Section 2.3 Tangent lines, rates of change, and derivatives

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1 Section.3 Tangent lines, rates of change, and derivatives (3/9/08) Overview: The derivative was developed in the seventeenth centur for determining tangent lines to curves and the velocit of moving objects in cases that could not be handled with geometr and algebra alone. The ancient Greeks had used reasoning similar to that in modern plane geometr to stud tangent lines to circles and other special curves. This approach could not be used, however, to find tangent lines to most curves. We will see in this section how tangent lines can be found as the limiting positions of secant lines and how instantaneous velocit and other instantaneous rates of change can be found as limits of average velocities and average rates of change. This leads to the definition of the derivative as the slope of a tangent line and as an instantaneous rate of change. Topics: Tangent lines, derivatives, and instantaneous rates of change Predicting a derivative b calculating difference quotients Finding eact derivatives Equations of tangent lines The f-formulation of the definition Finding a rate of change from a tangent line A secant-line program Tangent lines, derivatives, and instantaneous rates of change Euclid (c. 300 BC) defined a tangent line to a circle at a point P to be the line that intersects the circle at onl that point (Figure ). The tangent line is outside the circle ecept at the point of tangenc. This implies that the tangent line is perpendicular to the radius at P because P is the closest point on the tangent line to the center O of the circle and consequentl is at the foot of the perpendicular line from O to the tangent line. This propert enables us to find the slope of the tangent line: if, for eample, the center of the circle is the origin in an -plane, as in Figure, and P has coordinates (h,k), then the slope of the radius OP is k/h, and consequentl the slope of the perpendicular tangent line is h/k. 0 h k P(h,k) f(a) f() a Tangent line at P Tangent line at a FIGURE FIGURE The use of the term tangent in tangent line comes from the Latin tangere, to touch. 89

2 p. 90 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives The graph f() in Figure does not have a similar geometric propert that could be used to find its tangent lines. Instead, we find a tangent line as a limit of secant lines. As we saw in the last section, the slope of the secant line through the points at a and b for b a (Figure 3) equals the rise f(b) f(a) from the point at a to the point at b, divided b the corresponding run b a: [Slope of the secant line] f(b) f(a). () b a f() [Slope of the secant line] f(b) f(a). b a f(a) b a f(b) f(a) FIGURE 3 a b Imagine what happens as the number b approaches a (Figure ). The secant line turns about the fied point at a. If its slope () has a finite limit, then the secant line approaches the line with that slope the dashed diagonal line in Figure and this line is the tangent line to the graph of f at a that is shown in Figure. f() Secant lines approaching f(a) the tangent line as b a FIGURE a b The limit as b tends to a of the slope of the secant line is called the derivative of f at a and is denoted f (a): f f(b) f(a) (a) lim. b a b a Consequentl, the tangent line at a to the graph of f is the line through the point at a that has slope equal to the derivative f (a).

3 Section.3, Tangent lines, rates of change, and derivatives p. 9 (3/9/08) How about rates of change? We also saw in the last section that the slope () of the secant line is the average rate of change of f with respect to from a to b. If b is ver close to a, then this average rate of change is a good measure of how rapidl the function is changing at a. We define its limit f (a) as b tends to a to be the instantaneous rate of change of the function with respect to at a. These definitions can be summarized as follows: Definition The derivative of f() at a is f f(b) f(a) (a) lim b a b a () provided this limit eists and is finite. The derivative f (a), if it eists, is the slope of the tangent line to the graph of f at a and is the instantaneous rate of change of f with respect to at a. If and f() have units, then the units used for the derivative are those used for f() divided b those used for. The derivative is usuall referred to as the rate of change rather than as the instantaneous rate of change as in Definition ecept when it is being discussed with average rates of change. The process of finding derivatives is called differentiation. The average rate of change () is often called a difference quotient because it is a quotient of the differences f(b) f(a) and b a. We then have three terms for average rate of change and three terms for (instantaneous) rate of change, which are related as shown the following diagram, where the vertical arrows indicate that b is approaching a: f(b) f(a) b a Average rate of change Slope of the secant line The difference quotient f (a) Instantaneous rate of change Slope of the tangent line The derivative Instantaneous velocit is the instantaneous rate of change with respect to time of an object s coordinate s s(t) on an s-ais and is the limit of average velocities.

4 p. 9 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives Predicting a derivative b calculating difference quotients Figure 5 shows the graph of f() and its tangent line at, whose slope equals the derivative f (). In the net eample we predict the value of this derivative b calculating the slope of the secant line (Figure ) for values of b close to. b Tangent line Secant line Slope f () Slope f(b) f() b FIGURE 5 FIGURE Eample Solution Predict the derivative f () for f() b calculating the average rate of change f(b) f() b for b, b.5, b, b 0., b 0.00, and b b b The average rates of change in the table below suggest that the derivative f b () lim is. The secant lines corresponding to the values in the table b b are shown in Figures 7 through. The secant lines for b 0.00 and b in Figures and are indistinguishable and look like the tangent line in Figure 5 because these values of b are ver close to. b b b

5 Section.3, Tangent lines, rates of change, and derivatives p. 93 (3/9/08) b b b b b.5 b.5 FIGURE 7 FIGURE 8 FIGURE 9 b b. b.00 b.0000 FIGURE 0 FIGURE FIGURE Finding eact derivatives To make the algebra in finding eact derivatives easier to follow, we use instead of b for the -coordinate of the variable point in definition (). Then the secant line passes through (a,f(a)) and (,f()), as shown in Figure 3, and definition () takes the form f f() f(a) (a) lim. (3) a a f() f() Secant line of slope f() f(a) f() f(a) a f(a) a FIGURE 3 a

6 p. 9 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives Eample Use definition (3) to find the derivative of f() at. Solution Since f(), definition (3) with a gives f f() f() () lim lim. () We cannot find this limit b taking the limit of the numerator and denominator because the both tend to zero and 0/0 has no meaning. Instead, as we saw with similar eamples in Section., we can factor the numerator to obtain ( + )( ) for. Canceling one factor of the numerator with the denominator then gives + for. Because the polnomial + is continuous at, its limit as is equal to its value at. The last equation and () ield f [ ] () lim lim ( + ) +. The derivative of f() at is, as we predicted in Eample. Eample 3 Use definition (3) to find g (0) for g() 3 3. Solution Since g(0) 3(0) 0 3 0, definition (3) with g in place of f and a 0 ields g g() g(0) (3 3 ) (0) lim lim lim To epress the last quotient as a function whose limit we can find, we factor from the numerator and cancel it with the denominator to have 3 3 (3 ) 3 for 0. Since the polnomial 3 is continuous at 0, its limit as 0 is its value at 0, and we obtain g (0) lim [ lim (3 ) 3 ]

7 Section.3, Tangent lines, rates of change, and derivatives p. 95 (3/9/08) Equations of tangent lines The tangent line to f() at a in Figure passes through the point (a,f(a)) and has slope f (a). B the point-slope formula, it has the equation, f(a) + f (a)( a). (5) f(a) f() a f(a) + f (a)( a) Tangent line at FIGURE FIGURE 5 Eample Solution Eample 5 Solution Give an equation of the tangent line to the graph of f() at and draw it with the graph of the function. The value f () of the derivative was found in Eample. Also f(), so b (5) with a, the tangent line has the equation + ( ), which can be rewritten as. The curve and tangent line are shown in Figure 5. Give an equation of the tangent line to the graph of H() / at and draw the curve with the tangent line. We need to rewrite the difference quotient H() H() for 0, () so we can find its limit H () as. We first simplif the fractions to obtain for 0,, H() H() ( ) ( ) ( ) ( ). Net, we cancel the factors in the numerator and denominator to have for 0,, H() H(). Then, because the rational function / is continuous at, its limit as is its value at and

8 p. 9 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives H H() H() () lim [ ]. lim The tangent line has the equation H() + H ()( ) and since H() /, the tangent line is ( ). We rewrite this equation as to see that it is the line of slope with -intercept. The curve and tangent line are drawn in Figure. 3 The tangent line FIGURE 3 The net eample involves a function with parameters (constants) in its formula. Eample What is the derivative of J() a Solution B Definition, + b at 0 if a and b are nonzero constants? J J() J(0) (0) lim. (7) 0 0 Because J(0) is zero, the difference quotient on the right of (7) can be written J() J(0) 0 [J()] [ a + b ] for 0. Cancelling the s in the numerator and denominator then gives J() J(0) 0 a + b for 0. Since b is not zero, the rational function at 0 and definition (7) ields [ ] J a (0) lim 0 + b a + b 0 a + b a 0 + b a b. is defined and continuous

9 Section.3, Tangent lines, rates of change, and derivatives p. 97 (3/9/08) The -definition of the derivative It is often convenient to use definition () of the derivative with different notation for the rise and run on the secant line. We let be the fied value where we want to find the derivative and denote the -coordinate of the variable point b +, where is the capital Greek letter delta. Then is the run from (,f()) to ( +, f( + )) on the secant line, as shown in Figure 7. We denote the corresponding rise f(+ ) f() b f. With this notation, the definition of the derivative becomes f () lim 0 f lim 0 f( + ) f(). (8) f() Secant line of slope f f( + ) f() f( + ) f() f FIGURE 7 + Formulation (8) of the definition is especiall convenient for finding a derivative at a variable point. Eample 7 Solution Use formulation (8) of the definition to find the derivative of f() / at an arbitrar 0. We fi 0 and set f() / in (8) to have f () lim 0 +. (9) To convert the three-laer fraction on the right of (9) into a product of two-level fractions, we replace division b with multiplication b /. Then we take a common denominator and simplif the result: + [ + [ ] ( + ) ] for 0. [ ] ( + ) ( + ) Net, we cancel the s in the numerator and denominator to have + ( + ) for 0. (0) Notice that each of and f denotes a single number and not a product of a and or f. Think of as the change in, so that is the change in and and f is the change in f.

10 p. 98 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives Since is not zero, the rational function of on the right of (0) is continuous at 0, so that b (0), f () lim 0 [ ] ( + ) ( + ) 0.. Finding a rate of change from a tangent line The function A A(t) whose graph is shown in Figure 8 gives the percentage of alcohol in a person s blood t hours after he has consumed three fluid ounces of alcohol. () As ou can see from the graph, the blood-alcohol level rises from 0% to about 0.% in about two hours, and then drops close to 0.0% after ten more hours A (percent of alcohol) A A(t) 0. FIGURE t (hours) Eample 8 The tangent line at t in Figure 9 passes through the points P (, 0.8) and Q (, 0.8). What is the (instantaneous) rate of change of the person s blood alcohol level with respect to time one hour after consuming the alcohol? A (percent of alcohol) Q A A(t) P 0. FIGURE t (hours) Solution The rate of change A () of the blood-alcohol level after one hour is the slope of the tangent line in Figure 9. We can find its eact value because we are given the eact coordinates of two points on the line. We obtain A () ( ) percent ( ) hour 0. percent per hour. The net eample uses the technique of rationalization of differences of square roots that was used to find limits in Section.. () Data adapted from Encclopædia Britannica, Vol., Chicago: Encclopædia Britannica, Inc., 95, p.58.

11 Section.3, Tangent lines, rates of change, and derivatives p. 99 (3/9/08) Eample 9 Find the derivative of at 5. Solution B definition (3) with a 5 and in place of f(), () (5) 5 (5) lim lim In order to cancel the denominator, we need to first rationalize the difference 5 of square roots in the numerator. We multipl and divide b the sum + 5 of the square roots to obtain for nonnegative 5, () (5) ( 5)( + 5) ( 5)( + 5) 5 ( 5)( + 5) + 5. ( ) ( 5) ( 5)( + 5) Then, because /( + 5) is continuous at 5, we obtain [ ] (5) lim C A secant line program A secant line program for several tpes of calculators is available on the web page for this book. The program illustrates the definition of the derivative b generating secant lines with graphs of the functions f(b) f(a) and calculating the difference quotients that are their slopes. b a Eample 0 Solution Predict the derivative f () for f() + / b using a secant-line program to f(b) f() calculate the difference quotient b + /b for b.5,., b b.,.00, and b Use the window 0.5, 8. Enter the program and store the function f() + / and define the window in our calculator, according to the instructions on the web page. Run the program with a and the specified values of b to obtain the secant lines in Figure 0 through and their slopes in the following table. Because these slopes seem to be approaching 3 as b approaches, we predict that f () 3. (In the net section we will derive formulas that could be used to verif this prediction.) f(b) f(a) b a ashenk/

12 p. 00 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives b b a, b.5 a, b.. [Slope] [Slope].537 FIGURE 0 FIGURE b a, b. a, b.00. [Slope] [Slope] FIGURE FIGURE 3 a, b [Slope] FIGURE

13 Section.3, Tangent lines, rates of change, and derivatives p. 0 (3/9/08) Interactive Eamples.3 Interactive solutions are on the web page http// ashenk/.. A woman s trip is described b the mathematical model in which she is s(t) 0t + 3 sin(πt) miles east of a town at time t (hours) for 0 t 3. Use an average velocit to estimate her instantaneous velocit toward the east at t.. Predict the derivative f () for f() 3 + b calculating the average rate of change of f() from to b for b.5,.,.,.00, and,0000. (If ou use a secant line program on a calculator or computer, set the window to 0.5.5,.) 3. Use the definition f f() f(a) (a) lim to find the derivative f () for f() a a. Use results from the previous eample to find an equation of the tangent line to the graph of f() 3 + at. Then draw the curve and the tangent line.. 5. Use the -formulation of the definition to find a formula for f () with f() 3 + as a function of. Eercises.3 A Answer provided. CONCEPTS: O Outline of solution provided.. Decribe how secant lines are related to instantaneous rates of change. C Graphing calculator or computer required.. In some cases a tangent line at a point is the one line that intersects the curve onl that point. Eplain wh this characterization of a tangent line does not appl to the tangent line at P in Figure 5. cos G() P π FIGURE 5 FIGURE 3. Figure shows the graph of the function, G() { for for < < + 3 for Eplain wh the tangent line to this curve at the origin cannot be characterized as the line that intersects the curve onl at one point. In the published tet the interactive solutions of these eamples will be on an accompaning CD disk which can be run b an computer browser without using an internet connection.

14 p. 0 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives. Figure 7 shows the graph of blood-alcohol level from Eample 9 and its tangent line at t 8. The tangent line passes through the points R (, 0.) and S (8, 0.08). (a) What is the (instantaneous) rate of change of the person s blood alcohol level eight hours after consuming the alcohol? (b) Wh is is negative? A (percent of alcohol) A A(t) 0. R S FIGURE t (hours) BASICS: 5. O Calculate the slope of the secant line through the points at a and b on the graph of f() / with a and b 0.5, 0.75, 0.9, and Use the results to predict the value of the derivative f (). If ou use the secant-line program, set the window to 0, 5.. O (a) Use formulation () of the definition to find the eact derivative of f() / at. (b) Give an equation of the tangent line to at. 7. O Use the -formulation (8) of the definition to find the derivative of f() / at an arbitrar O Figure 8 shows the graph of a function f() and two points on its tangent line at. What is the derivative f ()? 5 (, 3) 0 (, 9) 5 f() FIGURE O Calculate the slope of the secant line through the points at a and b on the graph of g() / with a and b 0., 0.3, 0.9, and Use the results to predict the value of the derivative g (). C If ou use the secant-line program, set the window to 0,. 0. Calculate the slope of the secant line through the points at a and b on the graph of h() 3 with a and b 0, 0.5, 0.9, and Use the results to predict the value of the derivative h (). C If ou use the secant-line program, set the window to 0.5,.

15 Section.3, Tangent lines, rates of change, and derivatives p. 03 (3/9/08) f(b) f() In Eercises through, calculate the difference quotient for b 0,,.9,.999 and b.99999, and use the results to predict the derivative f (). C The given windows are to be used with the secant-line program.. O f() + + (, 8). f() ( ) /3 ( , 5) 3. A f() + ( 3.5, 0). f() 0 + ( ) ( , 5 50) Use the definition f (a) lim a f() f(a) a 5. O g () for g() + /. A G () for G() /(5 ) 7. O P (0) for P() ( + )/( + ) 8. A (0) for () /3 9. g () for g() 0. F () for F() + 3. f (3) for f() to find the derivatives in Eercises 5 through. In Eercises through use the definition f f() f(a) (a) lim to find the equation of the tangent a a line to the graph of the function at the given value of a. Then draw the graph of the function with the tangent line.. O f() / at a 3. A f() at a. f() 3 at a 0 In Eercises 5 through 9 use the -formulation of the definition to find the derivatives. 5. O f (3) for f() /( + ). A g () for g() 3/( + ) 7. O W () for W() /( + 0) 8. A Z () for Z() 8/ 9. () for () 8 5

16 p. 0 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives 30. O Figure 9 shows the graph of the rate of ogen consumption r r(v) (liters per minute) of a swimmer as a function of his velocit v (meters per second) and the tangent line to the graph at v. () (a) Wh is r(.) greater than r()? (b) What is the rate of change of his ogen consumption with respect to velocit at v? (Give the units.) r (liters per minute) T (degrees ) h (feet) 3 (.,.) (,.) 5 0 (3, 3.5) (, 9.) (, 70) (, 0) v (meters per second) 3 9 t (months) 3 t (hours) r r(v) T T(t) h h(t) FIGURE 9 FIGURE 30 FIGURE 3 3. A The temperature of the ocean T T(t) (degrees Celsius) at 30 south in the southern hemisphere at time t (months) with t 0 on Januar is given b the function of Figure 30, which also shows the tangent line at t 3. (3) (a) Wh is plausible that T() less than T(0)? (b) What is the rate of change of the temperature with respect to time at t 3? (Give the units.) 3. Figure 3 shows the graph of the height h h(t) (feet) of a balloon as a function of the time t (hours) and the tangent line to the graph at t. (a) During approimatel what portions of the time interval 0 t.5 is the balloon rising? (b) What is the upward velocit of the balloon at t? 33. The function s s(t) of Figure 3 gives the s-coordinate of an object on an s-ais as a function of the time. The drawing also shows the tangent line to the graph at t, the point (, ) of tangenc, and the point (8, 0) where the curve and tangent line also intersect. (a) What is the instantaneous rate of change of s with respect to t at t? (b) What is the average rate of change of s with respect to t for t 8? s (meters) s s(t) 75 5 (, ) (8, 0) FIGURE 3 8 t (minute) () Data adapted from The Human Machine b R. Aleander, New York, NY: Columbia Universit Press, 99, p. 7. (3) Data adapted from Monsoons b P. Webster, Scientific American, August, 98, New York, NY: Scientific American, Inc, 98, p. 0.

17 Section.3, Tangent lines, rates of change, and derivatives p. 05 (3/9/08) EXPLORATION: Use definition (3) or (7) to find the derivatives in Eercises 3 through 39. The letters a,b, c, and k are constant parameters. 3. O f () for f() a + b 35. O g (3) for g() /( + k) 37. O () for a + b/ 38. A z () for z ( + k) 3. A h (0) for h() a 5 + b 3 + c 39. w () for w a + b + c Use definition (3) or (7) and rationalization of differences of square roots as necessar to find the derivatives in Eercises 0 through O h () for h() 0. A () for / /. f (9) for h() / 3. A g () for g() + 5. O f () for f() 5. g () for f(). A A walker has gone s(t) miles past Reklaw, Teas, t hours after dawn one summer da. What do each of the following values tell about the hike: (a) s(3) 7, (b) s (3) 3, (c) s (5) 0.5, s(5) s(0) and (d)? 5 7. A The median price N N(t) (thousand dollars) of new houses and E E(t) (thousand dollars) of eisting houses on the market in the U.S. in ear t AD satisf (a) N(99) N(970) + 97, (b) N(99).5E(99), (c) E(99) 5E(970), (d) E (99) > 0, and (e) N (99) 0.9E (99). () Rephrase each of these equations as an English sentence without using the smbols N and E for the functions. 8. The amount of tar T T(t) (milligrams) and the amount of nicotine N N(t) (milligrams) in an average cigarette that was manufactured in the United States in ear t AD has the properties (5) (a) T(957) 38, (b) T (957), (c) N(980) N(957), and (d) N (980) 3 N (957). Rephrase each of these equations as an English sentence without using the smbols T and N for the functions. 9. A small airplane whose engine is running at 500 revolutions per minute has air speed v v(h) nautical miles per hour (knots) and uses g g(h) gallons of gasoline per hour when it is at an altitude of h thousand feet above the ground. () Epress the following statements with equations involving v(h), g(h), and their derivatives, under the assumption that the plane s engine is running at 500 revolutions per minute. (a A ) The plane s air speed is nautical miles per hour when it is at an altitude of 000 feet. (b A ) The plane s air speed is decreasing at the rate of 0.5 knots per thousand feet and its rate of gasoline consumption is decreasing 0. gallons per hour per thousand feet when the plane is at an altitude of 000 feet. (c) The rate at which the plane uses gasoline at an altitude of,000 feet is 8% of the rate at 000 feet. () Data adapted from 993 Statistical Abstract of the United States, Washington, DC: U.S. Department of Commerce, 993, p.70. (5) Data adapted from Drugs Societ and Human Behavior, b O. Ra and C. Ksir, Saint Louis, MO: Mosb College Publishing, 990, p.98. () Data adapted from Cessna Skhawk Information Manual, Wichita, KA: Cessna Aircraft Compan, 978, pp. 5-.

18 p. 0 (3/9/08) Section.3, Tangent lines, rates of change, and derivatives 50. In a stud of vehicle sales in the United States, passenger vehicles were classified into two groups, trucks (including pickups, mini-vans, and sport-utilit vehicles) and cars. (7) Epress the following three statements as equations involving values of the percent T T(t) of passenger vehicles sold in ear t AD that were trucks, the percent C C(t) that were cars, and their derivatives: (a) Twent percent of the passenger vehicles that were sold were trucks in 970. (b) Thirteight percent were trucks in 99. (c) The percent of passenger vehicles sold that were trucks was increasing 0. percent per ear in 99. (d) What are the values of C(970), C(99), and C (99)? 5. A The derivative of F() at is F (). Find, b trial and error, a difference quotient (b )/(b ) that approimates the derivative with an error that is less than 0.0 and greater than The derivative of P() at is P (). Find, b trial and error, a difference quotient ( b )(b ) that approimates the derivative with an error that is less than and greater than A Figure 33 shows the graph of the volume V V (t) of air in a balloon as a function of the time t with the point of tangenc (5, 0) and a second point (8, k) on the tangent line to the graph at t 5. What is the value of the constant k if V (5) 5 cubic feet per hour? V (cubic feet) V V (t) G() 00 (8,k) 0 50 (5, 0) t (hour) 7 FIGURE 33 FIGURE 3 5. The tangent lines to the graph G() at and 7 in Figure 3 are perpendicular. Also G (). What is the value of the derivative G (7) at 7? 55. Use Definition to find the derivative of /3 at. ( Rationalize the difference quotient b multipling the numerator and denominator of the difference quotient b /3 + /3 + and then using the identit (a b)(a + ab + b ) a 3 b Find the derivative of at 5 as the limit of 5 5 as 5. (End of Section.3) (7) Data adapted from The New York Times, August, 993, New York, NY: The New York Times Compan, p. C. Source: Ford Motor Compan.

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