Assignment 21

course Mth 174

Thanks for all the help this semster. I have put fourth my best effort in all that i have sent you this semster. I hope that i maintain my C average so i can graduate.Thanks

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assignment #021

021. `query 21

Cal 2

12-04-2007

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19:03:11

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

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The horizontal nullclines; y = 0, y = 2(1-x)

Vertical nullclines; x = 0, y = 1-x/3

the points that were seen here in these equations were at points (0,0), (0,2), (3,0) and (3/5,4/5)

The graph decription here is howned to have 4 sectors.

THere are two lines on the graph one is starting at (0,2) and ending at (1,0) with it crossing the other line at (3/5,4/5).

The other line started at (0,1) and ends at (3,0)

the 4th sector started at (1,0) and ends at (3,0)

Good. Here's a bit more detail:

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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19:03:25

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

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shown in previous screen

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19:03:27

describe the trajectories that result

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19:13:50

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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In the first part of the question when x(0) = 0 and x' (0) = v0

when the angle is increased or decreased or just plan 0 like in this case then the pendulum is not moving at all because there is no angle here. Now when the derivitive is taken from the equation below

d^2*x/dt^2 = - (g/l)*x

if we take the derivative here we will get the velocity of the pendulum which is

dx/dt = -g/l

which when solved will give us the velocity at which the pendulum is going but when it has no angle like here it will not have any velocity.

The variable here is t, not x.

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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19:13:52

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

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19:13:55

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

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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19:17:42

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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no when we look at this problem we are asked to find two equations and these two equations must be when Q(0) = 0 and I(0)=2 and this equation will be

Q = 36sin(1/18)t

the other equation will be in the case of Q(0)=6 and I(0) = 0 which will be

Q = 6cos(1/18)t

Very good.

The details:

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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19:17:44

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

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19:17:47

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

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19:17:49

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

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19:17:51

how did you evaluate your integration constants?

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19:23:41

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

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we will look at the equation of P'' + 2P' + P = 0

and we need to find the general solution for it

we will use the order of r^2 + 2r +2 = 0 when r = -1plus or minus i

P(t) = A1e^(-1+i)t + A2e^(-1-i)t

now we will use Euler's formula and we will end up finding the solution which will be 2e^-tcos(t) + 2e^-tsin(t)

Very good for the equation P '' + 2 P ' + 2 P = 0, which is actually a better problem that the problem quoted here.

The characteristic equation for P '' + 2 P ' + P = 0 is r^2 + 2 r + 1, with a repeated real root r = -1.

This gives us a general solution which is a linear combination of the functions e^-t and t e^-t so that

P(t) = c1 * t e^-t + c2 * e^-t.

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19:23:43

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

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19:23:44

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

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19:23:46

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

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19:34:21

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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we must analyze the equation of s'' + bs' - 16s = 0

to determine if it is overdamped, underdamped, or critically damped

r^2 + 2r +2 = 0 when r = -1 plus or minus i

s(t) = A1e^(-1+i)t + A2e^(-1-i)t when s(0) = 2

2 = A1e^(-1+i)0 + A2e^(-1-i)0 = A1 + A2

s'(t) = A1 (-1+i)e^(-1+i)t + A2(-1-i)e^(-1-i)t

s'(0) = 0

0 = A1(-1+i) + A2 (-1-i)

A1 = 1-i A2 = 1+i

This equation yields characteristic equation

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

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

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19:34:23

for what values of c is the general solution underdamped?

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s"" + bs' -16s = 0

b^2 - 4c = b^2 - 4(-16) = b^2 + 64. Since b^2 is always positive for real b, b^2 - 4c is always > 0. Therefore, therefore there is no value of b for which the general solution is underdamped.

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19:34:24

for what values of c is the general solution overdamped?

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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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19:34:26

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

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19:35:07

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

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i didnt really understand the last question i did the best i could like i have done all semster. I have put fourth my best in all my work this semster and i just hope it works out that i get my C average so i can graduate.

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Good work on this assignment.

You have indeed been putting forth a commendable effort. I sent you a note with results from your last test and comments on your progress and status.

Let me know if you have any questions regarding the final. As usual, feel free to send me a practice test.