Assignment 21 Query

course Mth 174

I am currently planning on taking Test 3 either Monday or Tuesday, and the Final by the end of the week. Thanks!

???s?????l??assignment #021021. `query 21

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Cal 2

01-11-2008

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22:19:08

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

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RESPONSE -->

The first thing I did was find the nullcines by taking the equation inside the parantheses, setting it equal to 0, and solving for x=0 then y=0:

Nullcines for dx/dt: (0,1), (3,0)

Nullcines for dy/dt: (0,2), (1,0)

Next I found the equilibrium points by checking for points in which dx/dt and dy/dt were equal to 0. I sketched a graph then looked for intercept points and points at which both dx/dt and dy/dt were equal to 0:

Eq. pts: (0,0), (0,2), (3,0), (3/5, 4/5)

Finally I needed to find the direction of the trajectories in each quadrant. After sketching a graph, I picked a point in each quadrant and then examined the value of dx/dt and dy/dt. I then created a vector based on those values:

Quadrant 1: (1,2) Direction: Southwest

Quadrant 2: (0.2), 1.1) Direction: Northwest

Quadrant 3: (0.5, 0.5) Direction: Northeast

Quadrant 4: (1.5, 0.2) Direction: Southeast

Good; however, for example, note that there are entire lines which comprise nullclines. Compare with the follwoing:

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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22:19:16

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

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RESPONSE -->

See above.

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22:23:17

describe the trajectories that result

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RESPONSE -->

I think I did.

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22:57:05

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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RESPONSE -->

a) To solve this I first set the equation equal to 0 by moving the right side over:

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

Then I set w^2 = g/l:

d^2x / dt^2 + w^2 x = 0

I then converted it into a harmonic equation based on the form C1 cos (wt) + C2 sin (wt) and substituting w=sqrt(g/l):

x(t) = C1 cos(sqrt(g/l) x) + C2 sin(sqrt(g/l) x)

Finally, I solved based on the given assumptions that x(0) = 0 and x'(0) = V0:

At x(0) = 0, C1=0.

At x'(0), C2 = V0.

So the equation becomes:

x(t) = V0 sin(sqrt(g/l) x)

b) I am not quite sure how this part is different.

The equation can be written

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).

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22:57:11

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

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RESPONSE -->

See above.

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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23:01:25

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

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RESPONSE -->

Hmm, does this mean V0 = 0 at t=0? But that doesn't make any sense, because then x(t) = 0...

If released from rest then the initial velocity is 0, so the x ' (0) = 0.

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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23:20:14

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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RESPONSE -->

Ah, this was a neat one.

a) First I substituted the values into the equation. I then rearranged the equation to get rid of the 36:

36 Q'' + 1/9 Q = 0

Q'' + 1/324 Q = 0

Next I made the substitution w^2 = 1/324, so w = 1/18, and inserted the values into the general form of the equation:

Q(t) = C1 cos(1/18 t) + C2 sin(1/18 t)

Q'(t) = -1/18 C1 sin(1/18 t) + 1/18 C2 cos(1/18 t)

I then solved this based on the given information that Q(0) = 0 and I(0) = 0. For I(t), I used the first derivative of Q:

C1 = 0

C2 = 36

The equation becomes:

Q(t) = 36 sin(1/18 t)

b) The assumptions are Q(0) = 6 and I(0) = 0, so the constants become:

C1 = 6

C2 = 0

And the equation becomes:

Q(t) = 6 cos (1/18 t)

Very good. However your final function doesn't give you Q(0) = 0. I believe the initial conditions dicate the sine, rather than the cosine function.

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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23:20:31

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

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RESPONSE -->

Q = 36 sin(1/18 t)

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23:20:55

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

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RESPONSE -->

Q(t) = 6 cos(1/18 t)

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23:21:01

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

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RESPONSE -->

See above.

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23:21:56

how did you evaluate your integration constants?

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RESPONSE -->

I evaluated by finding the equation and then entering the values given to solve for the value of the constants.

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23:26:31

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

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The first thing I did was to check the values of b and c:

b = 2

c = 1

And then check the value of b^2 - 4c to find out which equation I should use. Since 2^2 - 4(1) = 0, I need to use the equation y = (C1 t + C2) e^( -bt / 2) to find the general solution, which is:

y = (C1 t + C2) e^( -2t / 2)

y = (C1 t + C2) e^(-t)

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23:26:36

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

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RESPONSE -->

See above.

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23:37:20

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

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RESPONSE -->

I believe the assumption rests on the idea that the derivatives of e^(rt) are only changes in the multiples of e^(rt).

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23:42:40

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

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I believe the basic idea is to take each possible solution from the quadratic equation and place the value of r into an exponential function with a constant attached. In this case there is only one possible solution, -b/2, so there is only one exponential function in the equation.

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23:57:41

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

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RESPONSE -->

This is an interesting one. Basically, I can't think of any value of b that will make the equation underdamped or critically damped, since -4(-16) = 64, which is always positive, and b^2 is positive.

Actually, b^2 is positive except when i is used, so the equation is overdamped for any value of b.

The equation is critically damped for b = 8i.

The equation is underdamped for b > 8i.

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23:58:09

for what values of c is the general solution underdamped?

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RESPONSE -->

See above. Basically for b > 8i.

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23:58:25

for what values of c is the general solution overdamped?

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RESPONSE -->

For any value of b without i, including 0.

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23:58:38

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

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RESPONSE -->

For b = 8i.

Excellent. There are no real-valued solutions for which the system is overdamped, underdamped or critically damped.

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23:58:59

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

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RESPONSE -->

Interesting stuff.

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Excellent work, but see my notes to be sure you understand a few mostly minor discrepancies.