slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
§ 1.8 PowerPoint Presentation
Download Presentation
§ 1.8

Loading in 2 Seconds...

play fullscreen
1 / 19

§ 1.8 - PowerPoint PPT Presentation


  • 113 Views
  • Uploaded on

§ 1.8. The Derivative as a Rate of Change. Section Outline. Average Rate of Change Instantaneous Rate of Change Average Velocity Position, Velocity, and Acceleration Approximating the Change in a Function. Average Rate of Change. Average Rate of Change. EXAMPLE.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about '§ 1.8' - cato


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

§1.8

The Derivative as a Rate of Change

section outline
Section Outline
  • Average Rate of Change
  • Instantaneous Rate of Change
  • Average Velocity
  • Position, Velocity, and Acceleration
  • Approximating the Change in a Function
slide4

Average Rate of Change

EXAMPLE

Suppose that f(x) = -6/x. What is the average rate of change of f(x) over the interval 1 to 1.2?

SOLUTION

The average rate of change over the interval is

slide6

Instantaneous Rate of Change

EXAMPLE

Suppose that f(x) = -6/x. What is the (instantaneous) rate of change of f (x) when x = 1?

SOLUTION

The rate of change of f(x) at x = 1 is equal to . We have

That is, the rate of change is 6 units per unit change in x.

slide7

Average & Instantaneous Rates of Change

EXAMPLE

Refer to the figure below, where f(t) is the percentage yield (interest rate) on a 3-month T-bill (U.S. Treasury bill) t years after January 1, 1980.

(a) What was the average rate of change in the yield from January 1, 1981 to January 1, 1985?

(b) How fast was the percentage yield rising on January 1, 1989?

(c) Was the percentage yield rising faster on January 1, 1980 or January 1, 1989?

slide8

Average & Instantaneous Rates of Change

CONTINUED

SOLUTION

(a) To determine the average rate of change in the yield from January 1, 1981 to January 1, 1985, we must first determine the coordinates of the two points that correspond to the two given dates. They are (1, 14) and (5, 7). Now we use the average rate of change formula.

Therefore, the average rate of change in the yield from January 1, 1981 to January 1, 1985 is -7/4.

(b) To determine how fast the percentage yield was rising on January 1, 1989, we must determine the instantaneous rate of change of f(t) when t = 9 (corresponding to January 1, 1989). This means that we must find the slope of the tangent line to the graph of f (t) where t = 9. The tangent line is on the graph and so we need only determine any two points on the tangent line. Using the coordinates of these two points, we will calculate the slope of the tangent line and that will be the instantaneous rate of change that we seek.

Notice that two of the points on the tangent line are (5, 5) and (11, 10). Using these points we will now calculate the slope of the tangent line.

slide9

Average & Instantaneous Rates of Change

CONTINUED

Therefore, the rate at which the percentage yield was rising on January 1, 1989 was 5/6.

(c) To determine if the percentage yield was rising faster on January 1, 1980 or January 1, 1989, we would need to know the slopes of the tangent lines corresponding to t = 0 (January 1, 1980) and t = 9 (January 1, 1989). Although we already have this information for t = 9 (see part (b)), we do not yet have this information for t = 0. Therefore, we would first need to draw a tangent line to the graph corresponding to t = 0. This is done below.

slide10

Average & Instantaneous Rates of Change

CONTINUED

Obviously, finding the coordinates of two points on this tangent line might prove a little difficult. However, notice that the slopes of the two tangent lines (all we’re really interested in are their slopes) are not remotely similar (that is, the tangent lines are not close to being parallel). Therefore, in this circumstance, it would be sufficiently appropriate to notice that the blue tangent line (corresponding to t = 0) has a steeper slope and therefore the rate of change was greater on January 1, 1980 than it was on January 1, 1989.

NOTE: Use this technique of “eye-balling” a graph only when absolutely necessary and only with great care.

slide12

Position, Velocity & Acceleration

s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.

slide13

Position, Velocity & Acceleration

EXAMPLE

A toy rocket fired straight up into the air has height s(t) = 160t – 16t2 feet after t seconds.

(a) What is the rocket’s initial velocity (when t = 0)?

(b) What is the velocity after 2 seconds?

(c) What is the acceleration when t = 3?

(d) At what time will the rocket hit the ground?

(e) At what velocity will the rocket be traveling just as it smashes into the ground?

SOLUTION

(a) To determine what the rocket’s initial velocity is, we must first have a velocity function and then evaluate t = 0 in that function.

This is the given position function.

Differentiate to get the velocity function.

slide14

Position, Velocity & Acceleration

CONTINUED

Now replace t with 0 and evaluate.

Therefore, the initial velocity of the rocket is 160 feet per second.

(b) To determine the velocity after 2 seconds we evaluate v(2).

This is the velocity function.

Replace t with 2 and evaluate.

Therefore, the velocity of the rocket after 2 seconds is 96 feet per second.

(c) To determine the acceleration when t = 3, we must first find the acceleration function.

This is the velocity function.

Differentiate to get the acceleration function.

Since the acceleration function is a constant function, the acceleration of the rocket is a constant -32 ft/s2. Therefore, the acceleration when t = 3 is -32ft/s2.

slide15

Position, Velocity & Acceleration

CONTINUED

(d) To determine at what time the rocket will hit the ground we must determine what we know about the position, velocity, or acceleration of the rocket when the rocket hits the ground. We know that at the time of impact, the position of the rocket will be 0 feet above the ground. Therefore, we will use the position function and replace s(t) with 0.

This is the given position function.

Replace s(t) with 0.

Factor 16 out of both terms on the right and divide both sides by 16.

Factor.

Solve for t.

Therefore, the rocket will be 0 feet above the ground at times t = 0 and t = 10. t = 0 corresponds to when the rocket first began its flight, so that would not be the solution. So, the rocket hit the ground after 10 seconds, when t – 10 = 0.

slide16

Position, Velocity & Acceleration

CONTINUED

(e) To determine at what velocity the rocket will be traveling just as it smashes into the ground, we must use the velocity function. The question is, what do we use for t? From part (d), we know that the rocket will hit the ground at t = 10 seconds. Therefore, we will find v(10).

This is the velocity function.

Replace t with 10 and evaluate.

Therefore, when the rocket hits the ground, it will be have a velocity of -160 ft/s. That is, it will be traveling 160 ft/s in the downward direction.

slide18

Approximating the Change in a Function

EXAMPLE

Suppose 5 mg of a drug is injected into the bloodstream. Let f(t) be the amount present in the bloodstream after t hours. Interpret f(3) = 2 and Estimate the number of milligrams of the drug in the bloodstream after 3½ hours.

SOLUTION

First, f(3) = 2 means that after 3 hours, 2 milligrams of the drug still remain in the bloodstream.

Next, means that after 3 hours, the rate at which the amount of the drug is diminishing (because of the minus sign) within the bloodstream is ½ of a milligram per 1 hour.

To estimate the number of milligrams of the drug in the bloodstream after 3½ hours, we will use the formula where a = 3 and h = ½ .

slide19

Approximating the Change in a Function

CONTINUED

Therefore, approximate number of milligrams of the drug in the bloodstream after 3½ hours is 1.75.