question 21

#$&*

course Mth 174

021. `query 21Cal 2

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Question:

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

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Your solution:

x` = x(1- y - x/3) = 0

x = 0, or y = 1 - x/3

y` = y(1-y/2-x) = 0

y = 0, or y = 2 - 2x

equilibrium points are:

(0, 0), (0, 1), (3,1), (0, 2), (1,0) and (3/5, 4/5)

from the area of y>2-2x and y>1-x/3 clockwise

the first: x<0, y<0,

the second: x<0, y>0

the third: x>0, y>0

the fourth: x>0, y<0

confidence rating #$&*: 3

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Given Solution:

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

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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.

STUDENT QUESTIONS

I am having a hard time understanding what exactly this means> I see that the y = 1 line seg (or nullclines)

corresponds to the x = 3 pt on the x axis and the same for the y = 2 and the x = 1 nullcline but don’t understand what this

is saying exactly except that it is an outline to draw conclusions from. I mean I guess I saying I don’t quite understand how

to interpret them?

You're on the right track.

Along a vertical nullcline the slope field is vertical. So for example along the line y = 1 - x/3, at every point, the slope field would be vertical.

The vertical nullclines are the y axis (i.e., x = 0) and the line y = 1 - x/3. In the first quadrant, the latter is a straight line from (0, 1) to (3, 0). Sketch each line and indicate short vertical slope segments along each.

The horizontal nullclines are the x axis and the line y = 2 - x, which in the first quadrant runs from the point (0, 2) to the point (2, 0). Sketch each line and indicate short horizontal slope segments.

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10:32:28

describe the trajectories that result

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question:

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

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

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Your solution:

d^2 x / dt^2 = - g / L * x

d^2 x / dt^2 + g / L * x = 0

s(t) = C1 cos(`sqrt(g/L)x) + C2 sin(`sqrt(g/L)x)

s(0) = C1 cos0 + C2 sin0 = 0

C1 = 0

s`(t) = -`sqrt(g/L)C1 sin (`sqrt(g/L)x) + `sqrt(g/L)C2 cos (`sqrt(g/L)x))

s`(0) = -`sqrt(g/L)C1 sin 0 + `sqrt(g/L)C2 cos 0 = v0

`sqrt(g/L)C2 = v0, C2 = v0/(`sqrt(g/L))

s(t) = v0 / (`sqrt(g/L)) sin(`sqrt(g/L)x)

confidence rating #$&*: 3

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Given Solution:

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RESPONSE -->

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

d^2x/dt^2 - (-g/l)x = 0

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

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

x(0) = 0 = C1*cos[sqrt (g/l)*0 + C2*[sqrt (g/l)*0]]

0 = C1*cos(0) + C2*sin(0)

0 = C1

x(t) = C2*sin[sqrt (g/l)*t]

x' = sqrt(g/l)*C2*cos[sqrt (g/l)*t]

x'(0) = v0 = sqrt(g/l)*C2*cos[sqrt (g/l)*0]

v0 = sqrt(g/l)*C2*cos(0)

v0 = sqrt(g/l)*C2

C2 = v0/sqrt(g/l)

x(t) = [v0/sqrt(g/l)]*sin[sqrt (g/l)*t]

d^2 x / dt^2 = - g / L * x x(0) = 0 x'(0) = v0

w = sqrt(g/L)

This is not w; is it designated by the Greek letter omega, which in typewriter notation can be simply spelled out: omega = sqrt(g / L).

d^2 x / dt^2 = -w^2 x

d^2 / dt^2 (sint) = -sint does not equal -w^2 sint

d^2 / dt^2 (sinwt) = d/dt (wcoswt) = -w^2 sin(wt)

d^2 x / dt^2 + g / L * x = 0

x = c1cos(sqrt(g/L x)) + c2sin(sqrt(g/L x))

The variable here is t (not, for example, 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 ).

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

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STUDENT SOLUTION:

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)

x0 = C1

x(t) = x0*cos[sqrt (g/l)*t] + C2*sin[sqrt (g/l)*t]

x' = -[sqrt(g/l)]*x0*sin[sqrt (g/l)*t] + [sqrt(g/l)]*C2*cos[sqrt (g/l)*t]

x'(0) = v0 = 0 = -[sqrt(g/l)]*x0*sin(0) + [sqrt(g/l)]*C2*cos(0)

0 = [sqrt(g/l)]*C2

C2 = 0

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

INSTRUCTOR RESPONSE:

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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Self-critique (if necessary): ok

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Self-critique Rating: ok

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Question:

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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Your solution:

L * d^2Q/dt^2 + Q / C = 0

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

d^2 Q/dt^2 + Q/(9*36) = 0

Q(t) = C1 cos(`sqrt(1/324)t) + C2 sin(`sqrt(1/324)t)

a. Q(0) = 0, I(0) = 2

C1(cos(0/18) + C2sin(0/18) = 0

C1 = 0

dQ / dt = I(0) = -C1sin(0/18)/18 + C2 cos(0/18)/18 = 2

C2 = 36

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

b. Q(0) = 6, I(0) = 0

dQ / dt = I(0) = -C1sin(0/18)/18 + C2 cos(0/18)/18 = 0

C2 = 0

C1(cos(0/18) + C2sin(0/18) = 6

C1 = 6

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

confidence rating #$&*: 3

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Given Solution:

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

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RESPONSE -->

L*(d^2Q/dt^2) + Q/C = 0

Dividing by L:

d^2Q/dt^2 + Q/(L*C) = 0

Substituting for L and C:

d^2Q/dt^2 + Q/(36*9) = 0

d^2Q/dt^2 + [(1/18)^2]*Q = 0

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

Q(0) = 0 = C1*cos[(1/18)*0] + C2*sin[(1/18)*0]

0 = C1

Q(t) = C2*sin[(1/18)t]

dQ/dt = (1/18)*C2*cos[(1/18)t]

dQ/dt = I

dQ/dt (0) = I(0) = 2 = (1/18)*C2*cos(0)

2 = (1/18)*C2

C2 = 36

Q(t) = 36*sin[(1/18)t]

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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12:54:09

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

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RESPONSE -->

Using the general solution from the previous response:

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

Q(0) = 6 = C1*cos[(1/18)*0] + C2*sin[(1/18)*0]

6 = C1

Q(t) = 6cos[(1/18)t] + C2*sin[(1/18)t]

Q'(t) = -(1/3)sin[(1/18)t] + (1/18)*C2*cos[(1/18)t]

Q'(0) = I(0) = 0 = -(1/3)sin[(1/18)*0] + (1/18)*C2*cos[(1/18)*0]

0 = -(1/3)sin(0) + (1/18)*C2*cos(0)

C2 = 0

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

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12:54:21

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

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RESPONSE -->

See previous response.

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12:54:33

how did you evaluate your integration constants?

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question:

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

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Your solution:

b^2 - 4c = 2^2 - 4 = 0

P(t) = (C1t+ Ct)e^(-tb/2) = (C1t+ Ct)e^(-t/2)

confidence rating #$&*: 3

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Given Solution:

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

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RESPONSE -->

Using test solution P = A e^(r t) the equation

P’’ + 2P' + P = 0

yields characteristic equation

r^2 + 2 r + 1 = 0 with repeated solution 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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13:13:05

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

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RESPONSE -->

Using the second-order differential equation for P with respect to t, and substituting e^(rt) for P and r for dP/dt, we get[e^(rt)]*(r^2 + br + c) = 0. Since e^(rt) cannot be 0, r^2 + br + c = 0, and we have our quadratic equation.

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13:25:45

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

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RESPONSE -->

Since b^2-4c = 0, the solution to the quadratic is r = -b/2 = -1. The two solutions for P are P = C2*e^(-bt/2) and P = C1*t*e^(-bt/2). Since the general solution is the sum of the two individual solutions, we get P = (C1t + C2)*e^(-bt/2). Since b = 2, we get P = (C1t + C2)*e^(-t).

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question:

Query problem 11.11.30 3d edition 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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Your solution:

the general solution is underdamped if b^2 - 4c < 0, where b^2 - (-)64 < 0, in which is always bigger or equal to 64, then it can’t be underdamped

Also, it cannot equal to 0, then it never critical damped, it is always overdamped

confidence rating #$&*: 3

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Given Solution:

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This equation yields characteristic equation

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

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

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13:31:29

for what values of c is the general solution underdamped?

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RESPONSE -->

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 + 64 is always > 0.

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

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13:36:46

for what values of c is the general solution overdamped?

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RESPONSE -->

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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13:37:54

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

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RESPONSE -->

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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13:57:17

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

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RESPONSE -->

Question 11.9.10 involves analyzing the phase plane for two first-order differential equations (A' = 2A - 2B and B' = B - AB). The last part of the problem asks for a sketch of the slope field--using a computer or calculator. Finding dB/dA by dividing dB/dt by dA/dt I got an equation in B and A (dB/dA = (B-AB)/(2A-2B). The only way that I found to use the calculator to find the slope field was to pick a value for A, substitute it into the equation for dB/dA, graph it on the calculator, and pick off the values of y (dB/dA) for various values of x (B). I then repeated this with other values for A. Is there a better way to do this on the calculator?

I really enjoyed the math in this chapter. Differential equations are very interesting.

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""

The DERIVE command direction_field, which is part of the ODE applications files, does a direction field. If you are using a TI calculator, which is based on DERIVE, it's possible there is a similar application..

Most computer algebra systems have a direction field capability.

Direction fields are also easy to program, if you know a programming language. This is my usual approach; it gives me complete control over the window size, lengths of segments, colors, etc..

some of these problems don't appear in previous queries, while some do:

*&$*&$

**** Query problem 10.6.20 rate of expansion of universe:

R is the radius of the universe, M0 is mass. C is a constant, assumed here to be 0. G is the gravitational constant.

(R')^2 = 2 G M0 / R + C; case C = 0

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12:34:39

I have not been able to locate this problem in my book. I would like to attempt it, so if you could tell me where I can find it in my edition of the book I'll give it a try. Thanks

*&*& Unfortunately this problem was omitted from the present edition *&*&

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12:34:40

12:34:42

12:34:45

12:38:05

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**** what is your solution to the differential equation R' = `sqrt( 2 G M0 / R ), R(0) = 0? **** How the you determine the nature of the resulting long-term expansion of the universe, and what is your conclusion? ****

The equation

R' = `sqrt( 2 G M0 / R )

is separable. Write R as dR/dt:

dR/dt = sqrt(2 G M0 / R), and rearrange to get

sqrt(R) dR = sqrt( 2 G M0) dt

Integrating both sides we get

2/3 R^(3/2) = sqrt( 2 G M0) * t, so that

R = ( 3/2 sqrt(2 G M0))^(2/3) * t^(2/3).

We get

R(0) = 0?

*&$*&$

Simple harmonic oscillator

Problem Number 2

A simple harmonic oscillator has restoring force constant .68 N/m, mass 9 kg and experiences drag force - 24.48 N s / meter * v, where v is its velocity. What is the differential equation for the motion of the oscillator? What is the solution to the equation?

C = -24.48

k = .68

m = 9

24.48 +- 'sqrt(24.48^2 - 4(.68)(9))/2(9)

r1 = .028067 r2 = 2.6919

x = c1e^(.028067t) + c2e^(2.6919t)

Newton's Second Law says that

m x '' = net force,

where x is the position of the oscillator relative to equilibrium.

The restoring force constant .68 N / m implies that the restoring force at position x is -.68 N / m * x; suppressing units we say that the restoring force is -.68 x.

Drag force is -24.48 * v, again suppressing units. v is the velocity, which is the derivative of the position: v = dx/dt. So the drag force is -24.48 dx/dt.

Thus net force is -.68 x - 24.48 dx/dt and the resulting equation is

m x '' = -.68 x - 24.48 dx/dt. Since m = 9 kg this is

9 x '' = -.68 x - 24.48 dx/dt.

We rearrange this to the form

9 x '' + 24.48 dx/dt + .68 x = 0

and use trial solution x = A e^(rt). For this x we have x ' = r A e^(rt) and x '' = r^2 A e^(rt) so our equation is

9 r^2 A e^(rt) + 24.48 r A e^(rt) + .68 A e^(rt) =0. Dividing through by rt we have

9 r^2 + 24.48 r + .68 = 0, a simple quadratic equation with solutions

r = (-24.48 +- sqrt( 24.48^2 - 4 * .68 * 9) ) / ( 2 * 9).

The discriminant is positive so this is the overdamped case of the harmonic oscillator, and the general solution of the equation is

x = C1 e^(r1 t) + D2 e^(r2 t),

where r1 and r2 are the two real solutions of the quadratic equation (roughly r1 = -.05 and r2 = -2.7 or so). Using the rough solutions we have

x = C1 e^(-.05 t) + C2 e^(-2.7 t).

Direction field

direction field

course Mth 174

Can you help me understand how to draw the followingSketch the direction field for the differential equation dy / dx = .49 e^x / y. Sketch some solution curves corresponding to this field.

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

I understand that it has something to do with the slope at a point. Do we need to plug in different numbers fo x and y and see what the slope is at them points.

First, as y -> 0 the value of dy/dx approaches +infinity; the slopes approach vertical. So a set of short vertical line segments through the x axis would be a good start.

For all x, e^x > 0 so whenever y is positive the slopes will be positive, and whenever y is negative the slopes will be negative.

As you move to the right, e^x becomes very large very fast, and e^x / y will also be very large (though division by y decreases the magnitude of the result, it's easy to make e^x much larger than y by just moving a little ways to the right).

So to the right, the slopes will again be nearly vertical.

As we move to the left (for negative x), e^x becomes very small very quickly, so except very near the x axis the slopes will approach zero. A set of horizontal lines toward the left edge of the graph would depict this behavior.

To see what happens in between these extremes, consider first the y axis, where x = 0. The slopes at the point (0, y) is .49 / y. So when y = .49, the slope is 1; when y = .98 the slope is 1/2; when y = 1.47 the slope is 1/3 and the slopes continue to decrease as you move up the y axis. Something similar happens along the negative y axis, but the slopes will be negative.

Along the vertical line x = 1 the slopes will be greater; when y = .49 the slope will be e = 2.718, approx., when y = 2 the slope will be 1.36 approx., and as y increases the slopes gradually approach 0.

If you similarly consider the vertical lines where x = -1, x = 2 and x = 3 you will get a pretty good idea how to map this slope field.

Another strategy might be to consider the set of points where .49 e^x / y is equal to a fixed value c:

If .49 e^x y = c, then y = c / (.49 e^x) = 2.04 c * e^-x.

For c = 1 this is the curve y = 2.04 e^-x. This curve can be easily sketched, and at points on this curve the slope is c = 1. A series of short segments with slope 1 would depict the direction field along this curve.

The c = 2 curve is y = 4.08 e^-x; this curve is also easy to sketch and a series of short segments with slope 2 can be placed at intervals along this curve.

It would be worthwhile to continue for the c = -1 and c = -2 curves, then consider the curves c = 3 and c = 4, and ask yourself what would happen for additional values of c. The direction field so constructed would be consistent with that previously constructed.

To answer a question of this nature you could either sketch the direction field or give a written description of the process.

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&#This looks very good. Let me know if you have any questions. &#