Rates of Change
It is useful to contemplate slopes in practical situations. For example, suppose the following graph in Figure 8.1 is for y = f(x), a function that gives the price y for various amounts x of cheese. Because the straight line goes through the points (1 lb.,$2) and (2 lbs.,$4), the slope per pound.
The slope is therefore the rate at which the cheese is sold, in dollars per pound. Because slope = , a slope will always be a rate measured in yunits per xunit.
For example, suppose a passenger on a bus writes down the exact time she passes each highway mile marker. She then sketches the graph shown in Figure 8.2 of the bus's position on the highway over time. The slope at any point on this graph will be measured in yunits per tunit, or miles per hour. The steepness of the slope represents the speed of the bus.
Because the derivative of a function gives the slope of its tangent lines, these practice problems show that the derivative of a function gives its rate of change.
An excellent example comes from position functions. A position function s(t) states where an object is at any given time. The derivative s'(t) states the rate at which that object's position is changing—that is, the speed or velocity of the function. Thus, s'(t) = v(t) . The second derivative s"(t) = v'(t) tells how the velocity is changing, or the acceleration. Thus, s"(t) = v'(t) = a(t) where s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.
Example 1
Suppose an object rolls along beside a tape measure so that after t seconds, it is next to the inch marked s(t) = 4t^{2} + 8t + 5. Where is the object after 1 second? After 3 seconds? What is the velocity function? How fast is the object moving after 2 seconds? What is the acceleration function?
Solution 1
The position function s(t) = 4t^{2} + 8t + 5 tells us where the object is. After 1second, the object is next to the s(1) = 17 inch mark on the tape measure. After 3 seconds, the object is at the s(3) = 65inch mark.
The velocity function is v(t) = s'(t) = 8t + 8. Thus, after 2 seconds, the object is moving at the rate of v(2) = 24 inches per second. Do realize that this velocity of 24 inches per second is an instantaneous velocity, the speed just at a single moment. If a car's speedometer reads 60 miles per hour, this does not mean that it will drive for 60 miles or even for a full hour. The car might speed up, slow down, or stop. However, at that instant, the car is traveling at a rate that, if unchanged, will take it 60 miles in one hour. A derivative is always an instantaneous rate, telling you the slope at a particular point, but not making any promises about what will happen next.
The acceleration function is a(t) = v' (t) = s" (t) = 8. Because this is a constant, it tells us that the object increases in speed by 8 inches per second every second.
The most popular example of constant acceleration is gravity, which accelerates objects downward by every second. Because of this, an object sec thrown with a velocity of b feet per second from a height of h feet above the ground will have (after t seconds) a height of s(t) = –16t^{2} + bt + h feet.
The starting time is t = 0, at which point the object is s(0) = h feet off the ground, the correct initial height. The velocity function is v(t) = s'(t) = –32t + b. At the starting time t = 0, the velocity is v(0) = b, the desired initial velocity. The function v(t) = – 32t + b means that 32 feet per second are subtracted from the initial velocity b every second. The acceleration function is a(t) = v'(t) = s" (t) = – 32 .This is the desired constant acceleration.

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