Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17
Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is clled the displcement, or net distnce trvelled, by the prticle over the time intervl [, b]. This is in contrst to the totl distnce trvelled, which is given by the integrl of speed. Tht is, the totl distnce trvelled by the prticle for t [, b], is v dt. Derivtion of Totl Distnce If vi is the velocity of the prticle t some time ti in the time intervl [t i 1, t i ], then the speed vi is pproximtely constnt on the time intervl. So on this time intervl the distnce trvelled is pproximtely vi t, since So the totl distnce trvelled is distnce = speed time. lim n n i=1 vi t = v dt.
Exmple 1 Suppose v = t 11t + 4 for t [, 8]. Then 8 v dt = 8 ( t 11t + 4 ) dt [ t 3 = 3 11 t + 4t = 3 3 This mens tht fter 8 seconds the prticle is 3/3 units to the right of where it strted from. ] 8 The totl distnce is much more complicted clcultion becuse v > for t < 3; but v < for t > 3. Thus the totl distnce trvelled is 8 v dt = = 3 v dt 8 3 v dt [ t 3 3 11 t + 4t ( 15 ) 6 = 63 = 157 3 ] 3 [ t 3 3 11 t + 4t ] 8 3
Integrl Formuls Often to find formul to describe something be it geometricl or physicl we use Riemnn sum to first set things up in simple, pproximte wy. These pproximtions usully strt with regulr prtition. Then s the limit of the norm of the prtition goes to zero, the Riemnn sum pproches n integrl. In Chpter 6, formuls for re, volumes, length, surfce re, work to nme few will be derived this wy. Are Between Two Curves: The Bsic Formul If f (x) g(x) for x [, b], then y (f (x) g(x)) A i f (x) g(x) is the re of the region below f, bove g, for x b. Proof: x i x b x A i = (f (x i ) g(x i )) x A A i = (f (x i ) g(x i )) x; A = lim Ai = x lim x (f (x i ) g(xi )) x = (f (x) g(x)).
Exmple 1 Find the re of the region bounded by y = x nd y = 6 x. Solution: find intersection points: 6 x = x = x + x 6 x = 3 or x =. A = = 3 ( 6 x x ) [ 6x 1 3 x 3 1 x ] 3 = 15 6 Exmple Find the re between the x-xis nd the curve y = sin(x) + sin x on [, π]. Solution: sin x cos x + sin x = x =, π, π/3. Then A = A 1 + A where π/3 A 1 = (sin(x) + sin x) = [ 1 ] π/3 cos(x) cos x = 9 4, A = = π π/3 ( sin(x) sin x) [ 1 cos(x) + cos x ] π π/3 = 1 4
Exmple 3: Integrting with Respect to y Find re of region bounded by the curves with eqution y = x nd y = 8 x. Solution: find intersection points: y = 8 y y + y 8 = y = 4 or y =. A = = 4 ( 8 y y ) dy [ 8y 1 3 y 3 y ] 4 = 16 8 3 4 + 3 64 3 + 16 = 36 Exmple 3, Integrting with Respect to x; the Hrd Wy A = = 4 8 4 8 ( x ( 8 x)) ( x + 8 x) + + 8 4 8 4 ( 8 x ( 8 x) ) 8 x [ ] 4 [ ] 8 x (8 x)3/ 4(8 x)3/ = + 4 3 3 8 4 = 4 16 18 16 + 3 3 + 3 3 = 36, s before.
The Method of Slicing y Let xi be ny point in subintervl of length x; let the re of cross-sectionl slice perpendiculr to the x-xis with bse x be A(xi ). Add up the volumes of ll the slices: (xi, f (x i )) V V = A(xi ) x V = lim A(xi ) = x x = x x = b x A(x i ) x A(x) Exmple 1: Volume of Pyrmid, V = 1 3 h Let l be the length of the side of cross-section of the pyrmid t l height y bove the bse. By similr tringles, h y = h. y The cross section is squre: A(y) = l. (, h) ( l/, y) l V = h A(y) dy = h h = (1 yh + y ) dy = [ y y h + y 3 3h ( 1 y h ) dy h ] h = 1 3 h
Solids of Revolution: Method of Disks y (x y = f (x) i, f (x i )) Let xi be ny point in subintervl of length x. The volume of the disc obtined by revolving bout the x- xis the rectngle with bse x nd rdius f (xi ) is V = πf (x i ) x. x b x V V = πf (xi ) x V = lim πf (x i ) x = x πf (x) Exmple : Volume of Sphere V = πy = π (r x ) = π [r x x 3 ] r 3 = π (r 3 r 3 ) 3 Figure: x + y = r = 4 3 πr 3
Exmple 3 Find the volume of the solid obtined by revolving round the line y = the curve y = x for x [, 1]. V = 1 1 π( x + ) = π (x + 4 x + 4) [ x = π + 8 ] 1 3 x 3/ + 4x = 43 6 π Exmple 4 Find the volume of the solid obtined by revolving bout the x-xis the region bounded by the curves y = x nd y = x for x [, 1]. V = = π 1 1 π(r o r i ) (x x ) [ x = π x 3 3 ] 1 = 1 6 π
Suppose y = f (x) on [, b] is Revolved Around the y-xis y (xi, f (x i )) y = f (x) Let x i be ny point in subintervl of length x. x b x The pproximte volume of the bove cylindricl shell is V C h w = πxi f (xi ) x. Volumes By Cylindricl Shells Then V V = πxi f (xi ) x V = lim πx i f (xi ) x = x πx f (x)
Exmple 1: Volume of Cone Figure: x r + y h = 1 ( V = πx h 1 x ) r = πh (x x ) r [ x = πh x 3 ] r 3r ( r = πh r ) = 1 3 3 πhr Exmple : Volume of Sphere Figure: x + y = r V = = π πx y x r x = π u ( du) = π r [ ] r = π 3 u3/ = 4 3 πr 3 u du
Exmple 3 Find the volume of the solid obtined by revolving round the y-xis the region bounded by y = x nd y = x, for x [, 1]. V = 1 = π = π πx ( x x) 1 (x 3/ x ) [ 5 x 5/ x 3 3 ] 1 = 15 π Exmple 4 Find the volume of the solid obtined by revolving round the line x = the region bounded by y = x nd y = x, for x [, 1]. V = 1 = π = π π(x + ) ( x x) 1 (x 3/ + x x x) [ 5 x 5/ + 4 3 x 3/ x 3 3 x ] 1 = 4 5 π
Summry Consider y = f (x) on [, b], ssume f is invertible, with f 1 = g, nd tht c = f () nd d = f (b). Wht do these four integrls represent? 1.. 3. 4. d c d c π (f (x)) π x f (x) π (g(y)) dy π y g(y) dy y d c d c R y-xis. d c x-xis. R 1 b x π (f (x)) is the volume of the region R 1 revolved bout the x-xis. π x f (x) is the volume of the region R 1 revolved bout the y-xis. π (g(y)) dy is the volume of the region R revolved bout the π y g(y) dy is the volume of the region R revolved bout the
The Length of Curve y y y = f (x) If s is the length of smll prt of the curve, then s x + y ( ) y s 1 + x x x b x ( ) dy Now let x : S = lim s = 1 +. S is x clled the rc length, or simply length, of the curve. Exmple 1 The expression ds = differentil. Then S = 1 + ( ) dy ds = 1 + = S = 1 ds = (u = 9x /3 + 4) = 1 18 ( dy ) is clled the rc length ds. Exmple: let f (x) = x /3 on [, 1]. ( ) 9x 1 + 3 x 1/3 = /3 + 4 3x 1/3 1 9x /3 + 4 3x 1/3 = 1 13 u du 18 4 [ ] 13 3 u3/ = 1 7 (133/ 8) 4
Exmple : Circumference of Circle Figure: x + y = r ( dy C = 4 1 + = 4 = 4 = 4 1 + ) ( x r x ) r x + x r x r, since r > r x Exmple, Continued To evlute this integrl, we shll use inverse trigonometric functions. In prticulr: ( r x = x ) sin 1 + C, r s you my check. Then C = 4 r [ ( r x = 4r sin 1 x )] r r = 4r(sin 1 (1) sin 1 ()) = πr, s you my check.
Surfce Are y s y = f (x) b x Prtition the intervl [, b] into n subintervls of equl length x. In ech subintervl pick vlue xi nd consider the strip of surfce obtined by revolving the curve y = f (x) on the intervl [x i, x 1 ] round the x-xis. It s pproximte rdius is r i = f (xi ), nd its pproximte width is s. So the pproximte re of the strip is A = πf (x i ) s. Surfce Are Formuls Then the surfce re of the solid of revolution obtined by revolving the curve y = f (x) on the intervl [, b] round the x-xis is given by SA = lim A = x πf (x) ds = πf (x) 1 + ( ) dy. If you revolve the curve round the y-xis, then the pproximte rdius of the strip becomes xi nd the formul for the surfce re is ( ) dy SA = πx ds = πx 1 +.
Exmple 1: Surfce Are of Sphere is 4πr Figure: x + y = r SA = πy ds = 4π r x 1 + = 4π = 4π r x + x r, since r > = 4πr[x] r = 4πr ( x r x ) Exmple : Surfce Are of Cone is πr r + h Figure: x r + y h = 1 SA = πx ds ( ) h = π x 1 + r r = π x + h r r = π + h [ ] x r r = πr r + h
Work in Physics The work required to move n object through distnce d by pplying constnt force F is given by W = F d. Suppose n object is moved from x = to x = b by pplying non-constnt force F (x). You cn pproximte the work on smll subintervl of length x, by picking point xi in the subintervl nd tking the force to be constnt, F (xi ), over tht subintervl. Then on ech subintervl the work done is W F (xi ) x. The totl work done is W = lim W = F (x). x Two Formuls from Physics To use the formul W = F (x) you hve to know the force in terms of x. Here re two exmples in which such force is known: 1. Hooke s Lw. If you stretch mss on spring the force required is proportionl to the displcement: F (x) = kx, where x = is the equilibrium position of the spring.. Newton s Lw of Grvity. If m 1 nd m re seprted by distnce x then the grvittionl force of ttrction between the two msses is given by F = Gm 1m x, where G is the grvittionl constnt.
Exmple 1 1. The work done to stretch spring from x = to x = b is W = [ x kx = k ] b = k (b ).. If R is the rdius of the erth, nd m its mss, then the work required to put stellite of mss m 1 into n orbit of height h bove the erth s surfce is W = R+h R Gm 1 m x = [ Gm ] R+h 1m x R = Gm 1m h R(R + h). Work Done in Filling Tnk y = b y = y A(y) y x Suppose fluid of density ρ is pumped from ground level y = up into tnk, with bse t y = nd top t y = b. Suppose the crosssectionl re of the tnk t height y is A(y). Consider thin shell of the fluid of thickness y. The volume of this shell is pproximtely V = A(y) y. Its mss is pproximtely ρ V nd the work required to pump this thin shell of liquid up to height y is pproximtely W = ρ V g y.
Formuls for Work Done in Filling or Emptying Tnk 1. Thus the work required to pump the tnk full of fluid is W = lim W = lim ρa(y) y g y y y = lim ρg A(y) y y = y ρg A(y) y dy. If the tnk is emptied by pumping ll the liquid up to pipe or conduit bove the tnk t height y = h then the work done in emptying the tnk is W = ρg A(y) (h y) dy. Exmple Find the work done in pumping fluid of density ρ from ground level into conicl tnk with rdius t the top m nd height 8 m. y (, 8) (x, y) By similr tringles, x y = 8 x = 1 4 y. A(y) = πx = π 16 y 8 8 W = ρg π 16 y y dy = πρg y 3 dy 16 = πρg [ ] y 4 8 = 64πρg (Joules) 16 4
Exmple 3 A hemisphericl tnk of rdius m is full of liquid with density ρ. How much work is required to empty the tnk by pumping ll the liquid up to pipe 1 m bove the top of the tnk? y A(y) = πx h = 3 (x, y) x x + (y ) = 4 = π ( 4 (y ) ) = π(4 y + 4y 4) = π(4y y ) Exmple 3, Concluded W = = = ρg A(y) (h y) dy ρg π(4y y ) (3 y) dy ρgπ(y 3 7y + 1y) dy [ y 4 = ρgπ 4 7 3 y 3 + 6y ] = 8 3 ρgπ
The Work-Energy Reltionship Recll: F = m = m dv dt. Suppose n object of mss m is moved by force F from x = t time t = t i to x = b t time t = t f. Then W = = (by substitution) = tf t i vf F = m dv dt m dv tf dt dt dt = t i [ ] mv vf m v dv = v i v i m v dv dt dt = mv f mv i, the chnge in kinetic energy Exmple 4 A mss of 1 kg is moving long the x-xis with speed 5 m/sec. At position x = force F (x) = 3x N begins to push the object. Wht is the speed of the object when it reches x = 1? Assume position long the x-xis is mesured in meters. Solution: W = 1 F (x) = 1 3x = [ x 3] 1 = 1. 1v f 1v i = 1 5v f = 1 + 15 v f = 5 v f = 15
Nturl Growth Eqution Let x be the mount of some substnce present t time t. The following differentil eqution dt = kx, k hs mny importnt pplictions. It cn be interpreted s dt }{{} the rte of chnge = k }{{} is proportionl to }{{} x. the mount present In this cse, the substnce is sid to be growing nturlly. Solution to the Nturl Growth Eqution It is esy to verify tht x = x e kt is solution to the nturl growth eqution, where the initil vlue of x is x t t = : nd x = x e kt dt = k x e kt = kx, t = x = x e = x. The solutions to the nturl growth eqution re clled exponentil models.
Exponentil Growth: k >, x > In this cse the vlue of x is lwys incresing. Two key fetures: 1. Doubling time: x = x x e kt = x. lim t x = e kt = kt = ln t = ln k Exmples: exponentil popultion growth; compound interest. Exmple 1 The popultion of town is growing exponentilly so tht its popultion doubles every 1 yers. The popultion of the town ws 1, in 1995; wht will its popultion be in the yer? Solution: Let x be the popultion of the town t time t, where time is mesured in yers since 1995. So t = corresponds to 1995, nd x = 1. Use the doubling time to find k : 1 = ln k k = ln 1.693. So x = x e kt = 1 e ln 1 t = 1 t 1, or x 1 e.693t. Now let t = 5 : x = 1 5 1 = 1.5 56 569.
Exponentil Decy: k <, x > In this cse the vlue of x is lwys decresing, nd lim t x =. Hlf Life: x = 1 x x e kt = 1 x e kt = 1 kt = ln t = ln k Exmple: rdioctive decy. Exmple : Crbon-14 Dting The hlf life of crbon-14 is 5,7 yers. If specimen of chrcol found in Stonehenge contins only 63% of its originl crbon-14, how old is Stonehenge? Solution: Let x be the mount of crbon-14 present in the chrcol t time t, with t in yers since the chrcol ws creted. Use the hlf life to find k : 5 7 = ln k k = ln 5 7.116. Then x = x e kt = x e.116t. Let x =.63x, nd solve for t :.63x = x e.116t.63 = e.116t ln.63 =.116t t 3 8. So the ge of Stonehenge is pproximtely 3,8 yrs.
Two Different Trigonometries As opposed to the six regulr trigonometric functions, sin θ, cos θ, tn θ, csc θ, sec θ nd cot θ, which cn be defined in terms of the circle x + y = 1, the six hyperbolic trigonometric functions re defined in terms of the hyperbol x y = 1. However, this is not pprent from their definitions: sinh θ = eθ e θ, cosh θ = eθ + e θ, tnh θ = sinh θ cosh θ, csch θ = 1 sinh θ, sech θ = 1 cosh θ, coth θ = cosh θ sinh θ. Bsic Hyperbolic Trigonometric Identity cosh θ sinh θ = ( e θ + e θ ) ( e θ e θ = eθ + + e θ 4 = 4 4 = 1. ) eθ + e θ 4 If x = cosh θ nd y = sinh θ, then the bsic hyperbolic trig identity sttes tht x y = 1, which is the eqution of n hyperbol.
Grphs of sinh θ nd cosh θ Figure: hyperbolic sine Figure: hyperbolic cosine Hyperbolic Trigonometric Indentities The following identities cn ll be proved by putting ll hyperbolic trig functions in terms of e θ nd e θ nd simplifying: 1. 1 tnh θ = sech θ. coth θ 1 = csch θ 3. sinh(α + β) = sinh α cosh β + cosh α sinh β 4. cosh(α + β) = cosh α cosh β + sinh α sinh β 5. sinh θ = sinh θ cosh θ 6. cosh θ = cosh θ + sinh θ 7. cosh θ = 1 (cosh(θ) + 1) 8. sinh θ = 1 (cosh(θ) 1)
Derivtives of sinh θ nd cosh θ sinh θ = d dθ = eθ + e θ = cosh θ ( e θ e θ ) Similrly, cosh θ = d dθ ( e θ + e θ ) = eθ e θ = sinh θ Derivtives of the Other Hyperbolic Trig Functions You cn lso prove tht 1. tnh θ = sech θ. coth θ = csch θ 3. sech θ = sech θ tnh θ 4. csch θ = csch θ coth θ Observe tht ll the formuls from the lst few slides re very similr to the corresponding regulr trig identities you re lredy fmilir with, except for the odd plus or minus sign. In fct, you shouldn t memorize ny of these formuls; you should simply be wre of wht the hyperbolic trig functions re, just in cse they show up in the homework, or in your other courses.
Inverse Hyperbolic Trig Functions Since the hyperbolic trig functions re expressed in terms of exponentils, it should come s no surprise tht the inverse hyperbolic trig functions cn be expressed in terms of logrithms. For exmple, y = sinh 1 x x = sinh y x = ey e y x = e y 1 e y xe y = e y 1 ( using the qudrtic formul ) e y = x ± 4x + 4 Formul for y = sinh 1 x So fr, we hve: e y = x ± 4x + 4 Since e y > for ll y, we must tke from which we get e y = x + x + 1, = x ± x + 1. sinh 1 x = y = ln(x + x + 1). There re similr formuls for the other five inverse hyperbolic trig functions.
Derivtives of Inverse Hyperbolic Trig Functions These cn be clculted implicitly, or directly. y = sinh 1 x sinh y = x cosh y dy = 1 Alterntely, d sinh 1 x dy = 1 cosh y d sinh 1 x = 1 1 + x = d ln(x + x + 1) = 1 + x/ x + 1 x + x + 1 = 1 1 + x. Six New Integrtion Formuls 1.. 3. 4. 5. 6. du u + 1 = sinh 1 u + C du u 1 = cosh 1 u + C, if u > 1 du 1 u = tnh 1 u + C, if u < 1 du 1 u = coth 1 u + C, if u > 1 du u 1 u = sech 1 u + C, if u < 1 du u 1 + u = csch 1 u + C