Fall 2012
Math 122
Practice Lab Exam 3
Give all answers to at least 4 signicant gures.
1. When a monoculture of an organism is grown in a limited (but renewed) medium, then the
population of that organism often follows the logistic growth model. Below is
Fall 2001 1.
Complete Solutions
Trigonometric Functions
The graphs for the following are in the short solutions. 3. period T =
2 3
5. period T =
2 (1/2)
= 4
7. period T =
2 (1/2)
= 4
9. period T =
2 3
11. a. The maximum in the heart is 140 ml of blood, an
Spring 2003
Complete Solutions
Separation of Variables
1. a. This is a separable differential equation, so dy = 3t2 y dt dy = 3t2 dt y ln |y| = t3 + C y(t) = et
3 +C
= Aet
3
c. This is a linear equation like Newton's Law of Cooling. We write so take z(t)
Fall 2000 1. b. Solve
Selected Solutions
Linear Differential Equations
dz = 0.1z - 2, dt Factor the coefficient leading z(t) to give
z(0) = 5.
z (t) = 0.1(z(t) - 20). Make the substitution, w(t) = z(t) - 20, so w(0) = z(0) - 20 = -15 and w (t) = z (t). Th
Fall 2001
Complete Solutions
Optimization
1. The parabola is symmetric about the y-axis and is show in the diagram below. The point
(x, y) in the diagram lies on the parabola and appears in the upper right corner of the rectangle. By the symmetry, we see
Spring 2003
Math 122
Riemann Sums and Numerical Integration
1 2
1. a. With n = 2 and x [0, 2], the midpoints of the subintervals are x1 = x = 1, so the midpoint rule gives
2
and x2 =
3 2
with
(4 + 2x2 )dx
0
4+2
1 2
2
+
4+2
3 2
2
1 = 13.
With n = 2, the t
Spring 2003 1. Solutions for the integrals:
Math 122
Integration by Substitution
b. For the integral below, we make the substitution u = 2x2 - 3, so du = 4x dx. It follows that: x 2x2 - 3 dx = = = =
1 1 2x2 - 3 2 (4x)dx 4 1 1 u 2 du 4 1 3 u2 + C 6 3 1 (2x
Fall 2000
Solutions
Newtons Method
1. a. For x3 = 5 with x0 = 1, then x1 = 2.333333, x2 = 1.861678, x3 = 1.722002, with the actual solution being x = 1.709976, which requires 5 Newton iterates for this accuracy. b. For x4 = 13 with x0 = 1, then x1 = 4, x2
Fall 2000
Complete Solutions
Numerical Differential Equations
1. a. The solution to the initial value problem, dy = 0.3y, dt is y(t) = 20e0.3t . It requires five steps to use Euler's method to approximate the solution using a stepsize of h = 0.2 for t [0,
Fall 2012
Math 122
Practice Lab Exam 3 - Solutions
1. a. The solutions to the logistic growth models are
X0 M
,
X0 + (M X0 )ert
Y0 N
.
Y0 + (N Y0 )est
X (t) =
Y (t) =
The best tting parameters to the data are given by:
X0 = 0.6734
Y0 = 0.6341,
r = 0.10987
Spring 2003
Solutions
Qualitative Analysis of Dierential Equations
1. a. This dierential equation can be written dy = 0.2(y 50). dt Substitute z(t) = y(t) 50, then dz = 0.2 z with z(0) = z0 (some initial value). Thus, dt z(t) = z0 e0.2 t = y(t) 50. It fol