Unit 4. Applications of integration

Transcription

1 SOLUTIONS TO 8. EXERCISES Unit 4. Applictions of integrtion 4A. Ares between curves. / 4A- ) (3x x )dx = (3/)x x (/3)x 3 / = /4 b) x 3 = x = x = ± or x =. There re two enclosed pieces ( < x < nd < x < ) with the sme re by symmetry. Thus the totl re is: 3 )dx = x (/)x 4 (x x = / (,) 3/ 5/ (/,/) / / 3/ b c d c) x + /x = 5/ = x + = 5x/ = x = or /. Therefore, the re is [5/ (x + /x)]dx = 5x/ x / ln x / = 5/8 ln / d) (y y )dy = y / y 3 /3 = /6 4A- First wy (dx): ( x )dx = ( x )dx = x x 3 /3 = 4/3 Second wy (dy): (x = ± y) ydy = (4/3)( y) 3/ = 4/3 COPYRIGHT DAVID JERISON AND MIT 996, 3

2 E. Solutions to 8. Exercises 4. Applictions of integrtion y = -x - x = - -y x = -y 4A-3 4 x = 3x = x = or 4. First wy (dx): 4 (4 x 3x)dx = 4x x 3 /3 3x / = 5/6 Second wy (dy): Lower section hs re 3 (y/3 + 4 ydy = y /6 (/3)(4 y) 3/ 3 = 7/6 4 x=- 4-y 4 (,3) x= 4-y x= y 3 (-4,-) Upper section hs re ydy = (4/3)(4 y) 3/ = 4/3 3 3 (See picture for limits of integrtion.) Note tht 7/6 + 4/3 = 5/6. 4A-4 sin x = cos x = x = π/4 + kπ. So the re is 5π/4 π/4 5π/4 (sin x cos x)dx = ( cos x sin x) = π/4

3 4. Applictions of integrtion E. Solutions to 8. Exercises π/4 5π/4 4B. Volumes by slicing; volumes of revolution 4B- ) πy dx = π( x ) dx = π ( x + x 4 )dx = π(x x 3 /3 + x 5 /5) = 6π/5 b) πy dx = π( x ) dx = π ( 4 x + x 4 )dx = π( 4 x x 3 /3 + x 5 /5) = 6π 5 /5 c) πx dx = π/3 d) πx dx = π 3 /3 e) π(x x ) dx = π(4x 4x 3 + x 4 )dx = π(4x 3 /3 x 4 + x 5 /5) = 6π/5 (Why (e) the sme s ()? Complete the squre nd trnslte.) b d f g h b x y = x y = x y = x x y= x y = b ( x / ) (for, set = ) (for c, set = ) (for e, set = ) f) π(x x ) dx = π(4 x 4x 3 + x 4 )dx 3

4 E. Solutions to 8. Exercises 4. Applictions of integrtion g) xdx = π 3 / = π(4 x 3 /3 x 4 + x 5 /5) = 6π 5 /5 (Why is (f) the sme s (b)? Complete the squre nd trnslte.) h) πy dx = πb ( x / )dx = πb (x x 3 /3 ) = πb /3 4B- ) π( y)dy = π/ b) π( y)dy = π 4 / c) π( y )dy = π/3 d) π( y )dy = π 3 /3 e) x x + y = = x = ± y. Using the method of wshers: π[( + y) ( y) ]dy = 4π ydy = (8/3)π( y) 3/ = 8π/3 (In contrst with (e) nd (), rottion round the y-xis mkes the solid in (e) different from ().) y b d, f ) g h (,) b x = y x = y x = + y x = y / x = ( y /b ) (for, set = ) (for c, set = ) (for e, set = ) f) x x + y = = x = ± y. Using the method of wshers: π[( + y) ( y) ]dy = 4π ydy 4 = (8/3)π( y) 3/ = 8π 4 /3

5 4. Applictions of integrtion E. Solutions to 8. Exercises g) Using wshers: π( (y /) )dy = π( y y 5 /5 ) = 4π 3 /5. h) b b πx dy = π ( y /b )dy = π( y y 3 /3b b ) = 4π b/3 b (The nswer in (h) is double the nswer in (h), with nd b reversed. Cn you see why?) 4B-3 Put the pyrmid upside-down. By similr tringles, the bse of the smller bottom pyrmid hs sides of length (z/h)l nd (z/h)m. The bse of the big pyrmid hs re b = LM; the bse of the smller pyrmid forms cross-sectionl slice, nd hs re L M z h (z/h)l (z/h)m = (z/h) LM = (z/h) b Therefore, the volume is h (z/h) bdz = bz 3 /3h h = bh/3 4B-4 The slice perpendiculr to the xz-plne re right tringles with bse of length x nd height z = x. Therefore the re of slice is x. The volume is z x side view of wedge long y-xis y x x top view of wedge long z-xis x x side view of slice long y-xis x dy = ( y )dy = 4/3 4B-5 One side cn be described by y = 3x for x /. Therefore, the volume is / πy dx / π( 3x) dx = π 3 /4 5

6 E. Solutions to 8. Exercises 4. Applictions of integrtion y = 3x / 4B-6 If the hypotenuse of n isoceles right tringle hs length h, then its re is h /4. The endpoints of the slice in the xy-plne re y = ± x, so h = x. In ll the volume is (h /4)dx = ( x )dx = 4 3 /3 4B-7 Solving for x in y = (x ) nd y = (x + ) gives the vlues x -x -x top view slice x = ± y nd x = ± y The hrd prt is deciding which sign of the squre root representing the endpoints of the squre. x = - + y x = - y - (x- y) Method : The point (, ) hs to be on the two curves. Plug in y = nd x = to see tht the squre root must hve the opposite sign from : x = y nd x = + y. Method : Look t the picture. x = + y is the wrong choice becuse it is the right hlf of the prbol with vertex (, ). We wnt the left hlf: x = y. Similrly, we wnt x = + y, the right hlf of the prbol with vertex (, ). Hence, the side of the squre is the intervl + y x y, whose length is ( y), nd the Volume = (( y) dy = 4 ( y + y)dy = /3. 6

7 4. Applictions of integrtion E. Solutions to 8. Exercises 4C. Volumes by shells 4C- ) b+ b+ Shells: (πx)(y)dx = 4πx (x b) dx b b b) (x b) = y = x = b ± y Wshers: π(x x )dy = π((b + y ) (b y ) )dy = π 4b y dy y = - (x - b) x = b - - y x = b + - y b- b+ b- b+ - Shells y = - - (x - b} - Wshers c) y dy = π /, becuse it s the re of semicircle of rdius. Thus (b) = Volume of torus = π b d) z = x b, dz = dx b+ 4πx (x b) dx = 4π(z + b) z dz = 4πb z dz b becuse the prt of the integrnd with the fctor z is odd, nd so it integrtes to. 4C- πxydx = πx 3 dx = π/ y = x y = x x = y 4C (shells) 4C 3 (shells) 4C 3b (discs) 7

8 E. Solutions to 8. Exercises 4. Applictions of integrtion 4C-3 Shells: πx( y)dx = πx( x)dx = π/5 Disks: πx dy = πy 4 dy = π/5 4C-4 ) πy(x)dy = 4π y ydy b) πy(x)dy = 4π y ydy c) πy( y)dy d) πy( y)dy e) x x + y = = x = ± y. The intervl y x + y hs length y = V = πy( y)dy = 4π y ydy f) x x + y = = x = ± y. The intervl y x + y hs length y = V = πy( ydy = 4π y ydy 4b 4d 4f 4g 4h (, ) (,) b x = - y (right) x = y x = + - y x = y / x = - x = - - y (left) - y x = - y /b g) πy( y /)dy 8

9 4. Applictions of integrtion E. Solutions to 8. Exercises b b h) πyxdy = πy( ( y /b )dy (Why is the lower limit of integrtion rther thn b?) 4C-5 ) πx( x )dx c) πxydx = πx dx b) πx( x )dx d) πxydx = πx dx e) πxydx = πx(x x )dx f) πxydx = πx(x 5b 5d 5f 5g b 5h - x y = - x y = x (for 5, set = ) for 5c, set = ) y = x - x (for 5e, set = ) y = x b y = b - x / y = -b - x / x )dx g) πxydx = πx xdx h) πx(y)dx = πx(b ( x / ))dx (Why did y get doubled this time?) 4C-6 b b πx(y)dx = πx( b x )dx = (4/3)π(b x ) 3/ b = (4π/3)(b ) 3/ 4D. Averge vlue 4D- Cross-sectionl re t x is = πy = π (x ) = πx 4. Therefore, 9

10 E. Solutions to 8. Exercises 4. Applictions of integrtion y= b - x Shells b y=- b - x bse of removed cylinder πx 5 6π verge cross-sectionl re = πx 4 dx = =. 5 ( ) dx 4D- Averge = = ln x ln = (ln ln ) = ln =. x 4D-3 Let s(t) be the distnce function; then the velocity is v(t) = s (t) Averge vlue of velocity = b s (t)dt = s(b) s() by FT b b = verge velocity over time intervl [,b] 4D-4 By symmetry, we cn restrict P to the upper semicircle. By the lw of cosines, we hve P Q = + cos θ. Thus π π verge of P Q = ( cos θ)dθ = [θ sin θ] = π π (This is the vlue of P Q when θ = π/, so the nswer is resonble.)) P θ Q x 4D-5 By hypothesis, g(x) = f(t)dt To express f(x) in terms of g(x), x multiply thourgh by x nd pply the Sec. Fund. Thm:

11 4. Applictions of integrtion E. Solutions to 8. Exercises x f(t)dt = xg(x) f(x) = g(x) + xg (x), by FT.. T A A 4D-6 Averge vlue of A(t) = A e rt dt = e rt T = T T r (e rt ) rt (rt ) If rt is smll, we cn pproximte: e rt + rt +, so we get A (rt ) rt A(t) (rt + ) = A ( + ). rt (If T, t the end of T yers the interest dded will be A rt ; thus the verge is pproximtely wht the ccount grows to in T/ yers, which seems resonble.) 4D-7 b b x dx = b /3 4D-8 The verge on ech side is the sme s the verge over ll four sides. Thus the verge distnce is / / x + (/) dx / x + (/) x Cn t be evluted by formul until Unit 5. The verge of the squre of the distnce is / / (x + (/) )dx = (x + (/) )dx = /3 / π/ π/ 4D-9 sin x dx cos(x) π/ π = /π 4D. Work 4D - According to Hooke s lw, we hve F = kx, where F is the force, x is the displcement (i.e., the dded length), nd k is the Hooke s lw constnt for the

12 E. Solutions to 8. Exercises 4. Applictions of integrtion spring. To find k, substitute into Hooke s lw:, = k (/) k = 4. To find the work W, we hve 6 6 W = F dx = 4x dx = x ] 6 = 7, inch-pounds = 6, foot-pounds. 4D - Let W (h) = weight of pil nd pint t height h. W () =, W (3) = W (h) = 5 h, since the pulling nd lekge both occur t constnt rte. 3 3 h h ] 3 work = W (h) dh = ( ) dh = h = 33 ft-lbs D -3 Think of the hose s divided into mny equl little infinitesiml pieces, of length dh, ech of which must be huled up to the top of the building. The piece t distnce h from the top end hs weight dh; to hul it up to the top requires h dh ft-lbs. Adding these up, 5 ]5 totl work = h dh = h = 5ft-lbs. g m m 4D -4 If they re x units prt, the grvittionl force between them is. x nd ] nd ( ) ( ) g m m work = dx = g m m = g m m = g m m n. x x nd d d d d n g m m The limit s n is. d 4E. Prmetric equtions 4E- y x = t, y x = t. Therefore, y x = (y x) = y 4xy + 4x y + x = (prbol) 4E- x = t + + /t nd y = t + /t. Subtrct, getting the hyperbol x y = 4 4E-3 (x ) + (y 4) = sin θ + cos t = (circle) 4E-4 + tn t = sec t = + x = y (hyperbol) 4E-5 x = sin t = sin t cos t = ± y y. This gives x = 4y 4y 4.

13 4. Applictions of integrtion E. Solutions to 8. Exercises 4E-6 y = x, so t = x nd x = t/, y = t /4 4E-7 Implicit differentition gives x + yy =, so tht y = x/y. So the prmeter is t = x/y. Substitute x = ty in x + y = to get Thus t y + y = = y = /( + t ) t y =, x = + t + t For < t <, this prmetriztion trverses the upper semicircle y > (going clockwise). One cn lso get the lower semicircle (lso clockwise) by tking the negtive squre root when solving for y, t y =, x = + t + t 4E-8 The tip Q of the hour hnd is given in terms of the ngle θ by Q = (cos θ, sin θ) (units re meters). Next we express θ in terms of the time prmeter t (hours). We hve { } π/, t = θ = θ decreses linerly with t π/3, t = π π π 3 (t ) π = π Thus we get θ = 6 = θ. t. Q P θ Finlly, for the snil s position P, we hve P = (t cos θ, t sin θ), where t increses from to. So, π π π π π π x = t cos( t) = t sin t, y = t sin( t) = t cos t F. Arclength 4F- ) ds = + (y ) dx = 6dx. Arclength = 6dx = 6. 3

14 E. Solutions to 8. Exercises 4. Applictions of integrtion b) ds = + (y ) dx = + (9/4)xdx. Arclength = + (9/4)xdx = (8/7)( + 9x/4) 3/ = (8/7)((3/4) 3/ ) c) y = x /3 ( x /3 ) / = x /3. Therefore, ds = x /3 dx, nd Arclength = x /3 dx = (3/)x /3 = 3/ d) y = x( + x ) /. Therefore, ds = + x + x 4 dx = ( + x )dx nd Arclength = ( + x )dx = x + x 3 /3 = /3 x 4F- y = (e e x )/, so the hint sys + (y ) = y nd ds = + (y ) dx = ydx. Thus, Arclength = (/) b (e x + e x )dx = (/)(e x e x ) b = (e b e b )/ 4F-3 y = x, + (y ) = + 4x. Hence, rclength = b + 4x dx. 4F-4 ds = (dx/dt) + (dy/dt) dt = 4t + 9t 4 dt. Therefore, Arclength = 4t + 9t 4 dt = (4 + 9t ) / tdt = (/7)(4 + 9t ) 3/ = (4 3/ 8)/7 4F-5 dx/dt = /t, dy/dt = + /t. Thus ds = (dx/dt) + (dy/dt) dt = + /t 4 dt nd Arclength = + /t 4 dt 4F-6 ) dx/dt = cos t, dy/dt = sin t. ds/dt = (dx/dt) + (dy/dt) = cos t (speed of the point) Forwrd motion (dx/dt) is lrgest for t n odd multiple of π (cos t = ). Forwrd motion is smllest for t n even multiple of π (cos t = ). (continued ) Remrk: The lrgest forwrd motion is when the point is t the top of the wheel nd the smllest is when the point is t the bottom (since y = cos t.) π π π b) cos tdt = sin(t/)dt = 4 cos(t/) = 8 4

15 4. Applictions of integrtion E. Solutions to 8. Exercises π 4F-7 sin t + b cos tdt 4F-8 dx/dt = e t (cos t sin t), dy/dt = e t (cos t + sin t). t ds = e t (cos t sin t) + e t (cos t + sin t) dt = e cos t + sin tdt = e t dt Therefore, the rclength is e t dt = (e ) 4G. Surfce Are 4G- The curve y = R x for x b is revolved round the x-xis. Since we hve y = x/ R x, we get y= R - x b ds = + (y ) dx = + x /(R x )dx = R /(R x )dx = (R/y)dx Therefore, the re element is nd the re is da = πyds = πrdx b πrdx = πr(b ) 4G- Limits re x /. ds = 5dx, so / da = πyds = π( x) 5dx = A = π 5 ( x)dx = 5π/ 4G-3 Limits re y. x = ( y)/, dx/dy = /. Thus ds = + (dx/dy) dy = 5/4dy; da = πyds = π( y)( 5/)dx = A = ( 5π/) ( y)dy = 5π/4 5

16 E. Solutions to 8. Exercises 4. Applictions of integrtion 4G 3 y = x x = ( y)/ / 4G 4 4G-4 A = πyds = πx + 4x dx 4G-5 x = y, dx/dy = / y, nd ds = + /4ydy A = πxds = π y + /4ydy = π y + /4dy = (4π/3)(y + /4) 3/ = (4π/3)((9/4) 3/ (/4) 3/ ) = 3π/3 4G-6 y = ( /3 x /3 ) 3/ = y = x /3 ( /3 x /3 ) /. Hence ds = + x /3 ( /3 x /3 )dx = /3 x /3 dx Therefore, (using symmetry on the intervl x ) - y = ( /3 /3 - x ) 3/ A = πyds = π( /3 x /3 ) 3/ /3 x /3 dx = (4π)(/5)( 3/) /3 ( /3 x /3 ) 5/ = (π/5) 4G-7 ) Top hlf: y = (x b), y = (b x)/y. Hence, ds = + (b x) /y dx = (y + (b x) )/y dx = (/y)dx Since we re only covering the top hlf we double the integrl for re: b+ xdx A = πxds = 4π b (x b) b) We need to rotte two curves x = b + y 6

17 4. Applictions of integrtion E. Solutions to 8. Exercises y = - (x-b) b y = - - (x-b) upper nd lower surfces re symmetricl nd equl nd x = b y round the y-xis. The vlue dx /dy = (dx /dy) = y/ y So in both cses, The integrl is ds = + y /( y )dy = (/ y )dy x = b + - y x = b - - y inner nd outer surfces re not symmetricl nd not equl dy A = πx ds + πx ds = π(x + x ) y But x + x = b, so dy A = 4πb y c) Substitute y = sin θ, dy = cos θdθ to get A = 4πb π/ cos θdθ = 4πb π/ dθ = 4π b π/ cos θ π/ 4H. Polr coordinte grphs 4H- We give the polr coordintes in the form (r, θ): 7

18 E. Solutions to 8. Exercises 4. Applictions of integrtion ) (3, π/) b) (, π) c) (, π/3) d) (, 3π/4) e) (, π/4 or 7π/4) f) (, π/ or 3π/) g) (, π/6 or π/6) h) (, 3π/4 or 5π/4) 4H- ) (i) (x ) + y = x x + y = r r cos θ = r = cos θ. (ii) OP Q = 9 o, since it is n ngle inscribed in semicircle. In the right tringle OPQ, OP = OQ cos θ, i.e., r = cos θ. b) (i) Anlogous to 4H-(i); ns: r = sin θ. (ii) nlogous to 4H-(ii); note tht OQP = θ, since both ngles re complements of P OQ. c) (i) OQP is right tringle, OP = r, nd P OQ = α θ. The polr eqution is r cos(α θ) =, or in expnded form, r(cos α cos θ + sin α sin θ) =, or finlly x y + =, A B since from the right tringles OAQ nd OBQ, we hve cos α =, sin α = A cos BOQ =. B d) Since OQ = sin θ, we hve: if P is bove the x-xis, sin θ >, OP = OQ QR, or r = sin θ; if P is below the x-xis, sin θ <, OP = OQ + QR, or r = + sin θ = sin θ. Thus the eqution is r = ( sin θ). e) Briefly, when P = (, ), P Q P R = =, the constnt. Using the lw of cosines, P R = r + r cos θ; P Q = r + r cos(π θ) = r + + r cos θ Therefore P Q P R = (r + ) (r cos θ) = ( ) which simplifies to r = cos θ. 4H-3 ) r = sec θ = r cos θ = = x = b) r = cos θ = r = r cos θ = x = x + y = x c) r = ( + b cos θ) (This figure is crdiod for = b, limçon with loop for < < b, nd limçon without loop for > b >.) 8

19 4. Applictions of integrtion E. Solutions to 8. Exercises r = r + br cos θ = r + bx = x + y = x + y + bx y b= c prbol hyperbol b< r θ r θ x r θ θ r r θ b> c ellipse limcon <b crdioid (=b) limcon >b 8 8b 8c 8d (d) r = /(b + c cos θ) = = = (e) r = sin(θ) = = r(b + c cos θ) = = rb + cx = rb = cx = r b = cx + c x cx + (c b )x b y = r = sin θ cos θ = xy/r r 3 = xy = (x + y ) 3/ = xy r = cos r = sin r = cos r θ θ = sin θ θ (x y ) x f) r = cos(θ) = ( cos θ ) = ( x + y ) = (x + y ) 3/ = xy g) r = sin(θ) = sin θ cos θ = xy = r 4 = xy = (x +y ) = r x h) r = cos(θ) = ( x + y ) = (x + y ) = (x y ) θ i) r = e = ln r = θ = ln x + y = tn y x 4I. Are nd rclength in polr coordintes 9

20 E. Solutions to 8. Exercises 4. Applictions of integrtion 4I- (dr/dθ) + r dθ ) sec θdθ b) dθ c) + b + b cos θdθ b + c + bc cos θ d) dθ (b + c cos θ) e) 4 cos (θ) + sin (θ)dθ f) 4 sin (θ) + cos (θ)dθ g) Use implicit differentition: rr = cos(θ) = r = cos(θ)/r = (r ) = cos (θ)/ sin(θ) Hence, using common denomintor nd cos + sin =, ds = cos (θ)/ sin(θ) + sin(θ)dθ = dθ sin(θ) h) This is similr to (g): ds = dθ cos(θ) i) + e θ dθ 4I- da = (r /)dθ. The min difficulty is to decide on the endpoints of integrtion. Endpoints re successive times when r =. cos(3θ) = = 3θ = π/ + kπ = θ = π/6 + kπ/3, k n integer. π/6 π/6 Thus, A = ( cos (3θ)/)dθ = cos (3θ)dθ. π/6 (Stop here in Unit 4. Evluted in Unit 5.) π 4I-3 A = (r /)dθ = (e 6θ /)dθ = (/)e 6θ π = (e 6π )/

21 4. Applictions of integrtion E. Solutions to 8. Exercises θ = π/6 three-lef rose three empty sectors θ = π/6 3π e 4I-4 Endpoints re successive time when r =. sin(θ) = = θ = kπ, k n integer. π/ π/ Thus, A = (r /)dθ = ( /) sin(θ)dθ = ( /4) cos(θ) = /. 4I-5 r = cos θ, ds = dθ, π/ < θ < π/. (The rnge ws chosen crefully so tht r >.) Totl length of the circle is π. Since the upper nd lower semicircles re symmetric, it suffices to clculte the verge over the upper semicircle: r = cos θ π/ π/ 4 4 cos θ()dθ = sin θ = π π π 4I-6 ) Since the upper nd lower hlves of the crdiod re symmetric, it suffices to clculte the verge distnce to the x-xis just for point on the upper hlf. We hve r = ( cos θ), nd the distnce to the x-xis is r sin θ, so π π ( cos θ) r sin θdθ = ( cos θ) sin θdθ = π π π π = π

22 E. Solutions to 8. Exercises 4. Applictions of integrtion P r Q (b) ds = (dr/dθ) + r dθ = ( cos θ) + sin θdθ = cos θdθ = sin(θ/)dθ, using the hlf ngle formul. π rclength = sin(θ)dθ = 4 cos(θ/) = 8 For the verge, don t use the hlf-ngle version of the formul for ds, nd use the intervl π < θ < π, where sin θ is odd: π Averge = r sin θ cos θdθ = sin θ ( cos θ) 3/ dθ 8 π 8 π π π 4 = ( cos θ) 3/ sin θdθ = 4 ( cos θ)5/ = 5 π 4I-7 dx = sin θdθ. So the semicircle y > hs re π ydx = sin θ( sin θ)dθ = sin θdθ π But π π sin θdθ = ( cos(θ)dθ = π/ So the re is π / s it should be for semicircle. π Arclength: ds = dx + dy = (ds) = ( sin θdθ) + ( cos θdθ) = (sin dθ + cos dθ)(dθ) = ds = dθ (obvious from picture). dθ ds θ - ds = π dθ = π

23 4. Applictions of integrtion E. Solutions to 8. Exercises 4J. Other pplictions 4J- Divide the wter in the hole into n equl circulr discs of thickness Δy. ( ) Volume of ech disc: π Δy π Energy to rise the disc of wter t depth y i to surfce: ky i Δy. 4 Adding up the energies for the different discs, nd pssing to the limit, n π π πk y ] πk 4 E = lim ky i Δy = ky dy = =. n J- Divide the hour into n equl smll time intervls Δt. At time t i, i =,..., n, there re x e kti grms of mteril, producing pproximtely rx e kti Δt rdition units over the time intervl [t i, t i + Δt]. Adding nd pssing to the limit, n 6 e kt ] 6 r x ( ) R = lim r x e kti Δt = r x e kt dt = r x = e 6k. n k k 4J-3 Divide up the pool into n thin concentric cylindricl shells, of rdius r i, i =,..., n, nd thickness Δr. The volume of the i-th shell is pproximtely π r i D Δr. k The mount of chemicl in the i-th shell is pproximtely π r i D Δr. + r i Adding, nd pssing to the limit, n k R r A = lim π r i D Δr = πkd dr n + r i + r ] R = πkd ln( + r ) = πkd ln( + R ) gms. 4J-4 Divide the time intervl into n equl smll intervls of length Δt by the points t i, i =,..., n. The pproximte number of heting units required to mintin the temperture t 75 o over the time intervl [t i, t i + Δt]: is [ ( )] πt i 75 6 cos k Δt. 3

24 E. Solutions to 8. Exercises 4. Applictions of integrtion Adding over the time intervls nd pssing to the limit: n [ ( )] πt i totl het = lim 75 6 cos k Δt n 4 [ ( )] πt = k 75 6 cos dt 4 ( ) [ ] 4 πt πt = k 5 + cos dt = k 5t + sin = 36k. π 4J-5 Divide the month into n equl intervls of length Δt by the points t i, i =,..., n. Over the time intervl [t i.t i + Δt], the number of units produced is bout ( + t i ) Δt. The cost of holding these in inventory until the end of the month is c(3 t i )( + t i ) Δt. Adding nd pssing to the limit, n totl cost = lim c(3 t i )( + t i ) Δt n 3 [ t 3 ] 3 = c(3 t)( + t) dt = c 3t + t = 9c. 3 4J-6 Divide the wter in the tnk into thin horizontl slices of width dy. If the slice is t height y bove the center of the tnk, its rdius is r y. This formul for the rdius of the slice is correct even if y < i.e., the slice is below the center of the tnk s long s r < y < r, so tht there relly is slice t tht height. Volume of wter in the slice = π(r y ) dy Weight of wter in the slice = πw(r y ) dy Work to lift this slice from the ground to the height h+y = πw(r y ) dy (h+y). r Totl work = πw(r y )(h + y) dy r r = πw (r h + r y hy y 3 ) [ r ] r = πw r hy + r y hy 3 y In this lst line, the even powers of y hve the sme vlue t r nd r, so contribute when it is evluted; we get therefore [ ] ( ) = πwh r y y 3 r = πwh r 3 r 3 = πwhr3. r. r 4

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