Homework 4 solutions

Homework 4 solutions - AMS 361: Applied Calculus IV (DE...

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Unformatted text preview: AMS 361: Applied Calculus IV (DE & BVP) Homework 4 Assignment Date: Wednesday (02/25/2009) Collection Date: Wednesday (03/11/2009) 8:20AM Grade: Each problem is worth 10 points Problem 4.1 (Prob 33, P. 76, 3E) The following differential equation is of two different types considered in Chapter 1-separable, linear, homogenous, Bernoulli, exact, etc. Hence, derive general solutions for the given equation in two different ways, and then reconcile your results . Solution: Method 1: Bernoullis Eq: , . Substitute , ρ x , e , x , : constant 2 , : constant 2 Therefore, Method 2: Exact Eq: 3 2 , ) +4 = , 2 The two results are consistent. 2 , 4 , 2 3 4 , =c, Therefore, 3 =0( 4 , : constant 4 ) 2 4 0, Problem 4.2 (Prob 34, P. 76, 3E) The following differential equation is of two different types considered in Chapter 1-separable, linear, homogenous, Bernoulli, exact, etc. Hence, derive general solutions for the given equation in two different ways, and then reconcile your results . Solution: Method 1 Exact Eq: 3 ) + 3 =0( 3 , = , 3 , 3, Therefore, ) 2 3 2 3 , = 3 3 , 3 c, 2 2 2 3 , 2 : Method 2: Homogeneous Eq: , , , , 3 3 3 1 1 ln 2 6 Therefore, 2 2 3 2 2 6 1 1 2 6 1 1 3 3 , ln 1, , : The two results are consistent. Problem 4.3 Consider a prolific breed of rabbits whose birth and death rates, β and δ , are each 0 . (a) Compute the proportional to the rabbit population P = P (t ) with β > δ and rabbit population as a function of time and parameters and initial condition given; (b) Find the time for doomsday; (c) Suppose that P0 = 6 and that there are 9 rabbits after 10 months, when is the doomsday? (d) If β < δ , what happens to the rabbit population in the long run? Solution: (a) Let β=bP(t), =dP(t), b and d :constants Let k=(b-d). Then the population equation is: β- P t , b‐d By separation variables, we get: 1 1 1 , . 0 1 0 (b) Time for doomsday is when the population reaches infinity. Seeing the equation, we observe that the population reaches infinity when the denominator becomes 0. That is , when 1 0 is the time for doomsday. (c) Putting P(10)=9 in the solution obtained above, we find that k=1/180. Hence time for doomsday =180/6=30 months. (d) If β< , k 0 and hence we see that the population goes to 0 as t goes to infinity. This makes sense, as this is the case when the birth rate is less than the death rate. Hence, this is the case for EXTINCTION. Problem 4.4 Consider a rabbit population P (t ) satisfying the logistic equation P ' = aP − bP 2 where B = aP is the time rate at which births occur and D = bP 2 is the rate at which deaths occur. If the initial population is 220 rabbits and there are 17 births per month and 14 deaths per month occurring at time t=0, compute and sketch P(t) with given parameters. Solution : The solution to the differential equation: is 0 0 We are given that 0 0 17 which gives a 17/220 and similarly b 14/220. 220 17 218.7858 0.085 Problem 4.5 (Modified from Quiz-1): A man with a parachute jumps out of a “frozen” helicopter at height . During the fall, the man's drag coefficient is (with closed parachute) and (with open parachute) and air resistance is taken as proportional to velocity. The total weight of the man and his parachute is . Take the initial velocity when he jumped to be zero. Gravitational constant is . Find the best time for the man to open his parachute after he leaves the helicopter . for the quickest fall and yet “soft” landing at touchdown speed Solution: This is a problem of Newtonian mechanics F F ma. Total weight of the man and his parachute m with acceleration a, gravitational force F and a drag force F . Let x be the distance above the earth’s surface, with positive direction downward, a , where v is the velocity and F mg. Since the drag force is assumed to be proportional to velocity, F kv with closed parachute and F nkv with open parachute The deployment occurs at time t , During initial fall with closed parachute, m dv dt mg kv, v 0 0 g this is an IVP problem. Solve this separable ODE. We obtain, or dv dt, kv g m ln g g C x 0 C e g. => v(t) = v g 1 0 => C g g e C e C , t = 0, v(0) = 0 , => x g t e C , g So the distance of close parachute fall is e g, g 1 e At , V( Starting with the opening parachute time, the differential equation becomes dv m mg nkv dt => m => dv dx dx dt mg nkv dv v dx mg nkv m whose solution is v(t) = g C e C = g C e => x , since v( 1 e · ng 1 n g t C e x x Which means g C e One can easily find the constant C C g g 1 e e C e g H x x m m m m m x g C e C H g e g nk nk k k k The falling time t is in principle solvable from the above equation, but it’s tedious and may require numerical solution technique. (Reaching this point without the actual solutions for C and t is sufficient to gain full marks.) g Therefore the total falling time is + , the speed he hits the ground is v( ) = , we can the best time for the man to open his parachute after he leaves the C e helicopter for the quickest fall and yet “ soft” landing at touchdown speed ≤ by adjust the . Problem 4.6 Suppose that the fish population P(t) in a lake is attacked by a disease at time t=0, with the result that the fish cease to reproduce (so that the birth rate is β = 0 ) and the death rate δ (deaths per week per fish) is thereafter proportional to P −1/ 2 . If there were initially 900 fish in the lake and 444 were left after 7 weeks, how long did it take all the fish in the lake to die? Solution: This problem has two different interpretations, as the language is unclear. First one is: The birth rate is 0 and the death rate is proportional to . So, the population which is separable. Separating variables, we get: 3 √ √ 2 Using this equation and the given initial conditions that P 0 900 and 444, we get k 1680 and that the time for all fish to die is 10.71428 weeks. equation is P 7 Second interpretation: The problem says that The population equation is δP t Solving this: 2 Using the initial conditions that P 0 and that the tie for all fish to die is 23.5 weeks. 900 and P 7 444, we get k 2.5511 THE SECOND WAY TO DO THE PROBLEM IS PREFERRED OVER THE FIRST ONE, ALTHOUGH PEOPLE WHO HAVE DONE IT USING THE FIRST WAY ARE ALSO AWARDED MARKS. KINDLY TAKE CARE TO USE SECOND METHOD ONLY. ...
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This note was uploaded on 05/11/2009 for the course AMS 361 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.

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