Related_rates

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.

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.

The "four corner" approach to solving related rates problems. Knowing the relationship between position A and position B, differentiate to find the relationship between rate A and rate B.

Example

Suppose that there is a 10-meter ladder 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?

Calling the distance of the base of the ladder from the wall x and the height of the ladder on the wall y, the ladder, the wall, and the ground represent the sides of a right triangle with side lengths x, y, and 10 (the hypotenuse). The object is to find the rate of change of y with respect to time when x = 6. It is given that when x = 6, the rate of change of x is 3 meters per second. This rate of change is positive because the distance x is increasing.

An equation relating the three sides of a right triangle is the well-known pythagorean theorem, a²+b² = c². In this case, the equation that relates x and y is x²+y² = 10². Differentiating both sides of this equation with respect to time (t) yields

$\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(100)$

which when solved for the wanted rate of change, dy/dt, gives us

$\frac{dy}{dt}=\frac{-x\frac{dx}{dt}}{y}$

It is given that when x = 6, dx/dt = 3. Due to the pythagorean theorem, y = 8. Plugging these values into the equation gives us the answer:

$\frac{dy}{dt}=\frac{-6\times3}{8}=-\frac{9}{4}.$

The top of the ladder is sliding down the wall at a rate of ⁹⁄₄ meters per second.