Diff Eqns, Quiz 1, Prof Joyner

8-28-98

Name _____________________________________

This exam is considered sensitive material according to USNA regulations and AcDean instructions. Closed book and notes. Math tables and TI-92s are allowed.

1. Sherlock Holmes is awoken by a phone call from a policeman at 4am. A body has been discovered and foul play is suspected. Sherlock tells the police to determine the temperature of the body and, when he arrives at the scene 30 minutes later, he takes the temperature again. Thetwo readings on that cold 50 degree F morning were 90 degrees F and 85 degrees F. Find thetime of death (use 98.6 degrees F for the normal body temperature).











2. A parachutist weighs 160 lbs (with chute). The chute is released immediately after the jump from a height of 1000 ft. The force due to air resistence is proportional to velocity and is given by Fres = -8v. Find the time of impact.














3 . Solve:

y' = (ex/x2) - (3/x)y













4. Solve:

(2xy3 - sin(x))dx + (3x2y2+(1/y))dy = 0





























Solution to problem 1:

 T(0) = Tm - Cekt

 Tm = 50

T(0) = 90

T(30) = 85

T(?) = 98.6

  T(0) = 90 = 50 - C, C = -40

T(30) = 85 = 50 + 40e30k, k = -0.004451

T(x) = 98.6 = 50 + 40e-0.004451x, x = -43.75 minutes

  ANSWER: The time of death is 3:16 AM.









Solution to problem 2:

 v = ((mg)/k) + ce (-kt)/m

 k=8

v(0) = 0

g = 32

 m*g = 160, m = 5

v = 20 + ce-8t/5

v(0) = 0, c = -20

v(t) = 20 - 20e-8t/5

x(t) - x(0) (by FTC) = integral(v(t),t,0,t) = 20t - 20 (integral(e-8t/5,t,0,t))

= 20t - 20(-5/8e-8t/5) evaluated from 0 to t

= 20t + (100/8)(e-8t/5 - 1)

x(t) = 20t - 25/2 + (25/2)e-8t/5

20t = 1012.5 - (25/2)e-8t/5

t = 1012.5/20 + (error at most about 10-37)

 ANSWER: He hits the ground 50.6 seconds after jumping.







Solution to problem 3:

 Standard form: y' + py = q

 y' = (ex)/(x2) - (3/x)y

y' +(3/x)y = (ex)/(x2)

p = (3/x), q = (ex)/(x2)

 y = (integral(eintegral(p)q) +c)/(eintegral(p))

 y = (integral(eintegral(3/x) (ex)/(x2)dx) +c)/(eintegral(3/x))

y = (integral(e3 lnx (ex)/(x2)dx) +c)/(e3 lnx)

y = (xex dx +c)/(x3)

y = (xex-ex+c)/x3

  ANSWER: y = (ex(x-1)+c)/x3









Solution to problem 4:

 (2xy3 - sin(x))dx + (3x2y2+(1/y))dy = 0

 (2xy3 - sin(x)) = M

partial derivative of M with respect to y = 6xy2

(3x2y2+(1/y)) = N

partial derivative of N with respect to x = 6xy2

They are equal, the stated function is exact.

Integral(M dx) = x2y3 + cos(x)

Integral(N dy) = x2y3 + ln y

Collect unique terms to find:

  ANSWER: potential(x, y) = x2y3 + cos(x) + ln(y) = c


Written by Midn Daniel Icil Doyle on 8-31-98.