Introduction. Calculus I. Calculus II: The Area Problem


 Morgan Jenkins
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1 Introuction Clculus I Clculus I h s its theme the slope problem How o we mke sense of the notion of slope for curves when we only know wht the slope of line mens? The nswer, of course, ws the to efine n etermine the erivtive of the curve (function) The reson tht this purely geometric question is interesting n importnt (for nonmthemticins) is tht we cn interpret mny other questions s slope problems For exmple, velocity, ccelertion, mrginl cost, mrginl profit, or ny rte of chnge is relly slope question in isguise Aitionlly points long curve where the slope is 0 re criticl becuse they re potentil extreme points for the curve, plces where the function obtins its mximum or minimum vlues This cn be pplie to whole host of problems For exmple, wht tuition shoul HWS chrge you to mximize its revenue? (, f ()) tngent Figure : If f is ifferentible t, then then the secnt lines through the points (, f ()) n ( + h, f ( + h)) pproch the tngent line t (, f ()) secnts Clculus II: The Are Problem Clculus II hs its own theme which lso consists of geometric question In its simplest form we cn stte it this wy: The Are Problem Let f be continuous (nonnegtive) function on the close intervl [, b] Fin the re boune bove by f (x), below by the xxis, n by the verticl lines x = n x = b For exmple, if we cn solve the generl re problem, we will be ble to fin the re of circle Yes, we know it is πr 2 But why is it πr 2?! Eventully we will be ble to nswer this question While the re problem is interesting in its own right, it is lso importnt becuse we cn interpret number of problems s re problems in isguise Other geometricl quntities such s volume or length of curve cn be interprete s re problems Problems such s istnce trvelle or work one (in the physics sense), n clcultion of probbilities cn lso be interprete s questions bout res We ll lso see tht fining the verge (or men) vlue of continuous function (like temperture) epens on solving n re problem We will strt to nswer these questions in couple of clsses For now, we nee to quickly review the prerequisites for the course y = f (x) R f b Figure 2: Fin the re R f uner nonnegtive continuous curve on the intervl [, b]
2 Preliminries n Prerequisites You shoul be fmilir with ll the mteril in this chpter (Chpter 0) from your previous clculus experience We will not spen ny substntil mount of time reviewing this mteril Note: There is lot more from Clculus I tht I ssume you know (eg, how to grph functions n fin their extrem) The mteril in Chpter 0 is not the only mteril tht I expect you to know 0 The Derivtive The erivtive is the centrl concept of Clculus I You shoul be very fmilir with both the erivtive rules n the efinition of the erivtive The Definition of the Derivtive DEFINITION 0 Let f be function efine in n open intervl contining x Then the erivtive f (x) is efine by f f (x + h) f (x) (x) = lim h 0 h if the limit exists When the limit oes exist, we sy tht f is ifferentible t x You my be fmilir with n lterntive efinition for the erivtive t point which uses slightly ifferent nottion f () = lim x f (x) f () x The erivtive represents n instntneous rte of chnge Geometriclly the erivtive cn be interprete s the slope of the function t the point in question (, f ()) tngent Figure 3: If f is ifferentible t, then then the secnt lines through the points (, f ()) n ( + h, f ( + h)) pproch the tngent line t (, f ()) secnts Derivtive Formulæ These bsic erivtive expressions shoul be fmilir In the lst few, ssume both f n g re ifferentible functions n tht is constnt
3 mth 3, prerequisites, prt one preliminries 3 Differentition Formulæ x (c) = 0 x (xn ) = nx n x (sin x) = cos x x (tn x) = sec2 x x (ex ) = e x x (rctn x ) = 2 +x 2 x [c f (x)] = c f (x) x [ ] f (x) = f (x)g(x) f (x)g (x) g(x) (g(x)) 2 x (kx) = k x (ln x ) = x x (cos x) = sin x x (sec x) = sec x tn x x (rcsin x ) = 2 x 2 x [ f (x) ± g(x)] = f (x) ± g (x) x [ f (x)g(x)] = f (x)g(x) + f (x)g (x) x [ f (g(x))] = f (g(x))g (x) Stop! Review YOU TRY IT 0 The theme of Clculus I ws ifferentition Stte the erivtives of ech of these functions Answers to you try it 0 : () x n (b) sin x (c) cos x () tn x (e) sec x (f ) c (g) 2e x (h) ln x (i) x n (j) x (k) n n x (l) x m (m) x 6 sin x (n) cos(6x) + (o) e x tn(4x) (p) 7e sec x (q) ln(2x 4 ) (r) rctn(x 2 ) (s) x2 + e 2x (t) sin 2 x YOU TRY IT 02 Stte the erivtives of ech of these functions () f (x) = cos 4 x (b) g(x) = xe sin x (c) g(x) = ln(6x 4 ) 02 The Men Vlue Theorem The Men Vlue Theorem is one of the most importnt theorems in elementry clculus It reltes the globl behvior of function (how it chnges over n entire intervl) to the locl behvior of function (the erivtive of the function t prticulr point) The MVT is use to prove number of importnt results in clculus For exmple, it is use to prove the first erivtive test: If f (x) > 0, then f is incresing You shoul be ble to stte the MVT n rw grph tht illustrtes it THEOREM 0 (MVT: The Men Vlue Theorem) Assume tht f is continuous on the close intervl [, b]; 2 f is ifferentible on the open intervl (, b) Then there is some point c between n b so tht f (c) = f (b) f () b This is equivlent to sying f (b) f () = f (c)(b ) () nx n (b) cos x (c) sin x () sec 2 x (e) sec x tn x (f ) 0 (g) 2e x (h) 6x (i) nx n (k) n x n n (l) m n x m n n (m) 6x 5 sin x + x 6 cos x (o) e x tn(4x) + e x (4 sec 2 4x) (p) 7e sec x sec x tn x (q) (n) 6 sin(6x) 4 x (r) 2x + x 4 (s) 2xe2x (x 2 + )2e 2x e 4x = 2x 2(x2 + ) e 2x (t) 2 sin x cos x Answers to you try it 02 (remember to simplify): () f (x) = 4 sin x cos 3 x (b) g (x) = ( + x cos x)e sin x (c) g (x) = 6x 4 24x3 = 4x tngent c b Figure 4: Prllel secnt n tngent lines exist when the Men Vlue Theorem pplies secnt
4 mth 3, prerequisites, prt one preliminries 4 Wht the MVT Sys Let s interpret the MVT geometriclly The verge or men rte of chnge of f on the intervl [, b] is just y x f (b) f () = b This is the fmilir secnt slope of the line through (, f ()) n (b, f (b)) The MVT sys if f is ifferentible, we cn fin point c between n b so tht the instntneous rte of chnge or tngent slope or the erivtive f (c) is the sme s the secnt slope When two lines (here the secnt n tngent) hve the sme slope, they re prllel (See Figure 4 bove) If we think of the erivtive s velocity, then the MVT sys tht if you clculte your verge velocity over time intervl, then t some point uring the intervl your instntneous velocity ctully equls your verge velocity Mostly the MVT gets use to prove other theorems But we cn look t n exmple or two to see how it works EXAMPLE 0 Show how the MVT pplies to f (x) = x 3 6x + on [0, 3] SOLUTION Check the two conitions (hypotheses) f is continuous on the close intervl [0, 3] becuse it is polynomil; 2 f is ifferentible on the open intervl (0, 3) gin becuse it is polynomil; So the MVT pplies: There is some point c between 0 n 3 so tht f (c) = f (3) f (0) 3 0 = 0 3 = 3 Since f (x) = 3x 2 6, then f (c) = 3c 2 6 = 3 3c 2 = 9 c = ± 3 Only c = 3 is in the intervl, so this is the vlue of c EXAMPLE 02 Show there oes not exist ifferentible function on [, 5] with f () = 3 n f (5) = 9 with f (x) 2 for ll x SOLUTION The MVT woul pply to such function f : So there shoul be some point c between n 5 so tht f (c) = f (5) f () 5 = 9 ( 3) 4 = 3 But supposely f (x) 2 for ll x Contriction So no such f cn exist EXAMPLE 03 It is lso importnt to unerstn why the hypotheses of the theorem re necessry Tke the function f (x) = x on the intervl [, ] Notice tht if we trie to pply the MVT here, the enpoints woul be = n b = So there shoul be point c between n so tht f (c) = f () f ( ) ( ) = + = 2 But you cn see from the grph of f (x) = x tht the slope is never 0; f (x) is either when x < 0 or when x > 0 The problem here is tht f (x) = x is NOT ifferentible t x = 0 x fils to stisfy the hypotheses of the MVT n for tht reson x oes not stisfy the conclusion of the MVT on the intervl [, ] = 0 secnt 0
5 mth 3, prerequisites, prt one preliminries 5 Using the MVT We will lmost never use the MVT s in the three exmples bove Tht oes not men you shoul not spen time unerstning these exples becuse they illustrte the key ies of the theorem However, the key vlue of the MVT is in proving other results We mentione erlier tht the MVT is use to prove tht functions with positive (negtive) erivtives re incresing (resp, ecresing) We will see further ppliction of the MVT when we we prove the Funmentl Theorem of Clculus relting ntiifferentition to the problem of fining the re uner curve 03 Antierivtives If you know the velocity of n object, cn you etermine the position of the object? This coul hppen in cr, sy, where the speeometer reings were being recore Cn the position of the cr be etermine from this informtion? Similrly, cn the position of n irplne be etermine from the blck box which recors the irspee? Remembering tht velocity is relly just erivtive, we cn sk this sme question more generlly: Given f (x) cn we fin the function f (x)? We usully stte the problem this wy DEFINITION 02 Let f (x) be function efine on n intervl I We sy tht F(x) is n ntierivtive of f (x) on I if F (x) = f (x) for ll x I EXAMPLE 04 If f (x) = 2x, then F(x) = x 2 is n ntierivtive of f becuse F (x) = 2x = f (x) But so is G(x) = x 2 + or, more generlly, H(x) = x 2 + c Are there other ntierivtives of f (x) = 2x besies those of the form H(x) = x 2 + c? We cn use the MVT to show tht the nswer is No The proof will require two steps THEOREM 02 If F (x) = 0 for ll x in n intervl I, then F(x) = k is constnt function This mkes lot of sense: If the velocity of n object is 0, then its position is constnt (not chnging) Here s the Proof To show tht F(x) is constnt, we must show tht ny two output vlues of F re the sme, ie, F() = F(b) for ll n b in I So pick ny n b in I (with < b) Becuse F is ifferentible on I, then F is both continuous n ifferentible on the smller intervl [, b] So the MVT pplies (Theorem 0 look t its equivlent sttement) There is point c between n b so tht F(b) F() = F (c)(b ) becuse we re given tht F is lwys 0 we cn substitute in n sy F(b) F() = 0(b ) = 0 Since F(b) F() = 0, this mens F(b) = F() In other wors, F is constnt THEOREM 03 If F(x) n G(x) re both ntierivtives of f (x) on n intervl I, then G(x) = F(x) + k Tht is, ny two ntierivtives of the sme function iffer by constnt
6 mth 3, prerequisites, prt one preliminries 6 Proof Assume F(x) n G(x) re both ntierivtives of f (x) on n intervl I Consier the function G(x) F(x) on I Then x [G(x) F(x)] = G (x) F (x) = f (x) f (x) = 0 Therefore, by Theorem 02 tht we just prove G(x) F(x) = k so G(x) = F(x) + k DEFINITION 03 If F(x) is ny ntierivtive of f (x), we sy tht F(x) + c is the generl ntierivtive of f (x) on I Nottion for Antierivtives Antiifferentition is lso clle inefinite integrtion f (x) x = F(x) + c is the integrtion symbol f (x) is clle the integrn x inictes the vrible of integrtion (here it is x) F(x) is prticulr ntierivtive of f (x) n c is the constnt of integrtion We refer to f (x) x s n ntierivtive of f (x) or n inefinite integrl of f Here re severl exmples cos t t = sin t + c e z z = e z + c x = rctn x + c + x2 Antiifferentition reverses ifferentition so F (x) x = F(x) + c s long s F (x) is continuous An ifferentition unoes ntiifferentition so [ f (x) x] = f (x) x Since ifferentition n ntiifferentition re reverse processes, ech erivtive rule hs corresponing ntiifferentition rule
7 mth 3, prerequisites, prt one preliminries 7 Differentition x (c) = 0 x (xn ) = nx n x (ln x ) = x x (sin x) = cos x x (cos x) = sin x x (tn x) = sec2 x x (sec x) = sec x tn x x (ex ) = e x x (rcsin x ) = 2 x 2 x (rctn x ) = 2 +x 2 Antiifferentition 0 x = c x n x = xn+ n+ + c, n = x x = x x = ln x + c cos x x = sin x + c sin x x = cos x + c sec 2 x x = tn x + c sec x tn x x = sec x + c e x x = ex + c 2 x 2 x = rcsin x + c 2 +x 2 x = rctn x + c 04 Generl Antierivtive Rules The key ie is tht ech erivtive rule cn be written s n ntierivtive rule We ve seen how this works with specific functions like sin x n e x n now we exmine how the generl erivtive rules cn be reverse THEOREM 04 (Sum Rule) The sum rule for erivtives sys (F(x) ± G(x)) = x x (F(x)) ± x (G(x)) The corresponing ntierivtive rule is ( f (x) ± g(x)) x = f (x) x ± g(x) x THEOREM 05 (Constnt Multiple Rule) The constnt multiple rule for erivtives sys x (cf(x)) = c x (F(x)) The corresponing ntierivtive rule is c f (x) x = c f (x) x Simple Exmples Be sure you unerstn how the bsic ntierivtive rules pply in ech of these problems EXAMPLE 05 Ech of these ntierivtives uses multiple rules Try to ientify them 8x 3 7 x x = 8x 3 x 7x /2 x = 8 x 3 x 7 x /2 x = 8x4 4 7x3/2 3/2 + c = 2x4 4x3/2 + c 3 6 cos 2x 7 x + 2x /3 x = 6 cos 2x x 7 x x + 2 x /3 x = 6 2x2/3 sin 2x 7 ln x + 2 2/3 + c = 3 sin 2x 7 ln x + 3x2/3 + c
8 mth 3, prerequisites, prt one preliminries 8 3e x/2 8 x = 3 6 x 2 e 2 x x x x = 3 e x/2 8 rcsin x c 2 = 6e x/2 8 rcsin x 4 + c YOU TRY IT 03 Determine ech of the ntierivtives of ech of the functions below x () + x 99 x (b) 5x /4 3e 2x x (c) 4 sec 2 x 3 + 3x 2 x Exmples with Rewriting Sometimes the ntiifferentition process is gretly simplifie by rewriting the integrn before ny ntiifferentition is ttempte This my involve rewriting powers s exponents, iviing out common fctors, or multiplying out proucts Answers to you try it 03 : () 2 3 x3/ x00 + c (b) 4x 5/ e 2x + c (c) 2 tn x 3 3x + c Integrl Rewritten Solution 5 4 t 2 t = 4 t 2/5 t = 20 7 t7/5 + c x 4 +2 x 2 x = x 2 + 2x 2 x = 3 x3 2x + c 6x 2 (x 4 ) x = 6x 6 6x 2 x = 6 7 x7 2x 3 + c x = 6x 4 6 x 4 x = 8 x 3 + c s s = 7 4 s /3 s = 2 8 s2/3 + c 4 x = 4 x = x x rctn x c = 3 4 rctn 3 x + c 2 x = 25 x 2 2 x = 5 2 x 2 2 rcsin 5 x + c 2 x+3 x x = 2 x /2 x + 3 x x = 4x /2 + 3 ln x + c YOU TRY IT 04 Determine these ntierivtives () 2 sec 3x tn 3x x (b) 7x 3 x 4 x x (c) 5 4 x 3 + 3x 2 x YOU TRY IT 05 Which generl rules hve we not yet reverse? Try to write own the corresponing ntierivtive rules similr to Theorem 04 n Theorem 05 Answers to you try it 04 : () 2 sec 3x + c 3 (b) 7x3 3 2x/2 4 ln x + c (c) 20x7/4 7 3x + c
9 mth 3, prerequisites, prt one preliminries 9 05 Evluting c (Initil Vlue Problems) In ech generl ntierivtive there is n unknown constnt c This is wht mkes the integrl inefinite If n object is moving long stright line, we know tht the erivtive of the position function s(t) is the velocity function s (t) = v(t) Wht this mens is if we know the velocity v(t) of the cr we re riving in (just look t the speeometer), we cn etermine the position s(t) of the cr, up to constnt c if we cn fin n ntierivtive for v(t) If we hve more informtion, the position of the cr t prticulr time sy, then we re ble to etermine the precise ntierivtive n exct position Let s see how this works in generl before we pply it to motion problems EXAMPLE 06 Suppose tht f (x) = e x + 2x n f (0) = 3 Fin f (x) In this context f (0) is sometimes clle the initil vlue n such questions re referre to s initil vlue problems [How coul you interpret this informtion in terms of motion of cr?] SOLUTION f (x) must be n ntierivtive of f (x) so f (x) = f (x) x = e x + 2x x = e x + x 2 + c Now use the initil vlue to solve for c: Therefore, f (x) = e x + x f (0) = e c = 3 + c = 3 c = 2 EXAMPLE 07 Suppose tht f (x) = 6x 2 2x 3 n f () = 4 Fin f (x) SOLUTION Agin f (x) must be n ntierivtive of f (x) so f (x) = f (x) x = 6x 2 2x 3 x = 2x 3 x4 2 + c Now use the initil vlue to solve for c: Therefore, f (x) = 2x 3 2 x f () = 2 /2 + c = 4 c = 5/2 EXAMPLE 08 Suppose tht f (t) = 6t 2 (think ccelertion) with f () = 8 (think velocity) n f () = 3 (think position) Fin f (t) SOLUTION First fin f (t) which is just the ntierivtive of f (t) So f (t) = f (t) t = 6t 2 t = 6t + c Now use the initil vlue for f (t) to solve for c: f () = 6() + c = 8 c = 4 Therefore, f (t) = 6t + 4 Now we re bck to the erlier problem f (t) = f (t) t = 6t + 4 t = 6 ln t + 4t + c Now use the initil vlue of f to solve for c: f () = 6 ln + 4() + c = 3 6(0) c = 3 c = So f (t) = 6 ln t + 4t
10 mth 3, prerequisites, prt one preliminries 0 More Prctice EXAMPLE 09 Fin f given tht f (x) = 6 x + 5x 3 2 where f () = 0 SOLUTION f (x) must be n ntierivtive of f (x) so f (x) = f (x) x = 6 x + 5x 3 2 x = 4x 3/2 + 2x 5/2 + c Use the initil vlue to solve for c: f () = c = 0 c = 4 Therefore, f (x) = 4x 3/2 + 2x 5/2 + 4 EXAMPLE 00 Fin f given tht f (θ) = sin θ + cos θ where f (0) = n f (0) = 2 SOLUTION First fin f (θ) which must be the ntierivtive of f (θ) So f (θ) = f (θ) θ = sin θ + cos θ θ = cos θ + sin θ + c Now use the initil vlue for f (θ) to solve for c: f (0) = cos 0 + sin 0 + c = c = c = 2 Therefore, f (θ) = cos θ + sin θ + 2 f (θ) = f (θ) θ = cos θ + sin θ + 2 θ = sin θ cos θ + 2θ + c Now use the initil vlue of f to solve for c: f (0) = sin 0 cos 0 + 2(0) + c = 0 + c = 2 c = 3 So f (θ) = sin θ cos θ + 2θ Motion Problems In Clculus I you interprete the first n secon erivtives s velocity n ccelertion in the context of motion So let s pply the initil vlue problem results to motion problems Recll tht s(t) = position t time t s (t) = v(t) = velocity t time t s (t) = v (t) = (t) = ccelertion t time t Therefore (t) t = v(t) + c = velocity v(t) t = s(t) + c 2 = position t time t We will nee to use itionl informtion to evlute the constnts c n c 2 EXAMPLE 0 Suppose tht the ccelertion of n object is given by (t) = 2 cos t for t 0 with v(0) =, this is lso enote v 0 s(0) = 3, this is lso enote s 0 Fin s(t)
11 mth 3, prerequisites, prt one preliminries SOLUTION First fin v(t) which is the ntierivtive of (t) v(t) = (t) t = 2 cos t t = 2t sin t + c Now use the initil vlue for v(t) to solve for c : v(0) = c = c = Therefore, v(t) = 2t sin t + Now solve for s(t) by tking the ntierivtive of v(t) s(t) = v(t) t = 2t sin t + t = t 2 + cos t + t + c 2 Now use the initil vlue of s to solve for c 2 : So s(t) = t 2 + cos t + 2t + 2 s(0) = 0 + cos 0 + c 2 = 3 + c 2 = 3 c 2 = 2 EXAMPLE 02 If ccelertion is given by (t) = 0 + 3t 3t 2, fin the exct position function if s(0) = n s(2) = SOLUTION First v(t) = (t) t = 0 + 3t 3t 2 t = 0t t2 t 3 + c Now s(t) = 0t t2 t 3 + c t = 5t t3 4 t4 + ct + But s(0) = = so = Then s(2) = c + = so 2c = 0 c = 5 Thus, s(t) = 5t t3 4 t4 5t + EXAMPLE 03 If ccelertion is given by (t) = sin t + cos t, fin the position function if s(0) = n s(2π) = SOLUTION First v(t) = (t) t = sin t + cos t t = cos t + sin t + c Now s(t) = cos t + sin t + c t = sin t cos t + ct + But s(0) = = so = 2 Then s(2π) = 0 + 2cπ + 2 = so 2πc = 2 c = π + 2 Thus, s(t) = cos t + sin t π t 07 Constnt Accelertion: Grvity In mny motion problems the ccelertion is constnt This hppens when n object is thrown or roppe n the only ccelertion is ue to grvity In such sitution we hve (t) =, constnt ccelertion with initil velocity v(0) = v 0 n initil position s(0) = s 0 Then v(t) = (t) t = t = t + c But v(0) = 0 + c = v 0 c = v 0
12 mth 3, prerequisites, prt one preliminries 2 So v(t) = t + v 0 Next, s(t) = v(t) t = t + v 0 t = 2 t2 + v 0 t + c At time t = 0, Therefore s(0) = 2 (0)2 + v 0 (0) + c = s 0 c = s 0 s(t) = 2 t2 + v 0 t + s 0 EXAMPLE 04 Suppose bll is thrown with initil velocity 96 ft/s from roof top 432 feet high The ccelertion ue to grvity is constnt (t) = 32 ft/s 2 Fin v(t) n s(t) Then fin the mximum height of the bll n the time when the bll hits the groun SOLUTION Recognizing tht v 0 = 96 n s 0 = 432 n tht the ccelertion is constnt, we my use the generl formuls we just evelope v(t) = t + v 0 = 32t + 96 n s(t) = 2 t2 + v 0 t + s 0 = 6t t The mx height occurs when the velocity is 0 (when the bll stops rising): v(t) = 32t + 96 = 0 t = 3 s(3) = = 576 ft The bll hits the groun when s(t) = 0 s(t) = 6t t = 6(t 2 6t 27) = 6(t 9)(t + 3) = 0 So t = 9 only (since t = 3 oes not mke sense) EXAMPLE 05 A person rops stone from brige Wht is the height (in feet) of the brige if the person hers the splsh 5 secons fter ropping it? SOLUTION Here s wht we know v 0 = 0 (roppe) n s(5) = 0 (hits wter) An we know ccelertion is constnt, = 32 ft/s 2 We wnt to fin the height of the brige, which is just s 0 Use our constnt ccelertion motion formuls to solve for v(t) = t + v 0 = 32t n s(t) = 2 t2 + v 0 t + s 0 = 6t 2 + s 0 Now we use the position we know: s(5) = 0 s(5) = 6(5) 2 + s 0 s 0 = 400 ft Notice tht we i not nee to use the velocity function YOU TRY IT 06 (Extr Creit) In the previous problem we i not tke into ccount tht soun oes not trvel instntneously in your clcultion bove Assume tht soun trvels t 20 ft/s Wht is the height (in feet) of the brige if the person hers the splsh 5 secons fter ropping it? Check on your nswer: Shoul the brige be higher or lower thn in the preceing exmple? Why? EXAMPLE 06 Here s vrition This time we will use metric units Suppose bll is thrown with unknown initil velocity v 0 m/s from roof top 49 meters high n the position of the bll t time t = 3 is s(3) = 0 The ccelertion ue to grvity is constnt (t) = 98 m/s 2 Fin v(t) n s(t)
13 mth 3, prerequisites, prt one preliminries 3 SOLUTION This time v 0 is unknown but s 0 = 49 n s(3) = 0 Agin the ccelertion is constnt so we my use the generl formuls for this sitution v(t) = t + v 0 = 98t + v 0 n But we know tht which mens s(t) = 2 t2 + v 0 t + s 0 = 49t 2 + v 0 t + 49 s(3) = 49(3) 2 + v = 0 3v 0 = 49(9) 49(0) = 49 v 0 = 49/3 So n Interpret v 0 = 49/3 v(t) = 98t s(t) = 49t t + 49 EXAMPLE 07 Mo Green is ttempting to run the 00m sh in the Genev Invittionl Trck Meet in 98 secons 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 shoul be no unknown constnts in your eqution t this point Wht is his velocity t the en of the rce? Do you think this is relistic? SOLUTION We hve: constnt ccelertion = m/s 2 ; v 0 = 0 m/s; s 0 = 0 m So n v(t) = t + v 0 = t s(t) = 2 t2 + v 0 t + s 0 = 2 t2 But s(98) = 2 (98)2 = 00, so = 200 (98) 2 = m/s 2 So s(t) = 20825t 2 Mo s velocity t the en of the rce is v(98) = 98 = 20825(98) = 204 m/s not relistic EXAMPLE 08 A stone roppe off cliff hits the groun with spee of 20 ft/s Wht ws the height of the cliff? SOLUTION Notice tht v 0 = 0 (roppe!) n s 0 is unknown but is equl to the cliff height, n tht the ccelertion is constnt = 32 ft/ Use the generl formuls for motion with constnt ccelertion: v(t) = t + v 0 = 32t + 0 = 32t Now we use the velocity function n the one velocity vlue we know: v = 20 when it hits the groun So the time when it hits the groun is given by v(t) = 32t = 20 t = 20/32 = 5/4 when it hits the groun Now remember when it hits the groun the height is 0 So s(5/4) = 0 But we know s(t) = 2 t2 + v 0 t + s 0 = 6t 2 + 0t + s 0 = 6t 2 + s 0 Now substitute in t = 5/4 n solve for s 0 s(5/4) = 0 6(5/4) 2 + s 0 = 0 s 0 = 5 2 = 225 The cliff height is 225 feet
14 mth 3, prerequisites, prt one preliminries 4 EXAMPLE 09 A cr is trveling t 90 km/h when the river sees eer 75 m he n slms on the brkes Wht constnt ecelertion is require to voi hitting Bmbi? [Note: First convert 90 km/h to m/s] SOLUTION Let s list ll tht we know v 0 = 90 km/h or = 25 m/s n s 0 = 0 Let time t represent the time it tkes to stop Then s(t ) = 75 m Now the cr is stoppe t time t, so we know v(t ) = 0 Finlly we know tht ccelertion is n unknown constnt,, which is wht we wnt to fin Now we use our constnt ccelertion motion formuls to solve for v(t) = t + v 0 = t + 25 n s(t) = 2 t2 + v 0 t + s 0 = 2 t2 + 25t Now use the other velocity n position we know: v(t ) = 0 n s(t ) = 75 when the cr stops So v(t ) = t + 25 = 0 t = 25/ n Simplify to get s(t ) = 2 (t ) t = 2 ( 25/)2 + 25( 25/) = = = 625 = = so = = 25 6 m/s (Why is ccelertion negtive?) EXAMPLE 020 One cr intens to pss nother on bck ro Wht constnt ccelertion is require to increse the spee of cr from 30 mph (44 ft/s) to 50 mph ( ft/s) in 5 secons? SOLUTION Given: (t) = constnt v 0 = 44 ft/s s 0 = 0 An v(5) = ft/s Fin But v(t) = t + v 0 = t + 44 So Thus = 88 5 v(5) = = = = 88 3 YOU TRY IT 07 A toy bumper cr is moving bck n forth long stright trck Its ccelertion is (t) = cos t + sin t Fin the prticulr velocity n position functions given tht v(π/4) = 0 n s(π) = YOU TRY IT 08 (Grphicl integrtion) Let F(x) be n ntierivtive of f (x) on [0, 6], where f in Figure 5 Since F is n ntierivtive of f, then F = f Use this reltionship to nswer the following questions () On wht intervl(s) is F incresing? (Give reson for your nswer) (b) At wht point(s), if ny, oes F hve locl mx? (c) On wht intervl(s) is F concve up? () At wht point(s), if ny, oes F hve points of inflection? (e) Assume tht F psses through (0, 2) inicte with ; rw potentil grph of F Answer to you try it 07 : v(t) = (t) t = cos t + sin t t = sin t cos t + c So v(π/4) = 2 + c = 0 c = 0 Thus, v(t) = sin t cos t Now s(t) = v(t) t = sin t cos t t = cos t sin t + c Since s(π) = ( ) 0 + c = c = 0 So s(t) = cos t sin t Answer to you try it 08 : Justify ech () [0, 5] (b) x = 5 (c) [0, 2] () x = 2
15 mth 3, prerequisites, prt one preliminries Figure 5: This is the grph of f (x) Cn you rw grph of its ntierivtive F(x)?
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