Solutions to Team Homework 5

Math 116 Solutions to Team Homework 5 (1a) We have x = r cos(θ) = 3 sin2 (θ), so as θ increases from 0 to π/2 we see that x will increase from 0 to 3. The vertical asymptote is thus x = 3, since its x-value is the limit of the x-values of points as θ approaches π/2 from below. To rewrite the equation x = 3 as a polar curve r = f (θ), we must choose f (θ) so that the corresponding x-coordinate is 3, namely so that 3 . Hence the equation in polar coordinates of the 3 = f (θ) · cos(θ), which means that f (θ) = cos(θ) 3 vertical asymptote is r = cos(θ) . √ (1b) The line y = 3x has the polar equation θ = π3 , since the angle θ between this line and the √ √ positive x-axis satisfies tan(θ) = 3 so that θ = tan−1 ( 3) = π3 . Thus the light blue region is sin2 (θ) 3 defined by the polar equations 0 ≤ θ ≤ π3 and 3cos(θ) ≤ r ≤ cos(θ) , so its area is Z 0 π/3 
2
2 Z Z 1  3 2 1  3 sin2 (θ) 2  9 π/3 1 − sin2 (θ) 9 π/3 1 − 1 − cos2 (θ) dθ = dθ − dθ = 2 cos(θ) 2 cos(θ) 2 0 cos2 (θ) 2 0 cos2 (θ)
Z Z 9 π/3 2 cos2 (θ) − cos4 (θ) 9 π/3 1 − 1 − 2 cos2 (θ) + cos4 (θ) dθ = dθ = 2 0 cos2 (θ) 2 0 cos2 (θ) Z
9 π/3 = 2 − cos2 (θ) dθ. 2 0 To integrate this, we use the fact that cos(2θ) = 2 cos2 (θ) − 1, which lets us solve for cos2 (θ) as sin(2θ) R R cos(2θ)+1 +θ 2 2 , so that cos (θ) dθ = dθ = + C. Thus the area of the light cos2 (θ) = cos(2θ)+1 2 2 2 blue region is √ √ 9 sin(2θ) θ  π/3 9  3θ sin(2θ)  π/3 9  3 π 3/2  9π 9 3 2θ − − = − = · − = − . 2 4 2 0 2 2 4 2 2 3 4 4 16 0 (1c) This region is defined by the equations 0 ≤ θ < π/2 and Z 0 π/2  3 sin2 (θ) cos(θ) ≤r≤ 3 , cos(θ) so its area is 1  3 2 1  3 sin2 (θ) 2  − dθ. 2 cos(θ) 2 cos(θ) (1d) Just as in (1b), the integral in (1c) can be written more simply as Z π/2
9 2 − cos2 (θ) dθ, 2 0 which equals 93 sin(2θ)  π/2 9 3 π 27π = · · = θ− . 2 2 4 2 2 2 8 0 (1c, alternate solution) A different way to get an integral for the area in (1c) is by first converting from polar coordinates to Cartesian coordinates. Points on the red curve have x = sin3 (θ) r cos(θ) = 3 sin2 (θ) and y = r sin(θ) = 3cos(θ) , where 0 ≤ θ < π/2. Since y is expressed in terms of sin(θ) and cos(θ), in order to write y in terms of x we’ll first write sin(θ) and cos(θ) in terms of x. We can do that because we have sin2 (θ) = x/3 and cos2 (θ) = 1 − sin2 (θ) = 1 − x/3. Since p p 0 ≤ θ < π/2, we have sin(θ) ≥ 0 and cos(θ) > 0, so that sin(θ) = x/3 and cos(θ) = 1 − x/3, whence p 3( x/3)3 3 sin3 (θ) x3/2 =p , y= =√ cos(θ) 3−x 1 − x/3 √ √ where in the last step we multiplied by 3/ 3. Thus the area in (1c) is the area under the curve R 3 3/2 3/2 y = √x3−x for 0 ≤ x ≤ 3, namely 0 √x3−x dx. But that integral is much more painful to compute than the polar one was. (2a) The positive values of θ for which r = 0 are precisely the values of θ for which sin(θ) = 0, namely θ = π, 2π, 3π, 4π, . . . . The 7 smallest are π, 2π, . . . , 7π. (2b) The polar curve traces out the first loop L1 when π ≤ θ ≤ 2π. It traces out L2 when 2π ≤ θ ≤ 3π. It traces out the n-th loop when nπ ≤ θ ≤ (n + 1)π. R (n+1)π 1  sin(θ) 2 (2c) The area An of the n-th loop is An = nπ dθ. 2 θ P∞ R (n+1)π 1  sin(θ) 2 P∞ R ∞ 1  sin(θ) 2 dθ = π 2 dθ. Since 0 ≤ sin2 (θ) ≤ 1, (2d) Note that n=1 An = n=1 nπ 2 θ θ  2 R∞ 1 1 we have 0 ≤ 21 sin(θ) < when θ > 0. Now dθ converges by the p-test (p = 2), so the 2 θ θ π θ2   2 R∞ P Comparison Test implies that π 12 sin(θ) dθ converges, which says that ∞ n=1 An converges. θ (2d, alternate solution) Since 0 ≤ sin2 (θ) ≤ 1, we have 0 ≤ Z (n+1)π 0≤ nπ 1  sin(θ) 2 dθ < 2 θ Z 1 2  sin(θ) θ (n+1)π nπ 2 < 1 θ2 when θ > 0, so that 1 dθ. θ2 P (n+1)π −1 1 Thus 0 ≤ An < −1 = (n+1)π − −1 = (n2 +n)π < n12 . We know that ∞ n=1 θ nπ nπ P∞ the p-test (p = 2), so the Comparison Test implies that n=1 An converges. (2e) Writing f (θ) := We compute sin(θ) , θ we know that the arclength of the n-th loop is 1 n2 converges by R (n+1)π p nπ f (θ)2 + f 0 (θ)2 dθ. sin2 (θ)  θ cos(θ) − sin(θ) 2 sin2 (θ) θ2 cos2 (θ) − 2θ cos(θ) sin(θ) + sin2 (θ) + = + f (θ) + f (θ) = θ2 θ2 θ2 θ4 2 1 2 cos(θ) sin(θ) sin (θ) = 2− + , θ θ3 θ4 2 0 2 so the arclength is Bn = R (n+1)π q 1 follows that θ2 − 2 cos(θ) sin(θ) θ3 2 sin(θ) 1 − 2 cos(θ) + sinθ4(θ) , then θ2 θ3 R∞q 1 eventually, so since π dθ 2θ2 (2f) Defining f (θ) := Here f (θ) > nπ 1 2θ2 R∞p f (θ) dθ π + sin2 (θ) θ4 dθ. R∞p P just as in (2d) we get ∞ f (θ) dθ. n=1 Bn = π R∞ 1 = π √2θ dθ diverges by the p-test (p = 1), it diverges by the Comparison Test. (2f, alternate solution) We have n we get 2 1 θ2 sin(θ) − 2 cos(θ) + sinθ4(θ) > θ3 Z (n+1)π r Bn > nπ 1 dθ = 2θ2 Z (n+1)π nπ Still assuming n is big enough, if nπ ≤ θ ≤ (n + 1)π then Z (n+1)π Bn > √ nπ 1 2θ2 1 θ eventually, so that for big enough 1 √ dθ. 2θ ≥ 1 , (n+1)π so that 1 (n + 1)π − nπ 1 dθ = √ =√ 2(n + 1)π 2(n + 1)π 2(n + 1) P∞ 1 and thus B > eventually. Since n n=1 2n P∞ B diverges by the Comparison Test. n n=1 1 2n diverges by the p-test (p = 1), it follows that Note: the situation here is very common for calculus problems involving sequences of smaller and smaller geometric objects: very often the sum of the areas will converge (ultimately by the p-test with p = 2, perhaps combined with the Comparison Test), while the sum of the perimeters will diverge (ultimately by the p-test with p = 1, perhaps combined with the Comparison Test). Be on the lookout for this! Note the first crucial step in the solutions to both 2d and 2f, namely to write an inequality comparing the function being integrated to a simple function of the form θ12 times a constant or 1θ times a constant. This makes it possible to use the p-test and Comparison Test. Make sure you do this first step! (3ai) For the slope field, the key observation is that when testing stability of an equilibrium solution, it suffices for the solution to be stable “eventually”, in the sense that when x is big enough the slopes point towards the equilibrium solution. Even if the slopes at points near the equilibrium point away from the equilibrium when x is negative (or even positive, but small), the solution will still be stable as long as the slopes point towards the equilibrium when x is big enough. Anyway, here the equilibrium solution is y = 0, and it is stable. (3aii) We have dy y 2 = −x dx; integrating both sides yields ln(|y|) = − x2 + C with C constant. x2 x2 x2 Raising e to both sides yields |y| = e− 2 +C = e− 2 · eC , so that y = ±eC · e− 2 , which means x2 that y = D · e− 2 where D is constant. The condition y(0) = the particular solution is y = √1 2π √1 2π says √1 2π 02 = D · e− 2 = D, so x2 · e− 2 . dy (3bi) We compute dx = (x + y) x=0 = 0 + 1 = 1. Taking the derivative with respect to x of x=0 dy d2 y dy d2 y dy both sides of dx = x + y yields dx 2 = 1 + dx , so that dx2 x=0 = 1 + dx x=0 = 1 + 1 = 2. Taking d2 y dy d3 y d2 y the derivative with respect to x of both sides of dx 2 = 1 + dx yields dx3 = dx2 , and taking further dn y dn−1 y dn y d2 y derivatives implies that dx for each n ≥ 3, so for n ≥ 3 we get dx n = n x=0 = dx2 x=0 = 2. dxn Note: The idea in (3bi) is important, namely that it’s easy to compute the Taylor series of the solution to an initial value problem. Actually in practical applications one usually just computes a low-degree Taylor polynomial, but the method is the same. P∞ dn y dxn x=0 (3bii) By definition, the Taylor series is n=0 n! xn . Since we’re given that y(0) = 1, which P∞ 2xn d0 y in this notation says dx 0 x=0 = 1, this Taylor series is 1 + x + n=2 n! . n n+1 2x x (3biii) Writing an := 2xn! , we compute limn→∞ an+1 = limn→∞ (n+1)! · 2xn!n = limn→∞ n+1 = 0. an Since this is less than 1 for every x, the Ratio Test implies that the Taylor series converges for every x, so the interval of convergence of the Taylor series is (−∞, ∞). (3biv) Note: A “closed form formula” means a formula that doesn’t involve any summations. A closed form formula for the Taylor series in (3bii) is 2ex − 1 − x. This formula is valid on the interval (−∞, ∞). Note: Here is how to find the above closed form formula. There are only five functions whose 1 Taylor series we know, namely ex , sin(x), cos(x), ln(1 + x), 1−x . So we need to interpret the Taylor series in (3bii) as a modified version of the Taylor series of one of these five functions. The key thing is the n! in the denominator P of the coefficient of xnPin the Taylor series in (3bii), ∞ 2xn xn x which singles out ex . In order to turn ex = ∞ n=2 n! , we first multiply e n=0 n! into 1 + x + P ∞ 2xn x by 2 to get 2e = n=0 n! , which matches the series we want except for the terms of degrees 0 and 1. Since the series for 2ex starts with terms 2 + 2x, and the series from (3bii) starts with terms 1 + x, in order to get from 2ex to the series in (3bii) we need to subtract 1 + x. Make sure you can do this type of analysis. (3bv) Note: This question mentioned “the function you found in part iii”, but that surely meant “the function you found in part iv”. dy For y = 2ex − 1 − x we compute dx = 2ex − 1, which indeed equals x + y. (4a) Note: Here p was written, but it should have been P . The number of susceptible people decreases at the rate that susceptible people get infected, which in turn equals the rate of interactions between susceptible and infected people, since each such interaction will result in a susceptible person becoming infected. Since each of the I(t) infected people meets b people per day, and the proportion of those people who are susceptible is roughly S(t) , we conclude that the number of interactions between susceptible and infected people per P day is roughly I(t) · b · S(t) , so that dS = −b S(t) I(t). P dt P (4b) Note: Here “rate of recovery” means the rate of change of the number of people who have recovered, or in other words, dR . dt dR The equation is dt = k · I(t). (4c) Since S + R + I = P is constant, differentiating yields with (4a) and (4b) this gives dI = − dS − dR = b PS I − kI. dt dt dt (4d) Note: Here the “chain rule” means the fact that dI dS = dS dt + dR dt + dR dt = dP dt = 0. Together dI/dt . dS/dt b S I−kI dI/dt dI P We compute dS = dS/dt = −b = −1 + PbSk . Since P, k, b are all constants, the right side is S I P just a function of S, so we can think of S as a variable and think of I as a function of S; then integrating both sides yields I = −S + Pbk ln(|S|) + C with C constant. This is the general solution to the differential equation. (You could also use separation of variables to get the same answer.) (4e) If dI dt is negative then the number of infected people decreases, so there is no epidemic. (4f) S(t) is decreasing because susceptible people can become infected (and hence non-susceptible), but there is no way for non-susceptible people to become susceptible. Since dI = (b PS − k)I, if dt is negative then (since I(0) > 0) we must have b S(0) − k < 0; but since b and P are positive P S(t) constants, and S(t) is decreasing, this implies that b P −k < 0 for each t > 0, so (since I(t) > 0) we see that dI = (b S(t) − k)I(t) is negative for each t > 0. dt P dI dt t=0 (4g) The argument is the same as (4f). Namely: if S(0) ≤ Pbk then, since S(t) is decreasing, we get S(t) < Pbk for all t > 0, so (since I(t) > 0) it follows that dI = (b S(t) − k)I(t) < 0 for all dt P t > 0, whence by (4e) there is no epidemic.