Net Change and Displacement

Transcription

1 mth 11, pplictions motion: velocity nd net chnge 1 Net Chnge nd Displcement We hve seen tht the definite integrl f (x) dx mesures the net re under the curve y f (x) on the intervl [, b] Any prt of the region below the x-xis contributes negtively-signed re to this net clcultion To find the totl re enclosed by f on [, b], one needs to evlute the definite integrl of the bsolute vlue of f (x): f (x) dx We cn pply this ide to other contexts In Clculus I we interpreted the first nd second derivtives s velocity nd ccelertion in the context of motion In prticulr, l s(t) position t time t s (t) v(t) velocity t time t s (t) v (t) (t) ccelertion t time t Reversing our point of view nd using ntiderivtives, since position s(t) is n ntiderivtive of veloctty v(t), by the FTC (II) we hve v(t) dt s(t) b s(b) s() But s(b) s() represents the net distnce trveled or displcement of the object during the time intervl becuse it is the difference in loctions t the end nd strt of the time period If we wnted the totl distnce trveled rther thn the net distnce trveled, just s with the re problem, we would compute the integrl of v(t) insted Tht is, totl distnce trveled This cn be summrized s follows v(t) dt DEFINITION 611 (Displcement versus Distnce) Assume tht the position of n object moving long stright line t time t is denoted by s(t) reltive to the origing nd tht its velocity is denoted by v(t) Then 1 The displcement or net chnge in position of the object between times t nd t b > is given by s(b) s() v(t) dt The totl distnce trveled by the object between times t nd t b > is given by v(t) dt EXAMPLE 61 Suppose n object moves with velocity v(t) t 1t + 16 km/hr long stright rod (1) Determine the displcement of the object on the time intervl [1, ] nd [, ] nd interpret your nswer () Determine the distnce trvelled on the time intervl [, ] Solution NetChngetex Version: Mitchell-15/9/7::5

2 mth 11, pplictions motion: velocity nd net chnge (1) The displcement is esy to clculte: For the intervl [, ], On [, ], Displcement on [, ] s() s() On [1, ], s() s(1) 1 t 1t + 16 dt t 6t + 16t t 1t + 16 dt t 6t + 16t ( ) ( ) () The distnce trvelled is hrder to determine since we need to integrte v(t) We must first determine where v(t) is positive nd negtive t 1t + 16 (t 6t + 8) (t )(t ) t or t The number line to the right shows tht t 1t + 16 only [, ] We cn now find the distnce trvelled (totl re) by splitting the intervl into two pieces [, ] nd [, ], chnging the sign of v(t) on the second piece to obtin the bsolute vlue of v(t) Dist Trv Totl Are t 1t + 16 dt [ t t 1t + 16 dt t 1t + 16 dt ] [ t ] 6t + 16t 6t + 16t EXAMPLE 61 Suppose n object moves with velocity t 5t + t m/s (1) Determine the displcement of the object on the time intervl [, 6] nd interpret your nswer () Determine the distnce trvelled on [, 6] t 1t Figure 61: The distnce trvelled on [, ] is the re under the bsolute vlue of the velocity curve Solution (1) For displcement on [, 6], s(6) s() 6 t 5t + t dt t 5t + t 6 ( 6 + 7) 6 () For the distnce trvelled we must first determine where v(t) is positive nd negtive t 5t + t t(t 5t + ) t(t 1)(t ) t, 1, t 5t + t The number line to the right shows tht t 5t + t only [1, ] We cn now find the distnce trvelled (totl re) by splitting the intervl into three pieces [, 1], [1, ] nd [, 6], chnging the sign of v(t) on the second piece to NetChngetex Version: Mitchell-15/9/7::5

3 mth 11, pplictions motion: velocity nd net chnge obtin the bsolute vlue of v(t) 6 Dist Trv [ t t 1t + 16 dt 1 t 5t + t dt 1 ] [ 1 t 5t + t t 5t + t dt + ] 5t + t 1 + t 5t + t dt [ t ] 5t + 6 t 5 1 Future Positions Suppose tht we know the velocity of n object moving long stright line is v(x) nd we know its position s() t some time t [Note: Often we know the initil position s()] We cn determine the future position t time generl time t using the displcement eqution Since s(x) is n ntiderivtive of v(x), gin FTC (II) tells us tht v(x) dx s(x) s(t) s() Solving for the position s(t), we see THEOREM 61 (Position from Velocity) Position t time t s(t) s() + v(x) dx In similr wy, if (t) represents ccelertion of the object, velocity t time t v(t) v() + (x) dx t EXAMPLE 615 Suppose tht the ccelertion of n object is given by (x) cos x for x with Find s(t) v() 1 s() Solution First find v(t) using Theorem 61 v(t) v() + (x) dx 1 + cos x dt 1 + [x sin x] t 1 + (t sin t) () 1 + t sin t Now solve for s(t) by using Theorem 61 s(t) s() + v(x) dx x sin x dx + (x + x + cos x) t + (t + t + cos t) ( + + 1) + t + t + cos t EXAMPLE 616 If ccelertion is given by (t) 1 + t t, find the exct position function, if s() 1 nd s() 11 Solution This time we don t hve velocity t time So let v() v be some unknown constnt We will see if we cn work it out lter Then v(t) v() (x) dx v x x dx v + 1x + x x t v + 1t + t t NetChngetex Version: Mitchell-15/9/7::5

4 mth 11, pplictions motion: velocity nd net chnge Now s(t) s() + v(x) dx 1 + v + 1x + x x + c dx 1 + (v x + 5x + 1 x 1 x ) t 1 + v t + 5t + 1 t 1 t So s(t) 1 + v t + 5t + 1 t 1 t To solve for v, evlute t s(t) t t s() 1 + v so v 1 v 5 Thus, s(t) 1 5t + 5t + 1 t 1 t Constnt Accelertion: Grvity In mny motion problems the ccelertion is constnt This hppens when n object is thrown or dropped nd the only ccelertion is due to grvity In such sitution we hve (t), constnt ccelertion with initil velocity v() v nd initil position s() s Then using Theorem 61 So Next, v(t) v() + (x) dx v + s(t) s() + Therefore v(x) dx s + ds v + x t v + (t ) t + v v(t) t + v s + v ds s + ( 1 x + v x) t s + ( 1 t + v t ) 1 t + v t + s s(t) 1 t + v t + s EXAMPLE 617 Suppose bll is thrown with initil velocity 96 ft/s from roof top feet high The ccelertion due to grvity is constnt (t) ft/s Find v(t) nd s(t) Then find the mximum height of the bll nd the time when the bll hits the ground Solution Recognizing tht v 96 nd s nd tht the ccelertion is constnt, we my use the generl formuls we just developed v(t) t + v t + 96 nd s(t) 1 t + v t + s 16t + 96t + The mx height occurs when the velocity is (when the bll stops rising): v(t) t + 96 t s() ft The bll hits the ground when s(t) s(t) 16t + 96t + 16(t 6t 7) 16(t 9)(t + ) So t 9 only (since t does not mke sense) NetChngetex Version: Mitchell-15/9/7::5

5 mth 11, pplictions motion: velocity nd net chnge 5 EXAMPLE 618 A person drops stone from bridge Wht is the height (in feet) of the bridge if the person hers the splsh 5 seconds fter dropping it? Solution Here s wht we know v (dropped) nd s(5) (hits wter) And we know ccelertion is constnt, ft/s We wnt to find the height of the bridge, which is just s Use our constnt ccelertion motion formuls to solve for v(t) t + v t nd s(t) 1 t + v t + s 16t + s Now we use the position we know: s(5) s(5) 16(5) + s s ft Notice tht we did not need to use the velocity function YOU TRY IT 61 (Extr Credit) In the previous problem we did not tke into ccount tht sound does not trvel instntneously in your clcultion bove Assume tht sound trvels t 11 ft/s Wht is the height (in feet) of the bridge if the person hers the splsh 5 seconds fter dropping it? Check on your nswer: Should the bridge be higher or lower thn in the preceding exmple? Why? EXAMPLE 619 Here s vrition This time we will use metric units Suppose bll is thrown with unknown initil velocity v m/s from roof top 9 meters high nd the position of the bll t time t is s() The ccelertion due to grvity is constnt (t) 98 m/s Find v(t) nd s(t) Solution This time v is unknown but s 9 nd s() Agin the ccelertion is constnt so we my use the generl formuls for this sitution v(t) t + v 98t + v nd But we know tht which mens s(t) 1 t + v t + s 9t + v t + 9 s() 9() + v + 9 v 9(9) 9(1) 9 v 9/ So nd Interpret v 9/ v(t) 98t 9 s(t) 9t 9 t + 9 More on Net Chnge nd Future Vlues We ve interpreted net re s displcement nd totl re s totl distnce when working with velocity function But we cn pply these sme ides nytime we hve rte of chnge function Exmples include when f (t) is flow rte of liquid (in which cse we cn compute net chnge in volume), or f (t) is popultion growth rte (in which cse we cn compute future popultion estimtes), or p(t) is NetChngetex Version: Mitchell-15/9/7::5

6 mth 11, pplictions motion: velocity nd net chnge 6 the growth of finncil ccount (in which cse we might compute the net chnge in vlue of the ccount) In generl, suppose we know the rte of chnge in some quntity is given by Q (x) on the intervl [, t] Then by integrting using FTC II nd the fct tht Q(x) is n ntiderivtive of Q (x), we get Q (x) dx Q(x) t Q(t) Q() Agin, s in Theorem 61, we cn rerrnge terms to get THEOREM 611 (Net Chnge nd Future Vlue) Assume tht quntity Q chnges over time t know rte Q Then the net chnge in Q between t nd t b > is Q(b) Q() Q (t) dt Further, if Q() is the initil vlue, then the future vlue of Q t time t is Q(t) Q() + Q (x) dx EXAMPLE 6111 (Popultion Growth) Suppose rfter of wild turkeys in Genev hs n initil vlue of P() nd the community grows t rte of P (t) t, where time is mesured in yers Determine the popultion in yers Then find the generl formul for the popultion, P(t) A group of turkeys is clled rfter A group of turtledoves is clled pitying Solution By the second prt of Theorem 611, 6 P() P() + P (t) dt + 15 t dt ( ) + t 6t/ More generlly, the popultion t time t is P(t) P() + P (x) dx + 15 x dx ( ) + x 6x/ + (8 16) () 87 t + t 6t/ EXAMPLE 611 (Svings) Grndprents of newborn deposit $1, in college svings ccount tht hs growth rte of 555e 55t How much will be vilble in the ccount when the child is 18 nd strting college? Solution By the second prt of Theorem 611, Q(18) Q() + Q (t) dt 1, + 555e 555t dt 1, e555t 18 1, + 1, e 555t 18 Not bd, but not worth semester t HWS 1, + (1, e 999 1, ) 7, NetChngetex Version: Mitchell-15/9/7::5

7 mth 11, pplictions motion: velocity nd net chnge 7 EXAMPLE 611 (Economics) The mrginl cost of product is dditionl the cost incurred by producing one more item of the product Typiclly, s production increses, the mrginl cost decreses, t lest up to point This is wht people men by the term "economy of scle" Mrginl cost is pproximted by the derivtive of the cost function C (x) tht depends on the number of units x being produced Suppose tht the mrginl cost is given by the function C (x) + 1x 1x Find the dditionl cost incurred when production rises from 5 to 55 units Then find the cost incurred when production rises from 55 to 6 units Solution By the first prt of Theorem 611, C(55) C(5) For the second prt C(6) C(55) C (x) dx C (x) dx x 1x dx x + 5x 1x 55 5 ((55) + 5(55) 1(55) 19, x 1x dx x + 5x 1x 6 55 ((6) + 5(6) 1(6) 17, 8 ) ) Notice tht the cost mking 5 more units hs decresed, illustrting the ide tht mrginl cost decreses s production increses ) ((5) + 5(5) 1(5) ) ((55) + 5(55) 1(55) NetChngetex Version: Mitchell-15/9/7::5

8 mth 11, pplictions motion: velocity nd net chnge 8 For Fun: Additionl Motion Problems with Constnt Accelertion The following problems ll involve motion with constnt ccelertion When ccelertion is constnt, we cn use the equtions we developed few pges erlier But remember, when ccelertion is not constnt, be sure to use Theorem 61 EXAMPLE 611 Mo Green is ttempting to run the 1m dsh in the Genev Invittionl Trck Meet in 98 seconds He wnts to run in wy tht his ccelertion is constnt,, over the entire rce Determine his velocity function ( will still pper s n unknown constnt) Determine his position function There should be no unknown constnts in your eqution t this point Wht is his velocity t the end of the rce? Do you think this is relistic? Solution We hve: constnt ccelertion m/s ; v m/s; s m So v(t) t + v t nd s(t) 1 t + v t + s 1 t But s(98) 1 (98) 1, so (98) 85 m/s So s(t) 85t Mo s velocity t the end of the rce is v(98) 98 85(98) 1 m/s not relistic EXAMPLE 6115 A stone dropped off cliff hits the ground with speed of 1 ft/s Wht ws the height of the cliff? Solution Notice tht v (dropped!) nd s is unknown but is equl to the cliff height, nd tht the ccelertion is constnt ft/ Use the generl formuls for motion with constnt ccelertion: v(t) t + v t + t Now we use the velocity function nd the one velocity vlue we know: v 1 when it hits the ground So the time when it hits the ground is given by v(t) t 1 t 1/ 15/ when it hits the ground Now remember when it hits the ground the height is So s(15/) But we know s(t) 1 t + v t + s 16t + t + s 16t + s Now substitute in t 15/ nd solve for s The cliff height is 5 feet s(15/) 16(15/) + s s 15 5 EXAMPLE 6116 A cr is trveling t 9 km/h when the driver sees deer 75 m hed nd slms on the brkes Wht constnt decelertion is required to void hitting Bmbi? [Note: First convert 9 km/h to m/s] Solution Let s list ll tht we know v 9 km/h or m/s nd s Let time t represent the time it tkes to stop Then s(t ) 75 m Now the cr is stopped t time t, so we know v(t ) Finlly we know tht ccelertion is n unknown constnt,, which is wht we wnt to find NetChngetex Version: Mitchell-15/9/7::5

9 mth 11, pplictions motion: velocity nd net chnge 9 Now we use our constnt ccelertion motion formuls to solve for v(t) t + v t + 5 nd s(t) 1 t + v t + s 1 t + 5t Now use the other velocity nd position we know: v(t ) nd s(t ) 75 when the cr stops So v(t ) t + 5 t 5/ nd s(t ) 1 (t ) + 5t 1 ( 5/) + 5( 5/) 75 Simplify to get so m/s (Why is ccelertion negtive?) EXAMPLE 6117 One cr intends to pss nother on bck rod Wht constnt ccelertion is required to increse the speed of cr from mph ( ft/s) to 5 mph ( ft/s) in 5 seconds? Solution Given: (t) constnt v ft/s s And v(5) ft/s Find But v(t) t + v t + So Thus v(5) YOU TRY IT 6 A toy bumper cr is moving bck nd forth long stright trck Its ccelertion is (t) cos t + sin t Find the prticulr velocity nd position functions given tht v(π/) nd s(π) 1 + c c Thus, v(t) sin t cos t Now s(t) v(t) dt sin t cos t dt cos t sin t + c Since s(π) ( 1) + c 1 c So s(t) cos t sin t nswer to you try it 6 v(t) (t) dt cos t + sin t dt sin t cos t + c So v(π/) webwork: Click to try Problems 71 through 7 Use guest login, if not in my course NetChngetex Version: Mitchell-15/9/7::5