Problems for Test 3 - Solution
Plot (a) does not correspond to the diff. eq. y'=a - x - y.
Reason: From the diff. eq. we see that as x and y get larger, the slope becomes smaller and eventually negative. But plot (a) shows slopes getting larger (steeper) as x and y get larger. So plot (a) cannot be the slope field of the differential equation.
We may get the equilibrium point by solving 2.5*y - 7.5 = 0 for y, which produces y=3. The slope field plot shown above confirms our result. Also, the plot shows that the equilibrium value y=3 is unstable, as solutions that are close to it do separate further from y=3 as x is increased.
a) N' = k N.
b) N = 5 e0.1682 t
a) T' = k (T - 22)
b) T = 22 - 27 e -0.811 t
Use the relation RATE OF CHANGE OF Q
= R_in - R_out. We get,
Q' = 2 - (Q/10)*2, that is, Q' = 2 - 0.2 Q. Solve this equation to get
Q = 10 - 10 e-0.2 t. Finally set Q = 5 and solve for t to get t = 1.735 min.
7b r = - ln | -0.5 x2 + x + 0.5 |
7c y = (6 - 2 e8 x ) / (3+e8 x )
7d A solution in implicit form is 0.5 y2 - y cos(y) + sin(y) = 0.5 x2 ;no explicit solution is possible.
a) The population is growing the fastest when dP/dt is the largest possible. Since dP/dt = 0.8 P - 0.001 P2, the vertex of the parabola y = 0.8 P - 0.001 P2 gives what we want. The vertex is at P = 400. So the population grows the fastest when it is 400 million.
b) The logistic model we are considering has carrying capacity L = 800, and in our particular example, the initial population is 25 million. Then population will never surpass 800 million and thus will never reach 900 million, according to the logistic model properties.
a) Considering "up" as the positive direction, we have , that is, dv/dt = -9.8.
b) which with M=1 and g=9.8 becomes
c) solving for v yields .The
initial velocity is 30 m/s, so
30 = -9.8/k + A, that is, A = 30 + 9.8/k.
a) dQ/dt = k Q
b) Solve to get Q = A e k t . Note that 0.9*A = Q(10). Hence
, that is , . Solving for k we get k = .
Also Q(0) = A = 3.0. Then we have the formula
We wish to know t when Q = 5. Solve for t to get t = -48.51, that is, it was 48.5 years before the present time that the mass was 5 Kg.
a) the particle will be in the second quadrant when x < 0 and y > 0, that is,
4 t - t2 < 0 and t - 5 > 0.
Now 4 t - t2 < 0 when t < 0 or 4 < t, and t-5 > 0 when t > 5.
Then when t >5 both requirements are met, that is, the particle is in the second quadrant.
b) dx/dt = 4 - 2 * t and dy/dt = 1, so the speed at time t is
s= , and the speed at time t=4 is
c) The distance traveled is =
Using simpson's rule with n = 30 gives distance = 9.2935
The differential equation is (1/P) * (dP/dt) = -0.006 P + 0.7.
The equation 20 y - 2 y2=0
yields y=0, y=10, so the carrying capacity is L=10.
By substituting y=5 into the right hand side of y' = 20 y - 2 y2 we get that the population is changing initially at the rate of 50 mill/yr.
By substituting y'=22 into the left hand side of y' = 20 y - 2 y2 we get
22= 20 y - 2 y2 . This equation has only one positive solution: y = 11.
a) x = 0 + -3 t , y = 2 + -3 t, where 0 < = t < = 1.
b) x = cos t, y = sin t , where pi < = t < = 2 pi
b) x = t, y = t 2 , where 1 < = t < = 2