
[edit] External links
- MIT Calculus lecture on Related Rates




