Math 2260 Written HW #8 Solutions
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1 Mth 60 Written HW #8 Soltions. Sppose nd b re two fied positive nmbers. Find the re enclosed by the ellipse + y b =. [Hint: yo might wnt to find the re of the hlf of the ellipse bove the -is nd then doble it.] Answer: Here is pictre of the ellipse, which hs -intercepts t = ± nd y-intercepts t y = ±b. I cn find the re of the top hlf of the ellipse nd then doble it. To do so, I ll solve for y s fnction of. Certinly, y b =, so nd hence ( ) y = b ( ) y = ± b = ±b. The top hlf of the ellipse corresponds to choosing the positive sqre root in the bove. For lter convenience, notice tht = ( ), nd hence the top hlf of the ellipse is given by y = b.
2 Since the ellipse etends from = to =, the re nder the bove crve (which is the re of the top hlf of the ellipse) is eql to b d. Therefore, the re of the whole ellipse is eql to b d. To evlte this integrl, mke the trig sbstittion = sin θ. Then d = cos θ dθ. Also, since sin θ = /, we see tht sin θ = / = when =, mening tht θ =. Likewise, when =, sin θ = / =, so θ = π/. Ptting this ll together, then, the bove integrl is eql to b π/ sin θ cos θ dθ = b π/ cos θ cos θ dθ Therefore the re of the ellipse is πb. = b = b = b π/ π/ π/ cos θ dθ + cos(θ) dθ ( + cos(θ)) dθ [ = b θ + sin(θ) ] π/ [( π ) ( π )] = b = πb.. Evlte the integrl e 4t + e t e t e t dt. + Answer: First, mke the -sbstittion = e t. Then d = e t dt nd so the bove integrl is eql to ( e 3t + e t ) e t dt 3 + e t = + d. + Dividing nmertor by denomintor yields )
3 so or integrl is eql to ( + ) d = + ( + + ) d = + d+ + d + d. The first integrl is esy. For the second, let v = +, then dv = d nd so + d = dv v = ln v + C = ln( + ) + C. For the third integrl, let = tn θ. Then d = sec θ dθ nd so + d = sec θ tn θ + sec θ dθ = sec θ dθ = θ + C = rctn() + C. Therefore, the bove -integrl is eql to + ln( + ) rctn + C, where I ve combined ll the constnts into the single constnt C. Now, since = e t, this mens tht the originl integrl is eql to e t + ln(et + ) rctn(e t ) + C. 3. Sociologists sometimes se the phrse socil diffsion to describe the wy informtion spreds throgh popltion. The informtion cold be nything: scientific brekthrogh, news of ntrl disster, litercy, etc. In sfficiently lrge popltion, the nmber of people who hve the informtion is treted s differentible fnction of time t, nd the rte of diffsion, d/dt, is ssmed to be proportionl to the nmber of people who hve the informtion times the nmber of people who do not. This leds to the eqtion d dt = k(n ), where N is the nmber of people in the popltion. The constnt k determines how fst the informtion spreds: yo wold epect k to be reltively lrge for, sy, celebrity gossip, which is ssimilted lmost instntly, wheres it wold be qite smll for informtion tht only dissemintes very slowly (like how to compte integrls!). Sppose t is in dys, k = /50, nd two people strt rmor t time t = 0 in popltion of N = 000 people. () Find s fnction of t. [Hint: Logrithm identities re yor friends!] Answer: First, we seprte vribles nd integrte: d (000 ) = dt. ( ) 50 3
4 The right hnd side will be esy, so let s focs on the left hnd side. Using the method of prtil frctions, we gess tht Clering denomintors yields or, eqivlently, which yields the system of eqtions The second eqtion implies tht A = Therefore, ( /000 d (000 ) = (000 ) = A + B 000. = A(000 ) + B, = (B A) + 000A, 0 = B A = 000A. 000, while the first implies tht B = A = /000 ) d = ln ln(000 ) 000 (we don t hve to worry bot bsolte vles, becse we re only interested in vles of between 0 nd 000). Sbstitting this into eqtion ( ) nd integrting the right hnd side gives s Mltiply both sides by 000 to get 000 ln ln(000 ) = t + C. ln ln(000 ) = 4t + 4C. Now here s where the logrithm identity ln ln b = ln ( ) b comes in hndy. We cn se this identity to re-write the left hnd side: ( ) ln = 4t + 4C. 000 Since we wnt to solve for, eponentite both sides to get 000 = e4t+4c = e 4C e 4t = Ae 4t, where I ve let A = e 4C. This is still little nplesnt, bt it gets nicer if I tke the reciprocl of both sides: 000 = Ae 4t. 4
5 The left hnd side is eql to 000, so I hve tht 000 = + Ae4t + = Ae4t Ae 4t (where I fond common denomintor nd dded the terms on the right hnd side). Tking the reciprocl gin yields nd so 000 = Ae4t + Ae 4t, Ae4t = Ae 4t. Since the rmor strted with two people, we know tht (0) =. We cn se this to solve for the constnt A: Dividing both sides by 000, we hve Ae4(0) (0) = Ae 4(0) = 000 A + A 500 = A + A. To solve for A, tke the reciprocl of both sides: so nd hence Finlly, then, this mens tht 500 = + A A = A +, A = 499 A = 499. (t) = 000 Here s grph s fnction of t: 499 e4t e4t + = e4t e 4t. 5
6 (b) When will hlf the popltion hve herd the rmor? (This is when the rmor will be spreding the fstest.) Answer: The qestion is: t wht time t 0 will we hve tht (t 0 ) = 500? Well, evlte the fnction t t = t 0 : Diving both sides by 000 yields Tke the reciprocl of both sides to get so we hve nd so (t 0 ) = = e 4t. 0 = e 4t = e4t 0 Tke the ntrl logrithm of both sides: = 499 = t 0 = ln(499), = 499 +, nd hence ln 499 t 0 = Therefore, it tke bot one nd hlf dys for hlf the popltion to hve herd the rmor. 6
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