Math 115 HW #8 Solutions

Transcription

1 Math 115 HW #8 Solutions 1 The function with the given graph is a solution of one of the following differential equations Decide which is the correct equation and justify your answer a) y = 1 + xy b) y = 2xy c) y = 1 2xy Answer: If b) were correct, then, when x = 0, we would have that y = 0 However, the slope of the tangent line is clearly positive when x = 0 in the picture, so b) cannot be correct If a) were correct, then y would be positive for all positive values of x However, the slope of the tangent line is clearly negative for most positive x, so a) cannot be correct c) must be the equation which is solved by the function whose graph we see 2 Find an equation of the curve that passes through the point 0, 1) and whose slope at x, y) is xy Answer: If the slope of the curve at the point x, y) is xy, then we have that the curve is the graph of a function which solves the differential equation dy dx = xy This is a separable equation, so we can find solutions implicitly by separating variables and integrating: dy y = xdx Exponentiating both sides yields for A > 0 a constant ln y = x2 2 + C y = e C e x2 /2 = Ae x2 /2 1

2 The functions y = Ae x2 /2 give all the curves whose slopes at x, y) are xy; however, only one of these curves can pass through the point 0, 1), namely the one for which 1 = Ae 02 /2 = A Therefore, the equation of the curve we re looking for is y = e x2 /2 3 Find the orthogonal trajectories of the family of curves y 2 = kx 3 An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family orthogonally, that is, at right angles) Sketch several members of each family on a common graph feel free to use Mathematica or any other graphing utility to help create this sketch) Answer: Since the orthogonal trajectories to this family are those curves whose tangent lines are orthogonal to the tangent lines of the family, we should determine the slopes of the tangent lines to the given family of curves To do so, differentiate implicitly: Solving for dy dx, we see that dy 2y dy dx = 3kx2 dx = 3kx2 2y Since y 2 = kx 3, we know that k = y2 x 3 ; plugging this into the above gives that dy dx = 3 y2 2y x 3 x 2 = 3y 2x Since the orthogonal trajectories should have perpendicular tangent lines, their slopes will be negative reciprocals of the above In other words, the orthogonal trajectories of the given family are the solutions of the differential equation dy dx = 2x 3y This is a separable equation, so we separate variables and integrate: 3ydy = 2xdx 3 2 y2 = x 2 + C, so the orthogonal trajectories are the curves x y2 = C 2

3 Figure 1: Original family in red, orthogonal trajectories in blue 4 A glucose solution is administered intravenously into the bloodstream t a constant rate r As the glucose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that time Thus a model for the concentration C = Ct) of the glucose solution in the bloodstream is where k is a positive constant dc = r kc a) Suppose that the concentration at time t = 0 is C 0 Determine the concentration at any other time t by solving the differential equation Answer: This is a separable equation, so separate and integrate: dc r kc = 1 ln r kc = t + D k Multiplying both sides by k and exponentiating gives r kc = e Dk e t = Ae kt Allowing A to be either positive or negative lets us get rid of the absolute value signs; solving for C gives Ct) = r Ae kt k 3 = r k A k e kt

4 If we plug in t = 0, then C 0 = r k A k e k 0 = r k A k, so A = r kc 0 Hence, the concentration at time t is given by the equation or, equivalently, Ct) = r k r kc 0 e kt k Ct) = r r ) k k C 0 e kt b) Assuming that C 0 < r/k, find lim t Ct) and interpret your answer Answer: We take the limit [ r r lim Ct) = lim 0) t t k k C e kt] = r k since e kt 0 as t 5 A vat with 500 gallons of beer contains 4% alcohol by volume) Beer with 6% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate What is the percentage of alcohol after one hour? Answer: Let at) be the volume of alcohol in gallons) in the vat after t minutes The vat starts out with = 20 gallons of alcohol in it, so a0) = 20 The rate of change of alcohol in the vat is We know that the rate in is da = rate in) rate out) = 03 gal/min On the other hand, the rate of alcohol going out of the vat is at) = at) 100 gal/min Therefore, da = 03 a 100 = 30 a 100 This is a separable equation, so we can separate variables and integrate: da 30 a = 100 ln 30 a = t C 4

5 If all the beer in the vat were 6% alcohol, then there would be 30 gallons of alcohol in the vat Hence, at) 30 for all t, so 30 a 0, meaning that the absolute value signs above are superfluous Multiplying by 1 and exponentiating yields 30 a = e C e t/100 = Ae t/100 Therefore, Since a0) = 20, we know that at) = 30 Ae t/100 so A = = 30 Ae 0/100 = 30 A, Therefore, the volume of alcohol in the vat after t minutes is In particular, after 1 hour there will be at) = 30 10e t/100 a60) = 30 10e 60/ = 245 gallons of alcohol in the tank Hence, the percentage of alcohol in the tank is = 49% 6 Biologists stocked a lake with 400 fish and estimated the carrying capacity the maximal population for the fish of that species in that lake) to be 10,000 The number of fish tripled in the first year a) Assuming that the size of the fish population satisfies the logistic equation = rp 1 P ), N find an expression for the size of the population after t years Answer: As we saw in class, solutions of the logistic equation are of the form P t) = P 0 N N P 0 )e rt + P 0, where P 0 is the initial population and N is the carrying capacity Hence, in this case P 0 = 400 and N = 10, 000, so the fish population is given by P t) = , , )e rt = e rt Since the fish population tripled in the first year, we know that P 1) = = 1200, so we have that 1200 = 9600e r

6 Therefore, 9600e r = 1200 = 40, so 10,000 e r = = = 11 Taking the natural log of both sides gives that ) 11 r = ln Hence, the population after t years is given by or P t) = e ln )t P t) = 9600 ) 11 t = 10, 000, 3 b) How long will it take for the population to increase to 5000? Answer: We want to solve for t 0 such that P t 0 ) = 5000 Therefore, so Hence, 9600 Taking the natural log of both sides, ) 11 t 0 ln 5000 = 9600 ) 11 t , ) 11 t = = ) 11 t0 = = 1 24 ) 1 = ln, 24 so t 0 = ln ) 1 24 ln ) so the population will increase to 5000 in about 27 years 7 a) Show that if P satisfies the logistic equation, then d 2 P 2 = r2 P 1 P N ) 1 2P N Answer: If P satisfies the logistic equation, then = rp 1 P ) N 6 )

7 Differentiating this equation gives, d 2 P 2 = r = r = r 1 P ) + rp 1 N N [ 1 P ) P ] N N 1 2P ) N ) Substituting = rp 1 P ) N into the right side of the above yields d 2 P 2 = r2 P 1 P ) 1 2P ), N N as desired b) Deduce that a population grows fastest when it reaches half its carrying capacity Answer: The population will be growing fastest when reaches its maximum To find the maximum, we note that the critical points of occur when d2 P = 0; ie when 2 0 = r 2 P 1 P ) 1 2P ) N N This can only occur when P = 0, 1 P N = 0 or 1 2P N = 0 The first two never happen unless the initial population is either 0 or N in which case the population never grows at all) In the third case, 2P N = 1, Meaning that P = N 2, so, indeed, the population is growing fastest when it reaches half its carrying capacity 8 Consider a population P = P t) with constant relative birth and death rates α and β, respectively, and a constant emigration rate m, where α, β, and m are positive constants Assume that α > β Then the rate of change of the population at time t is modeled by the differential equation = rp m where r = α β a) Find the solution of this equation that satisfies the initial condition P 0) = P 0 Answer: This is a separable equation, so separate variables and integrate: rp m = 1 ln rp m = t + C, r 7

8 so rp m = e rc e rt = Ae rt where A is some constant Therefore, P t) = A r ert + m r When t = 0, P 0) = P 0, so P 0 = A er 0 + m r r = A r + m r Hence, so we see that or, equivalently, A = rp 0 m, P t) = rp 0 m e rt + m r r P t) = P 0 m ) e rt + m r r b) What condition on m will lead to an exponential expansion of the population? Answer: The above equation for P t) yields exponential growth when In other words, if P 0 m r > 0 m < rp 0, then there will be an exponential expansion of the population c) What condition on m will result in a constant population? A population decline? Answer: The condition P 0 m r = 0 will yield constant population, so the condition on m is that m = rp 0 Likewise, when m > rp 0, the population will decline d) In 1847, the population of Ireland was about 8 million and the difference between the relative birth and death rates was 16% of the population Because of the potato famine in the 1840s and 1850s, about 210,000 inhabitants per year emigrated from Ireland Was the population expanding or declining at that time? Answer: In this model, P 0 = 8, 000, 000, r = 0016 Hence, rp 0 = , 000, 000 = 128, 000 Since m = 210, 000, we see that m > rp 0, so the population was declining 8