# Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.8 The Derivative as a Rate of Change.

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.8 The Derivative as a Rate of Change

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 115  Average Rate of Change  Instantaneous Rate of Change  Average Velocity  Position, Velocity, and Acceleration  Approximating the Change in a Function Section Outline

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 115 Average Rate of Change

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 115 Average Rate of ChangeSOLUTION Suppose that f (x) = -6/x. What is the average rate of change of f (x) over the interval 1 to 1.2? EXAMPLE The average rate of change over the interval is

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 115 Instantaneous Rate of Change

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 115 Instantaneous Rate of ChangeSOLUTION Suppose that f (x) = -6/x. What is the (instantaneous) rate of change of f (x) when x = 1? EXAMPLE 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 115 Average & Instantaneous Rates of Change 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. EXAMPLE (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?

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 115 Average & Instantaneous Rates of ChangeSOLUTION (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. CONTINUED 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 115 Average & Instantaneous Rates of ChangeCONTINUED 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 115 Average & Instantaneous Rates of ChangeCONTINUED 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 115 Average Velocity DefinitionExample Average Velocity: Given a position function s(t), the average velocity from time t = a to t = a + h is Suppose a car is 3 miles from its starting point after 5 minutes and 7 miles from its starting point after an additional 6 minutes (after a total of 11 minutes). The average velocity of the car between the two given locations is miles per minute where a = 5 and h = 6.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 115 Position, Velocity & Acceleration s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 115 Position, Velocity & AccelerationSOLUTION A toy rocket fired straight up into the air has height s(t) = 160t – 16t 2 feet after t seconds. EXAMPLE (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? (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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 115 Position, Velocity & Acceleration Therefore, the initial velocity of the rocket is 160 feet per second. Now replace t with 0 and evaluate. CONTINUED (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/s 2. Therefore, the acceleration when t = 3 is -32ft/s 2.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 115 Position, Velocity & AccelerationCONTINUED 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. (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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 115 Position, Velocity & AccelerationCONTINUED 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. (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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 17 of 115 Approximating the Change in a Function

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 18 of 115 Approximating the Change in a FunctionSOLUTION 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. EXAMPLE 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 19 of 115 Approximating the Change in a Function Therefore, approximate number of milligrams of the drug in the bloodstream after 3½ hours is 1.75. CONTINUED

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