Related rates: Information from Answers.com

Fundamental theorem
Limits of functions
Continuity
Mean value theorem

Differential calculus
Derivative Change of variables Implicit differentiation Taylor's theorem Related rates Identities Rules: Power rule, Product rule, Quotient rule, Chain rule

Integral calculus
Integral Lists of integrals Improper integrals Integration by: parts, disks, cylindrical shells, substitution, trigonometric substitution, partial fractions, changing order

Vector calculus
Gradient Divergence Curl Laplacian Gradient theorem Green's theorem Stokes' theorem Divergence theorem

Multivariable calculus
Matrix calculus Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian

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 variables, including 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

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.

Step 1:

x = 6

h = 10

$\frac{dx}{dt}=3$

$\frac{dh}{dt}=0$

$\frac{dy}{dt}=?$

Step 2: From the Pythagorean theorem, the equation

$x^2+y^2=h^2\,$ ,

describes the relationship between x, y and h, for a right triangle. Differentiating both sides of this equation with respect to time, t, yields

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

Step 3: When solved for the wanted rate of change, dy/dt, gives us

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

$(2x)\frac{dx}{dt}+(2y)\frac{dy}{dt}=(2h)\frac{dh}{dt}$

$x\frac{dx}{dt}+y\frac{dy}{dt}=h\frac{dh}{dt}$

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

Step 4 & 5: Using the variables from step 1 gives us:

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

$\frac{dy}{dt}=\frac{10\times0-6\times3}{y}=-\frac{18}{y}.$

Solving for y using the Pythagorean Theorem gives:

x 2 + y 2 = h 2

62 + y 2 = 10 2

y 2 = 64

y = 8

Plugging in 8 for the equation:

$-\frac{18}{y}=-\frac{18}{8}=-\frac{9}{4}$

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

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