10 a The change in amount of phosphate P t is found by adding the amount

10 a the change in amount of phosphate p t is found

This preview shows page 8 - 11 out of 14 pages.

10. a. The change in amount of phosphate, P ( t ), is found by adding the amount entering and subtracting the amount leaving. dP dt = 200 · 10 - 200 · c ( t ) , where c ( t ) is the concentration in the lake with c ( t ) = P ( t ) / 10 , 000. By dividing the equation by the volume, the concentration equation is given by dc dt = 0 . 2 - 0 . 02 c = - 0 . 02( c - 10) , c (0) = 0 . With the substitution z ( t ) = c ( t ) - 10, the equation above reduces to the problem dz dt = - 0 . 02 z, z (0) = - 10 , which has the solution z ( t ) = - 10 e - 0 . 02 t . Thus, the concentration is given by c ( t ) = 10 - 10 e - 0 . 02 t . b. The differential equation describing the growth of the algae is given by dA dt = 0 . 5(1 - e - 0 . 02 t ) A 2 / 3 . By separating variables, we see Z A - 2 / 3 dA = 0 . 5 Z (1 - e - 0 . 02 t ) dt 3 A 1 / 3 ( t ) = 0 . 5( t + 50 e - 0 . 02 t ) + C A ( t ) = 0 . 5( t + 50 e - 0 . 02 t ) + C 3 ! 3 From the initial condition A (0) = 1000, we have 1000 = 25+ C 3 3 . It follows that C = 5, so A ( t ) = t + 50 e - 0 . 02 t + 10 6 ! 3 . 11. a. Write the differential equation dw dt = - 0 . 2( w - 80), then z ( t ) = w ( t ) - 80. It follows that dz dt = - 0 . 2 z, z (0) = - 80 , with the solution z ( t ) = - 80 e - 0 . 2 t = w ( t ) - 80. Thus, w ( t ) = 80 1 - e - 0 . 2 t .
Image of page 8

Subscribe to view the full document.

For a 40 kg alligator, w ( t ) = 40 = 80 ( 1 - e - 0 . 2 t ) or 40 = 80 e - 0 . 2 t , so e 0 . 2 t = 2 or 0 . 2 t = ln(2). Thus, t = 5 ln(2) 3 . 47 years. b. The pesticide accumulation is given by dP dt = 600 80 1 - e - 0 . 2 t , P (0) = 0 . The solution is given by P ( t ) = 48 , 000 Z 1 - e - 0 . 2 t dt = 48 , 000 t + 5 e - 0 . 2 t + C. The initial condition gives P (0) = 0 = 240 , 000 + C , so C = - 240 , 000. Hence, P ( t ) = 48 , 000 t + 5 e - 0 . 2 t - 240 , 000 . The amount of pesticide in the alligator at age 5 is P (5) = 48 , 000 ( 5 + 5 e - 1 ) - 240 , 000 = 240 , 000 e - 1 88291 μ g. c. The pesticide concentration for a 5 year old alligator is c (5) = P (5) 1000 w (5) = 88 , 291 80 , 000 (1 - e - 1 ) 1 . 75 ppm . 12. a. The differential equation can be written: dc dt = - 0 . 004( c - 15) , so we make the substitution z ( t ) = c ( t ) - 15. Since c (0) = 0, it follows that z (0) = - 15. The solution of the substituted equation is given by: z ( t ) = - 15 e - 0 . 004 t = c ( t ) - 15 c ( t ) = 15 - 15 e - 0 . 004 t . The limiting concentration satisfies: lim t →∞ c ( t ) = 15 mg / m 3 . b. We begin by separating variables, which gives: Z dc c - 15 = - 0 . 001 Z (4 - cos(0 . 0172 t )) dt ln( c ( t ) - 15) = - 0 . 001 4 t - sin(0 . 0172 t ) 0 . 0172 + C c ( t ) = 15 + Ae - 0 . 001 ( 4 t - sin(0 . 0172 t ) 0 . 0172 ) It is easy to see that the initial condition c (0) = 0 implies that A = - 15. Thus, the solution to this problem is given by: c ( t ) = 15 - 15 e - 0 . 001(4 t - 58 . 14 sin(0 . 0172 t ))
Image of page 9
13. a. We separate variables, so Z M - 3 / 4 dM = - k Z dt or 4 M 1 / 4 = - kt + 4 C M ( t ) = C - k 4 t 4 From the initial condition, M (0) = 16 = C 4 , it follows that C = 2. From the information that M (10) = 1 = (2 - 10 k/ 4) 4 , we have k = 0 . 4, so M ( t ) = (2 - 0 . 1 t ) 4 . The fruit vanishes in 20 days. b. We separate variables again to find: Z M - 3 / 4 dM = - 0 . 8 Z e - 0 . 02 t dt or 4 M 1 / 4 = 0 . 8 0 . 02 e - 0 . 02 t + 4 C M ( t ) = 10 e - 0 . 02 t + C 4 . From the initial condition, M (0) = 16 = (10 + C ) 4 , it follows that C = - 8, so M ( t ) = 10 e - 0 . 02 t - 8 4 . Solving 10 e - 0 . 02 t = 8, which is when the fruit vanishes, we find t = 50 ln(5 / 4). Thus, the fruit vanishes in 11.157 days. 14. a. The general solution to the Malthusian growth problem with the initial condition P (0) = 60 is P ( t ) = 60 e rt . We are given that 2 weeks later P (2) = 80 = 60 e 2 r , so it follows that r = 1 2 ln 4 3 = 0 . 14384. This gives the solution: P ( t ) = 60 e 0 . 14384 t .
Image of page 10

Subscribe to view the full document.

Image of page 11
  • Fall '08
  • staff

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes