Honors Calculus Quiz 9 Solutions 12/2/5

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1 Honors Calculus Quiz Solutions //5 Question Find the centroid of the region R bounded by the curves 0y y + x and y 0y + 50 x Also determine the volumes of revolution of the region R about the coordinate axes We first find where the two parabolas meet: So y, x 8 or y 8, x 8 0y y + y 0y + 50, 0 3y 30y + 48, 0 y 0y + 6, 0 y y 8 Plotting we see the region R lies between the two parabolas with the curve x y 0y + 50 on the left and x + 0y y + on the right Also we see that the region is symmetrical about the line y 5, so the centroid C of R lies on that line, so C [X, 5] for some X To verify the symmetry we put y t + 5, then we have: x + 0y y + 0t + 5 t t + 50 t + 0t t, x y 0y + 50 t + 5 0t t + 0t + 5 0t t Since clearly both of these parabolas are symmetrical under t t, the original graphs are symmetrical about the line y 5, as required

2 The area A of R is, using horizontal strips: 8 A 8 x + x dy 0y y + y 0y + 50dy 8 30y 3y 48dy [5y y 3 48y] If V x is the volume obtained by rotating about the x-axis and V y is the volume obtained by rotating around the y-axis, then we have by Pappus Theorem: [V y, V x ] πac 6π[X, 5], V y 6πX, V x 080π 33 Finally, using horizontal washers, we get for V y : V y π π 3 3 6π 8 3 x + x dy π 7 t t dt t 3t 4 dt 6π [ 43t 6t 3 t5 5 ] 3 0 6π t t 4 dt π π705 π Then we have: X V y 6π 555π 56π So the centroid of the region R is at C [7, 5] [44, 5] 5

3 Question Discuss the convergence of the following series and see if you can find a formula for the sum explicitly when the series converges nn + n We use partial fractions: nn + A n + B n +, An + + Bn, nn + B, A, n n + Writitng out the series we get for the first n terms: n n + This telescopes, with each negative term cancelling with the next but one positive term, so the n-partial sum, s n is: s n + n + n + 3n + n + n + n + 4n + n + 4n + n + 3n + n + 6 n 4 n n3n + 5 4n + n + Then the series sum s is the limit of s n as n so is s 3 4 3

4 n4 n 3 n n The ratio test gives: a n n4n 3 n, a n+ r n a n+ a n n + 4n+ 3 n+ n + 4 n+ 3 n 3 n+ 4n + n As n, we get the limiting ratio r n r 4, which is less than, so the series converges by the ratio test For the sum, see below n x n+ 4 n n +! n0 The standard series of siny is, valid for any y real or complex, in fact: n y n+ siny n +! n0 n4 n Substitute y x and we get: x sin n0 n x n+ n+ n +! Since n+ 4 n, this is one-half of the given sum, so the given sum x is sin, for any x 4

5 fx n0 n + x n n Define Gy by the formula: Gy n + y n + y + 3y + 4y 3 + n0 Then the given series is fx G x Note that if y, the individual terms in the series Gy have sizes larger than n, which goes to infinity as n goes to infinity, so the series diverges So we may assume that y < We see that for y, the n-th partial sum is: G n y + y + 3y + + ny n d dy + y + y + + y n d dy y n+ y y n + yn y y n+ y nyn+ n + y n + Since y <, the terms ny n+ and n + y n go to zero as n goes to infinity, so the series converges with sum: Gy lim n y nyn+ n + y n + y So the series for fx converges iff x <, so on the open interval,, with limit: x fx G x 4 x 5

6 We return to the first series of this problem, the convergent series: A n n4 n 3 n If we write out the terms of this series, we get: A G

7 Question 3 Solve each of the following differential equations and discuss the behavior of each solution: dy dt + 3y cost, y0 4 We first find a particular solution We try y A cost + B sint Then we need: cost y +3y A sint+b cost+3a cost+b sint, 0 sint A + 3B + costb + 3A So we want A + 3B 0 and B + 3A 0 The first equation gives 6A + B 0, the second 4B + 6A 4 0 Adding these equations gives: 3B 4 0, so B 4 3 Then A 3B 6 3 So we have the particular solution: y p 3 cost + sint 3 The associated homogeneous problem is: dy dt + 3y 0, dy dt 3y We recognize this as a special case of the standard equation for exponential growth/decay, with general solution: y h Ce 3t So the general solution of the given problem is: y y h + y p Ce 3t + 3 cost + sint 3 7

8 For the initial condition, y0 4, we put t 0 and y 4, giving: 4 C + 6 3, So the required solution is: C 46 3 y 3 3e 3t + 3 cost + sint Plotting, we see that for t negative the graph is always concave up and decreasing and y goes to infinity as t For t positive, the exponential term quickly dies off and the solution is then an almost perfect sinusoidal curve, of amplitude and period π Indeed already by the first local maximum, which occurs at t gives the y-value y 05548, which is very close to the sinusoidal amplitude 8

9 dy dt y3, y ty + t This is separable, we separate and integrate: ln t dy dt y3 ty + t y 3 ty +, y + dy dt y 3 t, y + dy y 3 y + y 3 dt t, dy C y y, ln t + y + y C The intitial condition y gives: C ln ln t y + y y ln t 3 + y + 0, 4 y ± 4 ln t 3 lnt ± ln t 4 ln t 3 When t, y is positive, which requires the negative square root, giving the required solution as: y + ln t 3 4 ln t

10 Plotting the graph of the solution we see that it is defined for the interval t < e The graph is symmetrical about the t-axis For positive t it has a cusp of infinite slope at the origin Since the slope is infinite there, even though y is well-defined, we should probably regard the solution as being strictly valid for t > 0 only, so on the open interval 0, e 3 4 Then y increases steadily from 0, going to infinity as t e 3 4 The graph is initially concave down and switches to concave up at the inflection point: , The y-value of the inflection point is the unique positive root y + of the cubic y 3 + 5y 4y 4 0, then the t-value is e We may give an explicit formula for y + : 3 4 y + y + y cosθ sinθ, Here θ is given as follows: 3θ arctan 76 8 The cubic arises by taking the logarithmic derivative of the differential equation: y y 3 ty +, y lny 3 lny lny + lnt, 3y y y y y + t 3y t y + y 3 y + ty + 3y y + y 3 y + ty + y3 + 5y 4y 4, y y 3 t y + 3 y3 + 5y 4y 4 So the inflection point arises, for y given by the positive root of the cubic, as described above 0