Assignment 21

course Mth 174

O}뉻۶ǝe|iassignment #021

021. `query 21

Cal 2

06-29-2009

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12:51:32

Query 11.9.7 (was page 576 #6) x' = x(1-y-x/3), y' = y(1-y/2-x)

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12:55:29

describe the phase plane, including nullclines, direction of solution trajectory in each region, and equilibrium points

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In the phaseplane, the horizontal nullclines are y = 0, y =2(1-x) and the vertical nullclines are x = 0, y = 1 - x/3. There are equilibrium points at (0,0), (0,2), (3,0). The first trajectory is when x is increasing and y is decreasing. The second is when both x and y are decreasing. The third is when both x nd y are increasing. The fourth is both x and y are decreasing.

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12:55:32

describe the trajectories that result

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dx/dt = x (1 y x/3); this expression is 0 when x = 0 or when 1 y x/3 = 0. Thus

dx/dt = 0 has 2 solutions: x = 0 and y = -(1/3)x + 1.

When dx/dt = 0 and dy/dt is not 0 (as is the case for both of these solutions) the nullcline has vertical slope.

Therefore, x = 0 (y-axis) is a nullcline with vertical slopes, and

the line segment (for x>=0 and y>=0) y = -(1/3)x + 1 is a nullcline with vertical slopes

dy/dt = 0, which yields horizontal nullclines, similarly has 2 solutions:

dy/dt = 0 for y = 0 and y = -2x+2.

Therefore

y = 0 is a nullcline with vertical slopes, and

the line segment (for x>=0 and y>=0) y = -2x + 2 is a nullcline with horizontal trajectories .

The lines x = 0, y = 0, y = -2x + 2 and y = -1/3 x + 1 divide the first quadrant into four regions.

There are 4 equilibrium points where the nullclines (with perpendicular trajectories) intersect:

x = 0 and y = -2x + 2 have perpendicular trajectories and intersect at (0,2)

x = 0 and y = 0 have perpendicular trajectories and intersect at (0, 0)

y = 0 and y = -1/3 x + 1 have perpendicular trajectories and intersect at (3, 0)

y = -2x + 2 and y = -1/3 x + 1 have perpendicular trajectories and intersect at (0.6,0.8)

Using test points in each region, the trajectories are:

Region bounded by (0,1), (0,0), (1,0), and (0.6,0.8): x and y both increasing.

Region bounded by (0.6,0.8), (1,0), and (3,0): x increasing and y decreasing.

Region bounded by (0,1), (0,2), and (0.6,0.8): x decreasing and y increasing.

Region outside (above and to the right of) segments from (0,2) to (0.6,0.8) and (0.6,0.8) to (3,0): x and y both decreasing.

Depending upon where the starting point at t=0 is in each region, all the trajectories will eventually move towards either y approaching 0 and x increasing without limit or towards x approaching 0 and y increasing without limit as t tends towards infinity. In other words for any starting point not at an equilibrium point, either x or y will go towards 0 while the other goes towards infinity.

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13:01:35

Query problem 11.10.22 (3d edition 11.10.19) (was 10.8.10) d^2 x / dt^2 = - g / L * x

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14:21:38

what is your solution assuming x(0) = 0 and x'(0) = v0?

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This one was a bit confussing for me but this is what I tried:

d^2x/dt^2 = -g/l x

= d^2s/dt^2 + g/l s = 0

w^2 = g/l

w = sqrt (g/)

s(t) = C1cos(wt) + C2 sin(wt)

= C1 cos(9.8 / change in x *t) + C2sin(9.8 / change in x *t)

v(0) = -wC1sin(w*0) + wC2 cos(w*0)

= -9.8 / change in x C1 sin(9.8 / change in x*0) + 9.8 / change in x C2 cos(w*0)

I wasn't sure what to do from here or if I set this up right.

You've got a good start.

The argument is sqrt(g/L) * t. This is expressed below as just omega * t, with omega = sqrt(g/L).

The equation is thus

x '' (t) = - omega^2 * x(t), where x(t) is a function of t and x '' (t) is the second derivative of that function.

The sine and cosine functions are characterized by second derivatives which are negative multiples of themselves. You can easily verify using the chain rule that the second derivative of B cos(omega t) is - omega^2 B cos(omega t) and the second derivative of C sin(omega t) is - C omega^2 sin(omega t). Thus, any function of the form B cos(omega t) + C sin(omega t) is a solution to the equation x ''(t) = - omega^2 x(t).

The general solution is usually expressed using c1 and c2 as the constants. So let's write it

x(t) = c1 cos(omega * t) + c2 sin(omega * t).

If x(0) = 0 then we have

c1 cos(omega * 0) + c2 sin(omega * 0) = 0 so that

c1 * 1 + c2 * 0 = 0, giving us

c1 = 0.

The solution for x(0) = 0 is thus

x = c2 sin(omega x) ).

Since v(0) = x ' (0) we have

v0 = c2 * omega cos(omega * 0)

so that c2 = v0 / omega, giving us solution

x = v0 / omega * sin(omega t ).

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14:21:54

What is your solution if the pendulum is released from rest at x = x0?

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We obtain the values of c1 and c2 using the given conditions x(0) = x0 and x ' (0) = 0:

Using general solution calculated in previous response:

x(t) = c1*cos[sqrt (g/L)*t] + c2*sin[sqrt (g/L)*t]

x(0) = x0 = c1*cos(0) + c2*sin(0) so

c1 = x0.

x ' (0) = 0 leads us to the conclusion that c2 = 0.

Thus for the given conditions

x(t) = x0*cos[sqrt (g/l)*t]

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14:34:45

Query problem 11.10.25 (3d edition 11.10.24) (was 10.8.18) LC circuit L = 36 henry, C = 9 farad

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14:39:17

what is Q(t) if Q(0) = 0 and *(0) = 2?

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36d^2Q/dt + Q/9 =0

Q(t) = C1cos(sqrt(Q/9)t) + C2sin(sqrt(Q/9)t)

Q(0)=C1cos(sqrt(Q/9)*0) + C2sin(sqrt(Q/9)*0)

LdI/dt + Q/C = 0

36dI/dt + Q/9 = 0

I'm not really sure what to do here. This section was confussing for me. I tried solving this problem like the example in the book, but I do not understand how to solve when you do not know what the value for Q is.

In LC, RC and RLC circuits Q stands for the charge on the capacitor, R for the resistance of the circuit, L the inductance and C the capacitance.

The voltage due to each element is as follows (derivatives are with respect to clock time):

capacitor voltage = Q / C

voltage across resistor = I * R

voltage across inductor = I ' * L,

where I is the current. Current is rate of change of charge with respect to clock time, so I = Q '. Note that since I = Q ', we have I ' = Q ''.

This circuit forms a loop, and the condition for any loop is that the sum of the voltages is 0. So the general equation for an RLC circuit is

L Q '' + R Q ' + Q / C = 0.

This equation is solved using the assumption that Q = A e^(k t) for arbitrary constants A and k. This assumption leads to the characteristic equation

L k^2 + R k + 1 / C = 0, a quadratic equation in k which is easily solved.

For an LC circuit, R = 0 and the equation is just

L Q '' + Q / C = 0 and the solutions to the characteristic equation are k = i / sqrt(L C) and k = -i / sqrt(L C), leading to the general solution

Q = c1 cos(omega * t) + c2 sin(omega * t), where omega = sqrt( 1 / (L C) ).

For the given conditions sqrt(1 / (LC) ) = sqrt( 1 / (36 * 9) ) = 1 / 18.

I = Q (t) = -c1 omega sin(omega * t) + c2 omega cos(omega * t). For the given condition I(0) = 0, this implies that c2 = 0 so our function is

Q(t) = -c1 omega sin(omega * t).

The condition Q(0) = 6 gives us

6 = -c1 sin(0) so that c1 = 6 and our function is therefore

Q(t) = 6 sin(omega * t), again with omega = 1/18.

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14:39:19

what is Q(t) if Q(0) = 6 and I(0) = 0?

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14:40:19

What differential equation did you solve and what was its general solution? And

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Q(t) = C1cos(sqrtQ/9)*t) + C2sin(sqrt(Q/9)*t)

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15:02:22

Query problem 11.11.12 (was 10.9.12)general solution of P'' + 2 P' + P = 0

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P'' + 2 P' + P = 0

r^2 + br + c = 0

= r^2 + 2r + r = 0

r= + or - 1

b^2 - 4c = 0 so the general solution is:

P(t) = (C1t + C2) e^-t

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15:02:26

what is your general solution and how did you obtain it?

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P'' + 2 P' + P = 0

r^2 + br + c = 0

= r^2 + 2r + r = 0

r= + or - 1

b^2 - 4c = 0 so the general solution is:

P(t) = (C1t + C2) e^-t

Good solution.

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15:04:35

If not already explained, explain how the assumption that P = e^(rt) yields a quadratic equation.

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P = e^(rt) yields a quadratic equation because it satisfies a differentail equation in which the second derivatie d^2/dt^2 is a sum of multples of dy/dt and y.

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15:05:53

Explain how the solution(s) to to your quadratic equation is(are) used to obtain your general solution.

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When you find the quadratic equation and what b and c equal, you must put it into the expression b^2 - 4c and compare it to 0. The general solution will differ depending on if the value is less than, greater than or equal to 0.

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15:20:35

Query problem 11.11.36 (was 10.9.30) s'' + 6 s' + c s NO LONGER HERE. USE 11.11.30 s '' + b s ' - 16 s = 0

This equation yields characteristic equation

r^2 + b r + - 16 = 0 with solutions

r = (-b +- sqrt(b^2 + 64) ) / 2.

for what values of c is the general solution underdamped?

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b^2 - 4c is less than 0

b^2 + 64 is less than 0

b must be less than or = to -65

s"" + bs' -16s = 0

b^2 4 a c = b^2 4(1)(-16) = b^2 + 64.

Since b^2 is always positive for real b, b^2 - 4c = b^2 + 16 is always > 0.

Therefore, therefore there is no value of b for which the general solution is underdamped.

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15:24:55

for what values of c is the general solution overdamped?

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b^2 - 4c is greater than 0

b^2 + 64 is greater than 0

b can be any number because any number squared is a positive number.

Since b^2-4c > 0 for all b, we check the values of r1 and r2.

Let r1 = -(1/2)b + sqrt(b^2 - 4c). This solution results in r1 > 0 for all b.

Let r2 = -(1/2)b - sqrt(b^2 - 4c). This solution results in r2 < 0 for all b.

Since r1 and r2 are never both < 0, there is no value of b for which the general solution is overdamped.

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15:27:13

for what values of c is the general solution critically damped?

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b^2 - 4c = 0

b^2 + 64 = 0

I do not know how this one can be possible since 8^2 = 64

Since b^2-4c is never = 0, there is no value of b for which the general solution is critically damped.

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15:27:38

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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&#Good work. See my notes and let me know if you have questions. &#