We know that a derivative is an instantaneous rate of change. We have already seen how useful this is in physics when studying distance, velocity, and acceleration in lesson 2-1, and lesson 2-6, and in sample problems in lesson 3-2, and in lesson 3-1. In this lesson we will also see applications for chemistry, biology, economics, geology, geography, meteorology and other sciences.

Sample physics problem: The position of a particle is given by the equation, .  where t is measured in seconds and s in meters. (a) Find the velocity at time t; (b) What is the velocity after 3 s? After 4 s?; (c) When is the particle at rest?
 

	Part (a). Our first step is to find the derivative of the position function. This will be our velocity formula.
	Part (b). Now that we have the velocity formula, we can find the velocity at any instant. In this case, we need to find the velocity at the instant when t = 4. Answer: 16 m/s. We also need to find the velocity when t = 3.
	Part (c). To find when the particle is at rest, we take the velocity (1st derivative formula), set it equal to zero, and solve for t.
	We find there are two times when this will happen.

Using your Tools for Enriching Calculus CD (that came with your book), load and run Module 3.3/3.4/3.5. This module will allow you to see an animation of a graph with distance, velocity, and acceleration on the same graph.

Sample economics problem: In economics the instantaneous rate of change of cost with respect to the number of items produced is called the marginal cost by economists. Suppose a company has estimated that the cost (in dollars) of producing item x is: . Find the instantaneous rate of change of the cost with respect to the number of items produced (marginal cost).
 

	Given cost function.
	We differentiate using the power rule in two separate terms.
	We substitute in the value of x given in the problem. Therefore the marginal cost when x = 500 is $26/item This then, is the predicted cost for the 501st item.

Rates of Change Teaching Video

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