Capter 4 Rates of Cange In tis capter we will investigate ow fast one quantity canges in relation to anoter. Te first type of cange we investigate is te average rate of cange, or te rate a quantity canges over a given interval. For exaple, if you ave 15 inutes to arrive at a destination tat is 10 iles away, you could calculate te average rate of cange by dividing 10 iles by ¼ our. Tus, you would need to travel 40 iles per our to arrive at your destination on tie. Te second rate of cange tat we investigate, and actually begins our study of differential calculus, is inseous rate of cange, or te rate a quantity canges at a given inst. Police officers parked on te side of a road, calculating te speed of cars wit teir radar gun is an exaple of inseous rate of cange. 4.1 Average Rate of Cange Table 4.1 below sows te daily low teperatures in College Station, Texas for te first 30 days in October 00. Tis data is graped as a scatter plot in Figure 4.1. Table 4.1 Daily low teperatures in October 00 for College Station, TX. Day 1 3 4 5 7 8 9 10 11 1 13 14 15 Low ( F ) Teperature 71 73 7 7 71 74 70 71 4 3 59 58 53 48 Day 1 17 18 19 0 1 3 4 5 7 8 9 30 Low ( F ) Teperature 48 49 0 3 1 0 3 4 4 58 57 3 3 58 54 Source: ttp://www.et.tau.edu/et/osc/cll/oct0.t
Figure 4.1 Scatter Plot of Table 4.1 data October 00 Daily Low Teperatures in College Station, TX Teperature (degree Fareneit) 80 70 0 50 40 1 11 1 1 Day of October We can use tis data to find te average cange in teperature between te 1st and 15t day. Since te teperature was 71 degrees on te 1st of October and 48 degrees on te 15t of October we can conclude tat over te 14-day period, te teperature canged 3 degrees. Representing tis as a ratio we get cange in teperature 71 48 3 = = 1.4. nuber of days passed 1 15 14 Tis tells us tat te teperature dropped at an average rate of 1.4 degrees per day between te 1st and 15t day of October. Figure 4. below sows a grap of te average rate of cange between 1st and 15t day. Figure 4. Average rate of cange in low teperatures between 1st and 15t day of October. October 00 Daily Low Teperatures in College Station, TX Teperature (degree Fareneit) 80 0 40 1 11 1 1 Day of October A siilar calculation can be done for te cange in teperature between te 1t and 19t day. cange in teperature 3 48 15 = = = 5. nuber of days passed 19 1 3 Tus, te teperature increased at an average rate of 5 degrees per day between te 1t and 19t day of October. Figure 4.3 sows a grap of te average rate of cange between te 1t and 19t days.
Figure 4.3 Average rate of cange in low teperatures between 1t and 19t day of October. October 00 Daily Low Teperatures in College Station, TX Teperature (degree Fareneit) 80 0 40 1 11 1 1 Day of October If we look closely at Figures 4. and 4.3 we notice tat te way we calculate te average rate of cange is te sae way we calculate te slope of a line. As a result we define te average rate of cange as follows. Average Rate of Cange Te average rate of cange between two points P1( x1, y 1) and P( x, y ) is cange in cange in y y y y = = x x x x 1 1 Te units for te average rate of cange are units of y units of x. Exaple 4.1 Te table below sows te ourly earnings (in dollars) for anufacturing plant eployees in te nited States fro te years 1997 to 001. Find te average rate of cange in te ourly earnings fro 1999 to 001. Year (x) 1997 1998 1999 000 001 Hourly Earnings (y) 13.17 13.49 13.90 14.37 14.83 Source: ttp://ftp.bls:gov/pub/suppl/epsit/ceseeb.txt Solution Te average rate of cange is y 14.83 13.9 0.93 = = = 0.45. x 001 1999 Terefore, te average ourly earnings were increasing at a rate of 4.5 cents per year between 1999 and 001. We can also find te average rate of cange over a given interval wen data is odeled by a function. Te average rate of cange between any two x values is found by calculating te slope of te secant line, a line tat intersects a curve at two points. Te secant line tat contains points P 1 and P is sown in Figure 4.4 below.
Figure 4.4 Te secant line, or average rate of cange, troug points P 1 and P. g(x) secant line P 1 P Exaple 4. Te grap in Figure 4.5 below represents te nuber of eters an object is above ground after t seconds ave elapsed. Find te average rate of cange between P 1 (3, 5) and P (8, 18). Figure 4.5 Grap of te disce an object traveled. Meters traveled P 1 P Tie elapsed (in seconds) Solution Te point P 1 (3, 5) eans tat after 3 seconds te object was 5 eters above te ground and P (8, 18) eans tat after 8 seconds te object was 18 eters above te ground. Te average rate of cange is te slope of te secant line containing P 1 and P. 5 18 7 = = 1.4. 3 8 5 Tis tells us tat between te 3rd and 8t second, te object was falling at an average rate of 1.4 eters per second. Exaple 4.3 Te nuber of annual layoff events tat occurred in te nited States fro 199 to 4 3 001 can be odeled by f ( x) = 0.0x 0.8x + 3.x 5.x+ 8.4, 1 x were x is te nuber of years since 199 and f(x) is te nuber of tousand of events. Find te average rate of cange in layoff events fro 199 to 001. (Source: ttp://data.bls.gov/servlet/surveyoutputservlet) Solution Since 199 corresponds to x = 1 and 001 corresponds to x =, te nuber of tousands of layoff events in 199 is f (1) and te nuber of tousands of layoff events in 001 is f (). Figure
4. sows a grap of f( x ) and te secant line tat contains te points at x = 1 and x =. Figure 4. Grap of f(x) and te secant line tat contains te points at x = 1 and x =. Nuber of Layoff Events (in tousands) 10 8 4 Layoff Events in.s. (1, 5.74) (, 7.31) 0 1 1 3 3 4 4 5 5 years since 1995 Te slope of te secant line is y f() f (1) 7.31 5.74 1.57 = = = = 0.314 x 1 1 5 Tis eans tat te nuber of layoff events was increasing at a rate of 0.314 tousand, or 314, events per year fro 199 to 001. It is useful to derive a general forula tat will calculate te slope of te secant line for any function f(x). To do tis let a and a+ be two points on a continuous function f(x) suc tat a < a+ as sown in Figure 4.7 below. (Notice tat is te disce between te two points a and a+.) Figure 4.7 Secant line on f( x ) f (a + ) f (a) a a + Te slope of te secant line tat contains (a, f(a)) and ( a+, f( a+ ) ) is Te forula above is known as te difference quotient. sec f ( a+ ) f ( a) f ( a+ ) f ( a) = =. a+ a
Exaple 4.4 Te 30-year ortgage rates during te ont of October 00 can be odeled by 3 f ( x) = 0.00017x + 0.0075x 0.07x+ 5.8 were x represents te day in October and f(x) represents te interest rate. se te difference quotient to calculate te average rate of cange in 30 year fixed ortgage rates fro October 15t to October 1st. Solution Since October 15t is represented by x = 15 and October 1st is represented by x = 1 we know tat = 1 15=. sing te difference quotient to find te slope of te secant line we get sec f ( a+ ) f ( a) f(15+ ) f(15) f (1) f (15).11 5.9088 = = = = = 0.03 So between October 15t and October 1st, te interest rates were canging at an average rate of 0.03 percent per day as sown in Figure 4.8 below. Figure 4.8 Average rate of cange of fixed ortgage rates fro te 15t to te 0t of October. 30 Year Fixed Mortgage Rates in October 00 Interest Rate..1 5.9 5.8 5.7 5. 0 5 10 15 0 5 30 Day in October (Source ttp://www.bankrate.co/ust/suboe/tg_1.asp) 4. Inseous Rate of Cange Te average rate of cange is a good calculation to use if we are looking for te rate of cange over an interval. If, owever, we want to find ow te interest rates were canging on te 0t of October, calculating te slope of te secant line is no longer possible because te 0t of October is represented by a single point. Tus, we need te slope of te line tat touces te grap at x = 0. Tis type of line is called a gent line. We can find an approxiation of te slope of te gent line by calculating te slopes of secant lines tat are close to x = 0 as sown in Figure 4.9. Figure 4.9 Slopes of secant lines were point b is getting closer and closer to point a ( is getting saller).
(Source ttp://www.bankrate.co/ust/suboe/tg_1.asp) 30 Year Fixed Mortgage Rates in October 00 Interest Rate. 5.8 5. 0 10 0 30 a b Day in October 30 Year Fixed Mortgage Rates in October 00 30 Year Fixed Mortgage Rates in October 00. a b. a b Interest Rate 5.8 5. 0 10 0 30 Interest Rate 5.8 5. 0 10 0 30 Day in October Day in October If we zoo in closer to point a, we can see, as sown in Figure 4.10 below, tat te slope of te secant lines are approxiately te sae as te slope of gent line at a. Figure 4.10 Graps of te secant lines wen we zoo in around point a and te grap of te gent line at point a. 30 Year Fixed Mortgage Rates in October 00 30 Year Fixed Mortgage Rates in October 00 Interest Rate.18 b.14 a.1.0.0 18 0 4 Day in October Interest Rate.18.14.1.0.0 a b 18 0 4 Day in October 30 Year Fixed Mortgage Rates in October 00 Interest Rate.18.14 a.1.0.0 18 0 4 Day in October
As te disce between a and b decreases, tat is as approaces zero ( 0 ), te slopes of te secant lines approac te slope of te line gent at x = a, as sown above in Figures 4.9 and 4.10. Te slope of te gent line at x = a gives te rate of cange in te ortgage rates at te inst x = a, terefore we call te slope of te gent line te inseous rate of cange of f(x) at x = a. Tis leads us to te following forula. If f( x ) is a continuous function, te rate of cange at a single point x = a, coonly called te inseous rate of cange at a or te slope of te line gent to f( x ) at a, can be found by coputing f '( a) = = li 0 f( a+ ) f( a) provided te liit exists. Note: Te notation f '( a ), read f prie of a, is coonly use to denote te inseous rate of cange of f( x ) at x = a. Exaple 4.5 Janice s Jewelry store as found tat te aount of profit er store akes eac day after Cristas can be odeled by Px ( ) = 4x + 40x 0 were x is te nuber of days after Deceber 5t and P(x) is te profit, in undreds of dollars. Find te inseous rate of cange in profit for Janice s Jewelry store for te following days and interpret eac answer. a. te 3rd day after Cristas. b. te t day after Cristas. Solution a. Te 3rd day after Cristas is represented by x = 3, tus we substitute a = 3 into te inseous rate of cange forula above we get = P'(3) = (3 ) (3) P + P li 0 It will be easiest if we first find P(3 + ). In doing so we get P (3 + ) = 4(3 + ) + 40(3 + ) 0= 4(9+ + ) + 10+ 40 0 = + + = + + 3 4 4 40 0 4 1 4 Now we need to find P (3) P (3) = 4(3) + 40(3) 0= 4(9) + 10 0= 4 Substituting tese expressions into te forula above we get P(3 + ) P(3) 4 + 1+ 4 4 ( 4+ 1) = li = li = li = li( 4+ 1) = 1 0 0 0 0
Tis eans tat te profit was increasing at a rate of $100 per day on te 3rd day after Cristas. Figure 4.11 below sows a grap of tis inseous rate of cange. Figure 4.11 A grap of te gent line, or inseous rate of cange, at x = 3, wic represents te tird day after Cristas. Profit of Janice's Jewlery Store Profit (in undreds of dollars) 40 0 0 0 4 8 10 nuer of days after 1/5 b. Te t day after Cristas is represented by x =, tus we substitute a = into te inseous rate of cange forula above we get First find P( + ). = P'() = ( ) () P + P li 0 P ( + ) = 4( + ) + 40( + ) 0= 4(3+ 1 + ) + 40+ 40 0 = + + = + 144 48 4 180 40 4 8 3 Ten find P (). Substituting tis into te forula above we get P () = 4() + 40() 0 = 4(3) + 40 0 = 3 P P = P'() = li = li 0 0 ( 4 8) = li = li( 4 8) = 8 0 0 ( + ) () 4 8 + 3 3 Tis eans tat te profit was decreasing at a rate of 8 undreds dollars per day on te t day after Cristas. A grap of tis is sown in Figure 4.1 below.
Figure 4.1 A grap of te gent line, or inseous rate of cange, at x =, wic represents te t day after Cristas. Profit of Janice's Jewlery Store Profit (in undreds of dollars) 40 0 0 0 4 8 10 nuer of days after 1/5 Exaple 4. Find te equation of te line gent to f( x) 1 = at x = 1. x Solution Te gent line will ave slope = f '(1) and will contain te point (1, f (1)) = (1,1). Tus, using te point-slope forula, te equation of te line gent to f( x) 1 = at x = 1 x will ave te for y f(1) = f '(1)( x 1). Te slope of te gent line is coputed as sown below. 1 1 1+ f (1 + ) f (1) 1 = f '(1) = li = li1+ = li1+ 1+ = li 1+ 0 0 0 0 1 1 = li = li = 1 01+ 01+ Te equation of te gent line is y 1= 1( x 1) wic is te sae as y = x+. A grap of tis is sown below in Figure 4.13.
Figure 4.13 A grap of te line gent to f( x ) = 1 x at x = 1. 10 8 f () x 1 = x 4 0 0 0.5 1 1.5.5 3 Exaple 4.7 Find te equation of te line gent to f ( x) = x+ 1 at x = 3. Solution Te slope of te line gent to f( x ) is f(3 + ) f (3) 3+ + 1 4 + 4 = f '(3) = li = li = li 0 0 0 To find tis liit we ust ultiply te nuerator and denoinator by te conjugate of + 4 wic is + 4 +. + 4 + 4 + + 4 4 = li = li = li 0 + 4 + 0 0 + 4 + + 4 + 1 1 1 = li = = = 0.5 0 + 4 + + 4 Te equation of te line gent to f( x ) at x = 3 is y f (3) = f '(3)( x 3) y = 0.5( x 3) ( ) ( ) y = 0.5x 0.75+ = 0.5x+ 1.5 A grap of tis is sown in Figure 4.14 below.
Figure 4.14 A grap of te line gent to f ( x) = x+ 1 at x = 3. 3.5 f ( x) = x+ 3 1.5 1 0.5 0 0 1 3 4 5 Exaple 4.8 Arlen s Air Service provides airline service for private individuals. Te cost for 3 flying fro Austin, TX to Denver, CO can be odeled by Cx ( ) = x 1x + 3x+ 50 were x is te nuber of round trips ade and C( x ) is te cost of te trip in undreds of dollars. Arlen as found tat te inseous rate of cange of C( x ) at any point x = a is C'( a) = 3a 4a + 3. Find slope of te line gent to C( x ) at x = 8 and interpret your answer. Solution We are given found by evaluating C '(8). C'( a) = 3a 4a + 3, tus te slope of te line gent to C( x ) at x = 8 is 3 C'(8) 8 1(8) 3(8) 50 8 = = + + =. We conclude tat te cost for flying fro Austin to Denver on te 8t round trip was increasing at a rate of $800 per round trip. Exaple 4.9 In Figure 4.15 below, gx ( ) represents te nuber of units a copany sells in one ont and x represents te day of te ont. Deterine weter te slope of te gent lines at points a, b, and c are positive, negative, zero, or undefined. Interpret your answers. Figure 4.15 Grap for Exaple 4.9 Nuber of units sold a b c Day of Mont Solution Te gent lines to eac point are sown in Figure 4.1 below and we ake te following conclusions.
Te gent line at a is positive wic eans tat on te a t of te ont te rate of cange of sells was increasing. Te slope of te gent line at b is zero, terefore, on te b t day of te ont te rate of cange of sells was not canging. Te slope of te gent line at c is negative, terefore, on te c t day of te ont te rate of cange of sells were decreasing. Figure 4.1 Grap for Exaple 4.9 a b c d Saple Quiz Question 4.1 Te table below sows te nuber of uneployed Texans in te San Marcos\Austin area fro January 001 to Deceber 001. Let x represent te ont, wit x = 1 representing January, and y represent te nuber of uneployed Texans. Find te average rate of uneployent fro April to Noveber. Mont Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec Nuber of 1189 17970 0010 0907 517 34055 3457 35 31 35453 37097 3545 neployed Source: ttp://data.bls.gov/servlet/surveyoutputservlet
Question 4. Te given grap represents te annual percentage rates for a 1 year certificate of deposit (CD) fro Septeber 1, 00 to Noveber 1, 00. se te grap to find te average rate of cange for a 1 year CD fro P 1 (1,.37) to P (8,.1). Source: ttp://www.bankrate.co.ust/publ/3otrend.asp % rate for 1 yr CD P 1 P Nuber of days since 9/01/0 Question 4.3 Te nuber of annual layoff events tat occurred in te nited States fro 199 to 001 4 3 can be odeled by f ( x) = 0.0x 0.8x + 3.x 5.x+ 8.4, 1 x were x is te nuber of years since 199 and f( x ) is te nuber of tousand of events. Find te average rate of cange in layoff events fro 1997 to 000. Question 4.4 Given f( x ) = 3 x, find te inseous rate of cange of f( x ) at x =. Question 4.5 Given f ( x) = x, find te slope of te line gent to f( x ) at x = 3. Question 4. Mike s Mean Macine sop sells otorcycles and as deterined tat Rx ( ) = 0.0x + x odels te aount of revenue (in tousands of dollars) after selling x otorcycles. Find and interpret te inseous rate of cange of te sop s revenue at x =. Question 4.7 Find te equation of te line gent to f x = at x =. ( ) 3x 1 Question 4.8 Find te equation of te line gent to gx ( ) = x at x = 9. Question 4.9 Katryn s Pottery sop as deterined tat gx ( ) = 0.45x 10 odels te nuber of pieces of pottery se can ake in x ours. Find te slope of te line gent to gx ( ) at x = 1 and interpret your answer.
Question 4.10 se te grap below to deterine weter te slope of te gent lines at points a, b, and c are positive, negative, zero, or undefined. a c b