Honors Calculus Homework 13 Solutions, due 12/8/5

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Aubrey Wilkinson
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1 Hoors Calculus Homework Solutios, due /8/5 Questio Let a regio R i the plae be bouded by the curves y = 5 ad = 5y y. Sketch the regio R. The two curves meet where both equatios hold at oce, so where: y = 5, = 5y y, = y + 5 = 5y y, y + = 5y y, y y + =, y y =, y =, = 4, or y =, = 5 9 = 5 9. So the regio is the regio bouded o the left by the lie = ad o the right by the parabola + = 5y y. The y-rage is from y = to y =. y + 5 Fid the area of the regio R. We use horizotal strips at fied y, of width dy. The legth of the strip is: + = 5y y y + 5 = 5y y y 5y+y = y y+ The area of the strip is the: da = + d = y y + d.

2 The total area of the regio is the: A = = [y y = = y y + d ] + y = = 4 6 = If the regio R is rotated about the -ais what is the volume of the solid geerated by it? Let this volume be V. We use horizotal cyliders of height + radius y ad width dy, so the cylider volume is: So the total volume is: V = = π = π = π dv = π + ydy. π + ydy = π = π = π 5 [ y y y + ydy 4 y yy y + dy + 5y ] = 4459π 486 =

3 If the regio R is rotated about the y-ais what is the volume of the solid geerated by it? Let this volume be V y. We use horizotal washers of ier radius ad outer radius + ad width dy, so the washer volume is: So the total volume is: 5 = π V y = π = π dv = π + dy. + dy = π y 5 y 4y dy dy = π y 5 y + y 4y + 5 dy 9 = π 5y y + y 4 4y + 5 dy 9 [ ] 5y = π 5y4 + y5 4y = π = π = π 6497π = Where is the cetroid of the regio R? By Pappus Theorem, the cetroid is at: πa [V y, V ] = 8 [ , 4459 ] = [479, 95] = [.66,.8]

4 Questio Plot the slope plot for each of the followig differetial equatios ad for each discuss the behavior of its solutios. I particular plot the poits where the slopes take the values ±, ±, ±, or. Also fid the geeral solutio of each equatio. dy dt = y y. We ote that all the poits o ay horizotal lie i.e. with the same y- value have the same plotted slope. We plot slope m whe y y = m, so whe y y + m =, or y = m, or y = ± m. So there are o poits with slope m = ad oe with slope m =. The poits with m = cosist of the horizotal lie y =. The poits with m = cosist of the lies y = ad y =. The poits with m = cosist of the lies y = + ad y =. The poits with m = cosist of the lies y = + ad y =. The poits with slope m = cosist of the lies y = + 5 ad y = 5. Plottig these slopes, we see a stadard logistic slope plot: The horizotal slopes alog y = ad y = give costat solutios y = ad y = respectively. Ay solutio through a poit with y > decreases steadily, is cocave up ad goes to y = + i the future ad i the past blows up to ifiity i fiite time. Ay solutio through a poit with y < decreases steadily, is cocave dow ad blows up to mius ifiity i fiite time ad goes to y = + i the future ad i the past blows goes to as t. Ay solutio through a poit with < y < icreases steadily, is cocave up whe y ad dow whe y ad goes to y = + i the future ad i the past goes to + as t. 4

5 We solve by writig: dt dy = = y y t = If y = y whe t = t, we get: y = y y = y + y, y + Ay dy = l, y y Ay y = et, Ay = ye t. Ay = y e t, A = y e t y, y e t y y = ye t, y y = y ye t t, yy e t t y + = y e t t y e t t y e t t y + = y y + y e t t. I this form the solutio is valid for ay give iitial poit t, y i the t, y plae. The blow up time t b is give by the formula: = y + y e t t b, e t b t = y, y t b = t + l y. This blow-up time is well-defied provided y < i which case t b > t, or y > i which case t b < t. I all the other cases: y, blow up does ot occur ad i those cases the y-value of the solutio lies i the iterval [, ] for all time. 5

6 dy dt = y t. Here the curve where slope m is plotted is the straight lie y t = m, or y = t + m. We plot slope alog the lie y = t +, which itself has slope, so the slopes are taget to this lie, correspodig to the solutio y = t + of the differetial equatio. If the slope m >, we plot slopes above the lie ad we see that the solutio goes quickly away from the lie, goig forwards i time ad rapidly apporoaches the lie, goig backwards i time. Plottig the other slope values, we see that the slopes of zero give local maima, ad the solutio icreases to a sigle local maimum, the decreases rapidly. Goig back i time we see that every solutio approaches the lie y = t + as t. Goig forward i time, we see that every solutio that lies above or o the lie y = t + goes to ifiity as t icrease ad every solutio that lies below the lie goes to mius ifiity, as t goes to ifiity. We ca solve by first oticig that y = t + C obeys the differetial equatio, provided: d t + C = t + C t, = C. dt So we have the particular solutio: y = y p = t +. To this we add the geeral solutio y = y h of the associated homogeeous equatio: dy dt = y This is the stadard epoetial growth equatio with growth costat k = ad the geeral solutio: y = y h = Ae t. The the full solutio of the give differetial equatio is: y = y h + y p = Ae t + t +. 6

7 Alteratively we use a itegratig factor, first re-writig the equatio i the stadard form: dy dt y = t. We use the itegratig factor e dt = e t, givig: d dy dt ye t = e t dt y = te t, ye t = te t dt. We itegrate by parts: u = t, du = dt, dv = e t dt, v = e t dt = e t. te t dt = udv = uv vdu = te t e t dt = e t t +C. So we get: ye t = e t t + C, y = e t e t t +C = t+ Ce t = t++ae t, where A = C. So our geeral solutio i either case is y = Ae t + t + with A costat. Whe A = the solutio is the straight lie y = t +. Whe A >, the solutio always lies above the straight lie y = t +, approachig it very closely from above for large egative t. I this case we have y = Ae t + > ad y = Ae t >, so the solutio icreases away from the lie y = t + as t icreases ad is cocave up, goig to ifiity as t. Whe A <, the solutio always lies below the straight lie y = t +, approachig it very closely from below for large egative t. I this case we have y = Ae t + ad y = Ae t <, so the solutio is always cocave dow. There is a local ad absolute maimum at the oly critical poit: l A [, ]. So the graph curves away from the lie, iitially icreasig, util it reaches its global maimum ad the decreases rapidly for all later times, goig to as t. Note that all solutios approach the lie y = t + arbitrarily closely for large egative t. Also every solutio is defied for all time ad has rage the whole real lie if A, whereas if A < has rage is, l l A ]. 7

8 Questio Let F = 5 l d. By itegratig by parts, determie a recursio relatio relatig F ad F. Hece determie F 4. We itegrate by parts: u = l, du = l d, dv = d, v = dv = d =. The we have, for ay : = =5 = F = = [l ] 5 5 l d udv = [uv] =5 = =5 5 = 5l5 5 = vdu l l d = 5l5 F. The we have, puttig = 4, = ad = : F 4 = 5l5 4 4F, F = 5l5 F, F = 5l5 F. 8

9 5 Fially to start the recursio, we eed to kow F = ld. We itegrate by parts with u = l, du = d, dv = d ad v = : The we have: F = = 5 =5 = ld udv =5 = [uv] =5 = vdu = 5 = [ l] 5 = 5 l5 5 d d = 5 l5 [] 5 = 5 l5 4. F = 5l5 F = 5l5 l5 + 8, F = 5l5 F = 5l5 5l5 + l5 4 F 4 = 5l5 4 4F = 5l5 4 l5 + 6l5 l =

10 Questio 4 Fid the iterval of covergece of each of the followig series. Ca you idetify the series i questio? f = = 5. This series is geometric with first term a, ratio r ad sum a = 5 9 f = =, r = 5, 9 a r = = The series coverges provided r <, so provided: 5 9 <, 5 < 9, 5 < 9, 9 < 5 < 9 4 < <. So the iterval of covergece is the ope iterval 4,. a r give by:

11 g = =!. We use the ratio test assumig : a =!, a + = + +! = + +!, a + +! = = a +! So we have: r = lim a + a = lim =. Sice the limitig ratio is less tha oe for all o-zero, by the ratio test, we see that the series coverges for all real actually it also coverges for all comple, by essetially the same argumet. We write out the series: g = +! + 4 4! + 6 6! +... We otice a resemblace with the stadard series for e which also coverges for all : e = +! +! +! + 4 4! + 5 5! + 6 6! We see that we wat oly the eve powers of the epoetial series. We replace by, givig the series for e which also coverges to e, for all : e =! +!! + 4 4! 5 5! + 6 6! Add the series for e to the series for e ad divide by to give eactly the series for g. So we have, for all real or comple : g = e + e = cosh.

12 h = = +. We use the ratio test assumig : a = a + = a + a = So we have: +, = + + +, r = lim a + a = lim = +. + =. So by the ratio test, we see that the series coverges whe < ad diverges whe >. I the borderlie case whe = we see that the absolute value of the term a = + is ad the series: + + coverges = with sum, by telescopig: k = + = k = = + k + as k. So the give series coverges absolutely whe =, so coverges. So the iterval of covergece is the closed iterval [, ].

13 We may sum the series as follows. We have by partial fractios, valid whe < < : h = + = + = = = + = = + = + = = + = + + = = = + = = = +. = =. = Now the series is recogized as the stadard Taylor series for = l, so our sum is, provided < < : h = + l. This has the correct limit of as, by L Hopital: lim + l = lim l = +l =. As, the term l goes to zero givig the correct value of h =, determied above by the telescopig series method. Whe +, we get the value l which ca be show to be correct also, so the formula for h suitably iterpreted, represets h o the etire iterval of covergece [, ].

14 Questio 5 Evaluate each of the followig itegrals, or eplai why the itegral i questio does ot eist. 5 t t t dt We have: 5 t 5 t t dt = t t t + dt. Near the poit t = the itegrad is approimately whose itegral 4t diverges logarithmically as t +, so the give itegral diverges. Alteratively we first use partial fractios to compute the idefiite itegral: t t t + = A t + B t +, t = At + + Bt, = A, = 4B, t 4 t t + dt = t + dt t + = l A t t + 4 Agai as t + this diverges, so, sice = lies i the itegratio regio, the give defiite itegral diverges. t 4 + t dt As t +, the itegrad is approimately 4t whose itegral coverges at, by direct itegratio or by the p-test p = <, which gives covergece at a fiite poit. As t, the itegrad is approimately t whose itegral coverges at, by direct itegratio or by the p-test p = >, which gives covergece at ifiity. 4

15 So the itegral coverges, so we have to calculate it. We substitute: u = t, so t = u. dt = udu. As t rages over the o-egative reals, so does u. So we get: = Now we substitute agai: = u = ta, so = arcta t 4 + t dt u4 + u 4 + u u. 4 + u = ta = 4 sec. du = sec d. udu du As u rages over the o-egative reals, goes from arcta = u to = lim arcta = π u. So the itegral becomes: t 4 + t dt = du 4 + u = π sec udu 4 sec u = π du = [u] π = π 4. 5

16 π cottdt π Near the origi, the itegrad cott = cost sit is approimately, so its t itegral diverges logarithmically. Alteratively the idefiite itegral of cott is la sit which blows up to ifiity as t. Sice t = is i the regio of itegratio, the give itegral diverges. te t dt The idefiite itegral of te t is e t + C, so we have: = lim a Alteratively, we chage variables: u = t, so t = u. du = tdt. a te t dt = lim te t dt a = lim ] a a [e t e a =. As t rages over the o-egative reals, so does u. So we get: te t dt = e u du = a lim e u du a = lim [e u ] a a = lim e u = a. 6

17 Questio 6 A wire i the plae is give i polar co-ordiates by the equatio r = 4 cosθ, for θ π where the polar ais is also the positive -ais. 4 Sketch the wire ad determie its legth ad cetroid, givig your reasos. If the desity at ay poit of the wire is, determie its ceter of mass. The curve r = 4 cosθ is a circle: usig the relatios = r cosθ, y = r siθ ad r = + y, we fid its Cartesia equatio as follows: r = 4 cosθ, r = 4r cosθ, + y = 4, y = 4, + y =. So the circle is of radius ad has ceter,. As θ goes from to π, the wire goes couter-clockwise aroud the circle from 4 the poit 4, to the poit with θ = π π 4 ad r = 4 cosθ = 4 cos = ; the 4 Cartesia co-ordiates of this poit are the: r[cosθ, siθ] = [ π π ] cos, si 4 4 = [, ] = [, ]. So the wire occupies the first quarter of the circle, goig aroud couter-clockwise from the poit 4,. 7

18 The legth of the wire is the L = π = π. 4 By symmetry is cetroid lies o the lie y =. The y compoet of its cetroid is: y = yds = yds. L Γ π Γ Here Γ deotes the wire, parametrized i a couter-clockwise directio. We parametrize the wire from its ceter: The t rage is the from t = to t = π. The we have ds = dt, givig: [, y] = [ + cost, sit]. = π y = π π Γ yds sitdt = π [ cost] π = 4 π. [ So the cetroid is at + 4 π, 4 ] = [.79544,.79544]. π 8

19 If the desity of the wire is ρ =, the the mass is: m = Γ ρds = 4 π Its vector momet is: = 8 = 8 = 6 π π = 8 + costdt = 8[t + sit] π = 4π + 8 = π The is ceter of mass is at: M = = 4 Γ π ρ[, y]ds [, y]dt [ + cost, sit + cost ] dt [ + 4 cost + cos t, sit + cost]dt [ + 4 cost + cost, sit + cost]dt [t + 4 sit + sit, + cost ] π [ ] π = 8 + 4, + 4 = 4[π + 8, 6]. m M = [π + 8, 6] = [ , ]. π + 9

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