qa_01

#$&*

course Mth 279

6/15 10:38 am

Q_A_QuestionsThere are three parts to this set of questions. 

• The questions in each part tend to be progressive in difficulty. 

• If you get bogged down on one question, move on to another. 

• If you get bogged down on one part, move the the next.

Part I, Part II and Part III follow below, in order.  The links below might or might not be useful in helping you navigate.

• Part I:  The Equation m x '' = - k x

• Part II:  Solution of Equations requiring only Direct Integration

• Part III:  Direction Fields and Approximate Solutions

Part I:  The equation m x '' = - k x

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Question:  `q001.  Show whether each of the following functions all satisfy the equation m x '' = -k x:

• x = cos(t)

• x = sin( sqrt( k / m) * t)

• x = 3 cos( sqrt( k / m) * t ) + 5 sin (sqrt(k / m) t)

• x = B sin(sqrt(k / m) * t) + C cos( sqrt( k / m) * t + 3)

If x = cos(t) then x ' = - sin(t) and x '' = - cos(t).  Substituting the expressions for x and x '' into the equation we obtain

m * (-cos(t)) = - k * cos(t).

Dividing both sides by cos(t) we obtain m = k.  If m = k, then the equation is satisfied.  If m is not equal to k, it is not.

 

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Your solution: 

 1. x' = -sin(t)

x'' = -cos(t)

m(-cos(t)) = -k(cos(t))

m = k

2. x' = sqrt(k/m)*cos(sqrt(k/m)*t)

x'' = -k/m sin(sqrt(k/m)*t)

m(-k/m sin (sqrt(k/m)*t)) = -k sin(sqrt(k/m)*t)

-k sin (sqrt(k/m)*t) = -k sin (sqrt(k/m)*t)

 0 =0 ?

I think this is a solution

 

 3. x ' = -3 sqrt(k/m) sin( sqrt(k/m)*t) +5 sqrt(k/m) cos(sqrt(k/m)*t)

x'' = -3k/m cos(sqrt(k/m)*t) - 5k/m sin (sqrt(k/m)*t)

m (-3k/m cos(sqrt(k/m)*t) - 5k/m sin (sqrt(k/m)*t)) = -k (3cos(sqrt(k/m)*t) + 5sin (sqrt(k/m)*t))

-k (3 cos(sqrt(k/m)*t) + 5 sin (sqrt(k/m)*t)) = -k ((3cos(sqrt(k/m)*t) + 5sin (sqrt(k/m)*t))

0 = 0

4. x' = B sqrt(k/m) cos (sqrt(k/m)*t) - C sqrt(k/m) sin( sqrt(k/m)*t + 3)

x'' = -B k/m sin(sqrt(k/m)*t) - C k/m cos (sqrt(k/m)*t +3)

m (-B k/m sin(sqrt(k/m)*t) - C k/m cos (sqrt(k/m)*t +3)) = -k ( B sin (sqrt(k/m)*t) + C cos(sqrt(k/m)*t + 3) )

 -k ( B sin (sqrt(k/m)*t) + C cos(sqrt(k/m)*t + 3) ) = -k ( B sin (sqrt(k/m)*t) + C cos(sqrt(k/m)*t + 3) )

 0 = 0

 

 

confidence rating #$&*:

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Given Solution: 

If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t).  Substituting this into the equation we have

m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t

Simplifying both sides we see that the equation is true.

 

The same procedure can and should be used to show that the third equation is true, while the fourth is not.

 

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Self-critique (if necessary):

 I don’t understand why the fourth equation is not true ???

 

 

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Self-critique rating: 0

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Question: 

`q002.  An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2.  After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2.

Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality.

Explain what is wrong with the reasoning given above.

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Your solution: 

 

 If we are integrating with respect to x, then t is just a constant.

integral(x' dx) = integral ( 2x + t dx)

x = x^2 + tx +C

However, if we were integrating both sides with respect to t, 2x would be considered a constant.

integral ( x' dt ) = integral( 2x + t dt)

x' *t = 2x*t + 1/2 t^2

You cannot integrate with respect to both terms at once

 

confidence rating #$&*:

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Given Solution: 

The given function is a solution to the equation, provided its derivative x ' satisfies x ' = 2 x + t. 

It would be tempting to say that the derivative of x^2 is 2 x, and the derivative of t^2 / 2 is t.

The problem with this is that the derivative of x^2 was taken with respect to x and the derivative of t^2 / 2 with respect to t.

We have to take both derivatives with respect to the same variable.

 

Similarly we can't integrate the expression 2 x + t by integrating the first term with respect to x and the second with respect to t.

Since in this context x ' represent the derivative of our solution function x with respect to t, the variable of integration therefore must be t. 

We will soon see a method for solving this equation, but at this point we simply cannot integrate our as-yet-unknown x(t) function with respect to t.

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Self-critique (if necessary):OK

 

 

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Self-critique rating:OK

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Question: 

`q003.  The general solution to the equation

m x '' = - k x

is of the form x(t) = A cos(omega * t + theta_0), where A, omega and theta_0 are constants.  (There are reasons for using the symbols omega and theta_0, but for right now just treat these symbols as you would any other constant like b or c).

Find the general solution to the equation 5 x'' = - 2000 x:

• Substitute A cos(omega * t + theta_0) for x in the given equation.

• The value of one of the three constants A, omega and theta_0 is dictated by the numbers in the equation.  Which is it and what is its value?

• One of the unspecified constants is theta_0.  Suppose for example that theta_0 = 0.  What is the remaining unspecified constant?

• Still assuming that theta_0 = 0, describe the graph of the solution function x(t).

• Repeat, this time assuming that theta_0 = 3 pi / 2.

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Your solution: 

 

General solution:

x'' = -400x, then integrate

x' = -400 (1/2) x^2 + C_0

@&

This is not valid.

The ' does not represent the derivative with respect to x, but the derivative with respect to a second variable. In this context, where x represents position, x ' velocity and x '' acceleration, that second variable is clock time t.

*@

x' = -200 x^2 + C_0, integrate again

x = -200 (1/3) x^3 + C_0 * x + C_1

x = -200/3 x^3 + C_0x + C_1

@&

So far your integrations are with respect to x, which per the previous note is not the appropriate variable.

Specifically if x was the variable of integration, then x '' would be zero.

*@

substituting x with Acos(omega*t + theta_0)

x = A cos (omega*t + theta_0)

x' = -A*omega sin( omega*t + theta_0)

x'' = -A * omega^2 cos(omega*t + theta_0)

5(-A * omega^2 cos (omega*t + theta_0) ) = -2000 (Acos (omega*t + theta_0))

5omega^2 = 2000

omega = 20

A is unspecified

The function x(t) will be a sine curve, oscillating between +A and -A. It goes through the point (0,0) and its period is 10/pi.

The new function x(t) will be the same as before except it will be shifted to the left by 3pi/2 units

confidence rating #$&*:

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Given Solution: 

If x = A cos(omega * t + theta_0) then x ' = - omega A sin(omega * t + theta_0) and x '' = -omega^2 A cos(omega * t + theta_0).

Our equation therefore becomes

m * (-omega^2 A cos(omega * t + theta_0) ) = - k A cos(omega * t + theta_0).

Rearranging we obtain

-m omega^2 A cos(omega * t + theta_0) = -k A cos(omega * t + theta_0)

so that

-m omega^2 = - k

and

omega = sqrt(k/m).

Thus the constant omega is determined by the equation.

The constants A and theta_0 are not determined by the equation and can therefore take any values. 

No matter what values we choose for A and theta_0,  the equation will be satisfied as long as omega = sqrt(k / m).

Our second-order equation

m x '' = - k x

therefore has a general solution containing two arbitrary constants.

 

In the present equation m = 5 and k = 2000, so that omega = sqrt(k / m) = sqrt(2000 / 5) = sqrt(400) = 20.

Our solution x(t) = A cos(omega * t + theta_0) therefore becomes

x(t) = A cos(20 t + theta_0).

If theta_0 = 0 the function becomes x(t) = A cos( 20 t ).  The graph of this function will be a 'cosine wave' with a 'peak' at the origin, and a period of pi / 10.

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Self-critique (if necessary):

 I said sine wave instead of cosine, but I understand the solution

 

 

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Self-critique rating:3

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Question: 

`q004.  In the preceding equation we found the general solution to the equation 5 x'' = - 2000 x.  Assuming SI units, this solution applies to a simple harmonic oscillator of mass 5 kg, which when displaced to position x relative to equilibrium is subject to a net force F = - 2000 N / m * x.  With these units, sqrt(k / m) has units of sqrt( (N / m) / kg), which reduce to radians / second.  Our function x(t) describes the position of our oscillator relative to its equilibrium position.

Evaluate the constants A and theta_0 for each of the following situations:

• The oscillator reaches a maximum displacement of .3 at clock time t = 0.

• The oscillator reaches a maximum displacement of .3 , and at clock time t = 0 its position is x = .15.

• The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0.

• The oscillator has a maximum velocity of 2, which occurs at clock time t = 0.  (Hint:  The velocity of the oscillator is given by the function x ' (t) ).

As seen in the preceding problem, a general solution to the equation is

x = A cos(omega * t + theta_0),

where omega = sqrt(k / m).  For the current equation 5 x '' = -2000 x, this gives us omega = 20.  In the current context omega = 20 radians / second.

So

x(t) = A cos( 20 rad / sec * t + theta_0 ).

Maximum displacement occurs at critical values of t, values at which x ' (t) = 0.

Taking the derivative of x(t) we obtain

x ' (t) = - 20 rad / sec * A sin( 20 rad/sec * t + theta_0).

The sine function is zero when its argument is an integer multiple of pi, i.e., when

20 rad/sec * t + theta_0 = n * pi, where n = 0, +-1, +-2, ... .

A second-derivative test shows that whenever n is an even number, our x(t) function has a negative second derivative and therefore a maximum value.

We can therefore pick any even number n and we will get a solution.

If maximum displacement occurs at t = 0 then we have

20 rad / sec * 0 + theta_0 = n * pi

so that

theta_0 = n * pi,  where n can be any positive or negative even number.

We are free to choose any such value of n, so we make the simplest choice, n = 0.  This results in theta_0 = 0.

 

Now if x = .3 when t = 0 we have

A cos(omega * 0 + theta_0) = .3

 

 

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Your solution: 

 1. I am unsure how to solve for two variables without a second equation

.3 = A cos(theta_0) * , I have two variables and only one equation.

at time of max displacement, deriv. should be 0

x' = -20A sin(20t + theta_0)

0 = -20A sin(theta_0) *

how do I combine these two equations (*) to solve for A and theta_0???

@&

You are told that maximum displacement occurs at t = 0.

Maximum displacement occurs when the cosine function takes its greatest possible value.

What is the maximum possible value of the cosine function?

If cos(theta_0) is equal to that maximum value, what therefore is the value of theta_0?

*@

2. I am confused. Isn't maximum displacement equal to it's x coordinate???

@&

The x coordinate does represent displacement from equilibrium. So max displacement represents the maximum possible value of x.

*@

3. max velocity of 2 means that x''(2) = 0, max displacement means x' = 0

@&

x is position.

x ' is therefore velocity, and x '' is acceleration.

Maximum velocity 2 implies that x ' (t) has maximum value 2.

Maximum displacement does occur when x ' (t) = 0. Physically, this means that the oscillator comes to rest for an instant before turning around and heading back toward the equilibrium position.

*@

4 ???

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 I do not understand this problem ???

 

 

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Self-critique rating: 0

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Question: 

`q005.  Describe the motion of the oscillator in each of the situations of the preceding problem.  SI units for position and velocity are respectively meters and meters / second.

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Your solution: 

 

 

The oscillator reaches a maximum displacement of .3 at clock time t = 0.

At t = 0, the oscillator is as far away from the equilibrium point as it will ever be. The velocity at this point is 0 m/s. the acceleration will be at it's maximum because at the point of max displacement, the acceleration will be going from negative to positive.

The oscillator reaches a maximum displacement of .3 , and at clock time t = 0 its position is x = .15.

I do not understand how the max displacement can be unequal to the position when it is at max displacement.

The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0

the maximum velocity occurs when the object is passing through it's equilibrium point. The equilibrium point occurs halfway through the period f the oscillation. At t = .3, the object is momentarily at rest, x' = 0.

at 10/pi, the object is going its max speed of 2 m/s, x'(10/pi) = 2 m/s.

The oscillator has a maximum velocity of 2, which occurs at clock time t = 0.  (Hint:  The velocity of the oscillator is given by the function x ' (t) ).

 at the start of time, the object has a velocity of 2 m/s.

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 I am still unsure about this and the previous question ???

 

 

@&

Hopefully my notes on the preceding will be helpful. You still might have additional questions, which are always welcome.

*@

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Self-critique rating:0

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Question: 

Part II:  Solutions of equations requiring only direct integration.

`q006.  Find the general solution of the equation x ' = 2 t + 4, and find the particular solution of this equation if we know that x ( 0 ) = 3.

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Your solution: 

 integrate both sides

Integral of 2t = t^2, integral of 4 = 4t

x = t^2 + 4t + C

Substitute 0 for t and 3 for x, solving for C

3 = 0 + 0 + C

C = 3

x = t^2 +4t +3

confidence rating #$&*:

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Given Solution: 

Integrating both sides we obtain

x(t) = t^2  + 4 t + c,

where c is an arbitrary constant.

The condition x(0) = 3 becomes

x(0) = 0^2 + 4 * 0 + c = 3,

so that c = 3 and our particular solution is

x(t) = t^2  + 4 t + 3.

 

We check our solution.

Substituting x(t) = t^2  + 4 t + 3 back into the original equation:

(t^2  + 4 t + 3) ' = 2 t + 4 yields

2 t + 4 = 2 t + 4,

verifying the general solution.

The particular solution satisfies x(0) = 3:

x(0) = 0^2 + 4 * 0 + 3 = 3.

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Self-critique (if necessary): OK

 

 

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Self-critique rating:OK

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Question: 

`q007.  Find the general solution of the equation x ' ' = 2 t - .5, and find the particular solution of this equation if we know that x ( 0 ) = 1, while x ' ( 0 ) = 7. 

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Your solution: 

 

 take the integral of both sides. integral of 2t = 2(1/2)t^2, integral of -.5 = -.5t

x' = t^2 -.5t +C

using the initial values, x' = 7, t = 0

7 = 0 + 0 + C, C = 7

x' = t^2 -.5t +7

Integrate both sides. integral of t^2 = (1/3)t^3, integral of -.5t = -(1/2).5t^2, integral of 7 = 7t

x = (1/3)t^3 - .25t +7t +C

plug in initial values

1 = 0 + 0 + 0 + C

C = 1

x = (1/3)t^3 -.25t +7t +1

 

 

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Given Solution: 

Integrating both sides we obtain

x ' = t^2 - .5 t + c_1,

where c_1 is an arbitrary constant.

Integrating this equation we obtain

x = t^3 / 3 - .25 t^2 + c_1 * t + c_2,

where c_2 is an arbitrary constant.

Our general solution is thus

x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2.

The condition x(0) = 1 becomes

x(0) = 0^3 / 3 - .25 * 0^2 + c_1 * 0 + c_2 = 1

so that c_2 = 1.

x ' (t) = t^2 - .5 t + c_1, so our second condition x ' (0) = 7 becomes

x ' (0) = 0^2 - .5 * 0 + c_1 = 7

so that c_1 = 7.

For these values of c_1 and c_2, our general solution x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2 becomes the particular solution

x(t) = t^3 / 3 - .25 t^2 + 7 t + 1.

You should check to be sure this solution satisfies both the given equation and the initial conditions.

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Self-critique (if necessary):

 OK

 

 

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Self-critique rating:OK

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Question: 

`q008.  Use the particular solution from the preceding problem to find x and x ' when t = 3.  Interpret your results if x(t) represents the position of an object at clock time t, assuming SI units.

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Your solution: 

Simply plugging in 3 wherever t appears,

 x(3) = (1/3)(3)^3 - .25(3)^2 +7(3) + 1 = 9 - 2.25 +21 +1 = 28.75 meters

x' (3) = (3)^2 - .5(3) +7 = 9 - 1.5 + 7 = 14.5 meters/second

At 3 seconds, the object is 28.75 meters from when t = 0. Also at that instant, it's velocity is 14.5 meters/second

confidence rating #$&*:

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Given Solution: 

Our solution was

x(t) = t^3 / 3 - .25 t^2 + 7 t + 1.

Thus

x ' (t) = t^2 - .5 t + 7.

When t = 3 we obtain

x(3) = 3^3 / 3 - .25 * 3^2 + 7 * 3 + 1 = 28.75

and

x ' (3) = 3^2 - .5 * 3 + 7 = 14.5.

A graph of x vs. t would therefore contain the point (3, 28.75), and the slope of the tangent line at that point would be 14.5.

 

x(t) would represent the position of an object.  x(3) = 28.75 represents an object whose position with respect to the origin is 28.75 meters when the clock reads 3 seconds.

x ' (t) would represent the velocity of the object.  x ' (3) = 14.5 indicates that the object is moving at 14.5 meters / second when the clock reads 3 seconds.

 

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Self-critique (if necessary):OK

 

 I like these questions better than part 1

 

 

@&

Part 1 requires the interpretation and manipulation of trigonometric functions, and benefits from some experience with oscillatory systems. Both are reasonable things to expect of a student who has had the prerequisites for a Differential Equations course, but prerequisite courses vary in coverage, topics and level of challenge, so in reality this is often not the case.

That is my purpose for introducing these ideas at the beginning of the course. Better to struggle a little bit with them now and be better prepared a little later on when you need them, than to struggle then.

*@

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Self-critique rating:OK

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Question: 

`q009.  The equation x '' = -F_frict / m - c / m * x ', where the derivative is understood to be with respect to t, is of at least one of the forms listed below.  Which form(s) are appropriate to the equation?

• x '' = f(x, x')

• x '' = f(t)

• x '' = f(x, t)

• x '' = f(x', t)

• x '' = f(x, x ' t)

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Your solution: 

 

 I think the form x'' = f(x', t) is the only appropriate one. It cannot be the first option because x is not in the equation. It cannot be the second option because x' is a part of the equation but not listed as an input. It cannot be option 3 because x is not in the equation. And for the same reason it cannot be the last one.

It has to be the one that includes t and x' as inputs.

 

 

confidence rating #$&*:

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Given Solution: 

The right-hand side of the equation includes the function x ' but does not include the variable t or the function x.

So the right-hand side can be represented by any function which includes among its variables x '.  That function may also include x and/or t as a variable.

The forms f(t) and f(x, t) fail to include x ', so cannot be used to represent this equation.

All the other forms do include x ' as a variable, and may therefore be used to represent the equation.

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Self-critique (if necessary):

 

 Why do we not take t into consideration? How can you take a derivative with respect to t if t is not an input?

???

 

@&

The derivative with respect to t would, in this case, be zero.

A function can certainly be constant with respect to a variable. For example

f(x) = 7

has no x dependence, but it's still a funciton of x, and can be graphed on the coordinate plane as a horizontal line.

Similarly any function f(x, x ') is also a function of f(x, x ', t), even though t does not appear in the form of the function.

When the number 0 was first introduced a lot of very smart people had a similar question of why we would want a symbol for something that isn't even there.

*@

 

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Question: 

`q010.  If F_frict is zero, then the function x in the equation

x '' = -F_frict / m - c / m * x '

represents the position of an object of mass m, on which the net force is - c * x '. 

Explain why the expression for the net force is -c * x '.

Explain what happens to the net force as the object speeds up.

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Your solution: 

 I'm not too sure what you are asking.

when F_frict = 0, the only thing that changes in the equation is that F_frict/m drops out and we get

x'' = -c/m*x'

and net force = -c*x'

I'm not sure why net force is that. Do we have an original equation for net force that includes F_frict?

???

@&

Newton's Second Law says that the net force on an object of mass m is

F_net = m x ''.

It follows immediately that x '' = F_net / m.

*@

 

 However, when the object speeds up, x' gets larger

if F_net = -c*x' and x' is increasing, F_net is getting more negative

 

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Given Solution: 

Newton's Second Law gives us the general equation

m x '' = F_net

so that

x '' = F_net / m.

It follows that

x '' = -F_frict / m - c / m * x '

represents an object on which the net force is -F_frict - c x '.

If F_frict = 0, then it follows that the net force is -c x '.

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Self-critique (if necessary):

 I didn't realize we were using the equation F_net = mx'', but I understand the problem.

What happens to F_net when the object speeds up? ???

 

@&

F_net = -c x '.

If x speeds up, what in this equation changes?

*@

 

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Question: 

`q011.  We continue the preceding problem.

• If w(t) = x '(t), then what is w ' (t)? 

• If x '' = - b / m * x ', then if we let w = x ', what is our equation in terms of the function w?

• Is it possible to integrate both sides of the resulting equation?

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Your solution: 

 If w(t) = x'(t), then w'(t) = x''(t)

Subbing w = x'

x'' = -b/m*w

it is possible to integrate both sides as long as the equation for w, which is x' = t^2-.5t+7

 

 

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Given Solution: 

If w(t) = x ' (t) then w ' (t) = (x ' (t) ) ' = x '' ( t ).

If x '' = - b / m * x ', then if w = x ' it follows that x '' = w ', so our equation becomes

w ' (t) = - b / m * w (t)

The derivative is with respect to t, so if we wish to integrate both sides we will get

w(t) = integral ( - b / m * w(t) dt),

The variable of integration is t, and we don't know enough about the function w(t) to perform the integration on the right-hand side.

[ Optional Preview:

There is a way around this, which provides a preview of a technique we will study soon.  It isn't too hard to understand so here's a preview:

w ' (t) means dw / dt, where w is understood to be a function of t.

So our equation is dw/dt = -b / m * w.

It turns out that in this context we can sort of treat dw and dt as algebraic quantities, so we can rearrange this equation to read

dw / w = -b / m * dt.

Integrating both sides we get

integral (dw / w) = -b / m integral( dt )

so that

ln | w | = -b / m * t + c.

In exponential form this is

w = e^(-b / m * t + c).

There's more, but this is enough for now ... ].

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Self-critique (if necessary):

 

 why can't we plug in x' for w and integrate???

@&

This is exactly what we do.

However we can't integrate

x '' = -b / m * w

because we have to functions of t, namely x and w.

Can x '' be written in terms of w? If so we might end up with an equation that we can integrate.

*@

 

 

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Self-critique rating:0

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Question: 

Part III:  Direction fields and approximate solutions

`q012.  Consider the equation x ' = (2 x - .5) * (t + 1).  Suppose that x = .3 when t = .2.

If a solution curve passes through (t, x) = (.2, .3), then what is its slope at that point?

What is the equation of its tangent line at this point?

If we move along the tangent line from this point to the t = .4 point on the line, what will be the x coordinate of our new point?

If a solution curve passes through this new point, then what will be the slope at the this point, and what will be the equation of the new tangent line?

If we move along the new tangent line from this point to the t = .6 point, what will be the x coordinate of our new point?

Is is possible that both points lie on the same solution curve?  If not, does each tangent line lie above or below the solution curve, and how much error do you estimate in the t = .4 and t = .5 values you found? 

 

At the point (.2, .3) in the (t, x) plane, our value of x ' is

x ' = (2 * .3 - .5) * (.2 + 1) = .12, approximately.

This therefore is the slope of any solution curve which passes through the point (.2, .3).

The equation of the tangent plane is therefore

x - .3 = .12 * (t - .2)

so that

x = .12 t - .24.

If we move from the t = .2 point to the t = .4 point our t coordinate changes by `dt = .2, so that our x coordinate changes by `dx = (slope * `dt) = .12 * .2 = .024.  Our new x coordinate will therefore be .3 + .024 = .324.

This gives us the new point (.4, .324).

At this point we have

x ' = (2 * .324 - .5) * (.4 + 1) = .148 * 1.4 = .207.

If we move to the t = .6 point our change in t is `dt = .2.  At slope .207 this would imply a change in x of `dx = slope * `dt = .207 * .2 = .041.  Our new x coordinate will therefore be .324 + .041 = .365.

Our t = .6 point is therefore (.6, .365).

 

From our two calculated slopes, the second of which is significantly greater than the first, it appears that in this region of the x-t plane, as we move to the right the slope of our solution curve in fact increases.  Our estimates were based on the assumption that the slope remains constant over each t interval.  We conclude that our estimates of the changes in x are probably a somewhat low, so that our calculated points lie a little below the solution curve.

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Your solution: 

 I believe you gave the solution before the part where I'm supposed to solve it.

Pretending I didn't see that:

x' = (2(.3)-.5)(.2+1) = (.6-.5)*1.2 = 0.12

this is the slope of the solution curve at (.3,.2)

The equation of tangent line at (.2, .3):

x = .12t + C

.3 = .12(.2) + C

.3 = .024 + C

C = .276

x = .12t + .276

plugging in t = .4

x = .12(.4) + .276 = .324

at (.4, .324)

x' = (2(.324) - .5) (.4 +1) = (.648 - .5)*1.4 = 0.2072

The slope of the solution curve at this point is .2072

If that is the new slope:

x = .2072t + C

.324 = .2072(.4) + C

C = .24112

x = .2072t + .24112, equation for new tangent line

At t = .6

x = .2072(.6) + .24112 = 0.36544

I'm not too sure how you calculate the error, although I know there is one.

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

This feels like Euler's method. However, I'm not really sure why we are doing this, and I'm having a hard time visualizing it

 

 

@&

Very good observation. This is in fact Euler's Method, which can be applied to direction fields in any number of dimensions.

Euler's Method is simply a means of following the direction field to obtain an approximate solution.

The direction field allows us to easily visulaize the entire family of solution curves.

*@

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Self-critique rating:OK

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Question: 

`q013.   Consider once more the equation x ' = (2 x - .5) * (t + 1). 

Note on notation:

The points on the grid

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

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

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

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

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

can be specified succinctly in set notation as

{ (t, x) | t = 0, 1/4, ..., 1, x = 0, 1/4, ..., 1}.

( A more standard notation would be { (i / 4, j / 4) | 0 <= i <= 4, 0 <= j <= 4 } )

Find the value of x ' at every point of this grid and sketch the corresponding direction field.  To get you started the values corresponding to the first, second and last rows of the grid are

-.5, -.625, -.75, -.875, -1

0, 0, 0, 0, 0

...

...

1.5, 1.875, 2.25, 2.625, 3

So you will only need to calculate the values for the third and fourth rows of the grid.

• List your values of x ' at the five points (0, 0), (1/4, 1/4), (1/2, 1/2), (3/4, 3/4) and (1, 1).

• Sketch the curve which passes through the point (t, x) = (.2, .3). 

• Describe your curve.  Is it increasing or decreasing, and is it doing so at an increasing or decreasing rate? 

• According to your curve, what will be the value of x when t = 1?

• Sketch the curve which passes through the point (t, x) = (.5, .7).  According to your curve, what will be the value of x when t = 1?

• Describe your curve and compare it with the curve you sketched through the point (.2, .3).

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Your solution: 

 The missing two rows:

.5, .625, .75, .875, 1

1, 1.25, 1.5, 1.75, 2

(0,0) = -.5

(.25, .25) = 0

(.5, .5) = .75

(.75, .75) = 1.75

(1, 1) = 3

at point (.2, .3)

x' = (2(.3) - .5)(.2+1) = .12, the curve at (.2, .3) is increasing at a steady rate

At t = 1

x ' = (2x-.5) (1+1) = 4x-1

How do I solve for the value of x without another number???

Curve at (.5, .7):

x' = (2(.7) - .5)((.5) +1) = 1.35

The slope is positive so the curve is increasing at a steady rate.

The curve at (.2, .3) is smaller than the curve at (.5, .7)

.12<1.35

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 Not sure how to solve for x in the fourth bullet ???

 

 

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Self-critique rating:

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Question: 

`q014.   We're not yet done with the equation x ' = (2 x - .5) * (t + 1). 

x ' is the derivative of the x(t) function with respect to t, so this equation can be written as

dx / dt = (2 x - .5) * (t + 1).

Now, dx and dt are not algebraic quantities, so we can't multiply or divide both sides by dt or by dx.  However let's pretend that they are algebraic quantities, and that we can.  Note that dx is a single quantity, as is dt, and we can't divide the d's.

• Rearrange the equation so that expressions involving x are all on the left-hand side and expressions involving t all on the right-hand side.

• Put an integral sign in front of both sides.

• Do the integrals.  Remember that an integration constant is involved.

• Solve the resulting equation for x to obtain your general solution.

• Evaluate the integration constant assuming that x(.2) = .3.

• Write out the resulting particular solution.

• Sketch the graph of this function for 0 <= t <= 1.  Describe your graph.

• How does the value of your x(t) function at t = 1 compare to the value your predicted based on your previous sketch?

• How do your values of x(t) at t = .4 and t = .6 compare with the values you estimated previously?

The equation is easily rearranged into the form

dx / (2 x - .5) = (t + 1) dt.

Integrating the left-hand side we obtain 1/2 ln | 2 x - .5 |

Integrating the right-hand side we obtain t^2 / 2 + 4 t + c, where the integration constant c is regarded as a combination of the integration constants from the two sides.

Thus our equation becomes

1/2 ln | 2 x - 5 | = t^2 / 2 + t + c. 

Multiplying both sides by 2, then taking the exponential function of both sides we get

exp( ln | 2 x - 5 | ) = exp( t^2 + 2 t + c ),

where as before c is an arbitrary constant. 

Since the exponential and natural log are inverse functions the left-hand side becomes | 2 x + .5 |.

The right-hand side can be written e^c * e^(t^2 + 8 t), where c is still an arbitrary constant.  e^c can therefore be any positive number, and we replace e^c with A, understanding that A is a positive constant.

Our equation becomes

| 2 x - .5 | = A e^(t^2 + 2 t).

For x > -.25, as is the case for our given value x = .3 when t = .2, we have

2 x - .5 = A e^(t^2 + 2 t)

so that

x = A e^(t^2 + 2 t) + .25.

Using x = .3 and t = .2 we find the value of A:

.05 = A e^(.2^2 + 2 * .2)

so that

A = .05 / e^(.44) = .03220, approx..

Our solution function is therefore

x(t) = .05 / e^(.44) * e^(t^2 + 2 t) + .25, or approximately

x(t) = .03220 e^(t^2 + 2 t) + .25

 

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Your solution: 

 

 I'm not sure why you're giving part of the solution before where we're supposed to solve

knowing that

x(t) = .03220 e^(t^2 + 2 t) + .25

plugging in points

t x

0 .2822

.25 .306

.5 .362

.75 .503

1 .897

The graph is increasing at an increasing rate.

In the problem above, you calculated the graph at (1, 1) to be 3. Our previous guess was less than this guess.

Previously I calculated that at t = .4 the graph is at .324, and at .6 it is at .3644

using x(t) = .0322 e^(t^2 + 2t) +.25

x(.4) = .0322 e^((.4)^2 + 2(.4)) + .25  =  .334, this estimate is larger than the last estimate

x(.6) = .0322e^((.6)^2 + 2(.6)) + .25 = .403, this estimate is larger than the last estimate

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 I'm not really sure what we're doing anymore. Is there a purpose to calculating this graph multiple times with different equations? which way is correct ???

 

 

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Self-critique rating:

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Question: 

`q015.  OK, this time we are really going to be done with this equation.  Again, x ' = (2 x - .5) * (t + 1)

• Along what line or curve is x ' = 1?

• Along what line or curve is x ' = 0?

• Along what line or curve is x ' = 2?

• Along what line or curve is x ' = -1?

• Sketch these three lines and/or curves for 0 <= t <= 1. 

• Along each of these lines x ' is constant.  Along each sketch 'slope segments' with slopes equal to the corresponding value of x '.

• How consistent is your sketch with your previous sketch of the direction field?

• Sketch a solution curve through the point (.2, .25), and estimate the coordinates of the t = 1 point on this curve.

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Your solution: 

 I'm not sure what you are asking. When x' = 1, we still have two variables to solve for.

1 = (2x -.5)(t + 1)

 Rearranging so that x is on one side

x = 1/(2t+2) + .25

@&

Good. This is a curve which when plotted on an x vs. t plane has a vertical asymptote at t = -1 and a horizontal asymptote at x = .25. The curve lies above the horizontal asymptote to the right of the vertical asymptote, and below to the left.

*@

when x' = 0

0 = (2x -.5)(t+1)

2x-.5 = 0

x = .25

when x' = 2

x = 1/(t+1) + .25

when x' = -1

x = -1/(2t+2) +.25

Sketch these three lines and/or curves for 0 <= t <= 1. 

 ??? aren't there four?

@&

There are in fact four.

*@

I don't understand what you are asking me to graph.

 

 

@&

The note should clarify this.

*@

confidence rating #$&*:

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Given Solution: 

x ' = 1 when

(2 x - .5) * (t + 1) = 1.

Solving for x we obtain

x = 1/2 ( 1 / (t + 1) + .5) = 1 / (2(t + 1)) + .25.

The resulting curve is just the familiar curve x = 1 / t, vertically compressed by factor 2 then shifted -1 unit in the horizontal and .25 unit in the vertical direction, so its asymptotes are the lines t = -1 and x = .25.  The t = 0 and t = 1 points are (0, .75) and (1, .5).

Similarly we find the curves corresponding to the other values of x ':

For x ' = 0 we get the horizontal line x = .25.  Note that this line is the horizontal asymptote to the curve obtained in the preceding step.

For x ' = 2 we get the curve 1 / (t + 1) + .25, a curve with asymptotes at t = -1 and x = .25, including points (0, 1.25) and (1, .75).

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Self-critique (if necessary):

 I do not understand this question.

Are you graphing four separate graphs on one xt plane? ???

Are we supposed to know what the graph looks like without graphing?

I don't know how you solved for asymptotes.

What is a 'slope segment' and how do you graph it?

@&

This terminology should be clear from the very first few Introductory Videos. The direction field consists of a large number of short line segments indicating the slopes of solution curves at various points.

*@

@&

Briefly, these functions are rational functions. Graphing and analysis of rational functions is a very standard topic in Algebra 2 and especially Precalculus courses, and is expanded when the derivative is used in the process of graphing in a first-year calculus course.

Here's what everyone should know and be able to do when coming into this course:

You should know from algebra and precalculus courses what the graph of y f(t) = 1 / t looks like.

You should also be able to construct, on paper, a reasonable graph of this function in a minute or less.

From this graph you should be able to graph by vertical and horizontal shifts and stretches the graph of any function

A f(c ( t - h ) ) + k,

given the values of A, t and k, where f(t) = 1 / t.

This is also a standard Algebra 2 and Precalculus procedure, which should also be utilized in first-year calculus courses.

Unfortunately it is not always the case that those courses cover the tools required to understand the basic functions when they are encountered in more advanced courses such as this one.

A very basic toolkit for graphing functions would consist of the following:

Ability to quickly graph the following functions:

y = x^2

y = x^3

y = 1 / x

y = 1 / x^2

y = 1 / x^3

y = 2^x

y = cos(x)

y = tan(x)

using a table for each consisting of not more than half a dozen well-chosen x values, accompanied by a little clear thinking about what happens as | x | or | y | approaches infinity.

Ability to construct, from the graph of any given function f(x), the graph of A f( c ( x - h ) + k

The information required to understand the process is in a typical precalculus text, and is all over the Internet.

I just entered the phrase 'graphing by transformations' into my search engine and got a page full of what appear to be excellent links, all saying pretty much the standard stuff.

One item from the page showed this blurb:

Graph Transformations

math.aa.psu.edu/~mark/Math40/shifts.pdf?

Graph Transformations. This is a list of the standard shifting and stretching operations. The point is to allow us to understand a variety of functions and their ...

The process is not difficult to understand, and I'm pretty confident it has been seen before by 99% of the people entering a Differential Equations course. However it is too often the case that students fall back on graphing calculators and never use these skills.

The functions in question on this particular problem are all rational functions which can be put into the form A f( c ( t - h) ) + k and hence graphed using simple transformations. The algebra of the process isn't trivial, but is only of moderate difficulty.

*@

 

 

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Self-critique rating:0

 

These questions can be challenging if you are rusty on first-year calculus and trigonometric functions.

Fortunately we won't be using these functions a lot near the beginning of the course, but they become very important later, and if you're rusty, or if you never really mastered these functions, you want to get a good start now.

If you are rusty, try to answer the following without the use of a calculator:

Quick refresher on trigonometric functions:

What are the maximum and minimum possible values of cos(theta)?

max = 1 min = -1

#$&*

For what values of theta does cos(theta)

take its maximum value, and for what values of theta is cos(theta) equal to zero?

the max is at 0 and the min is at pi

@&

One of your answers is correct, the other is not.

Suggestion: Sketch a unit circle and apply the unit-circle definition of the cosine function.

*@

#$&*

For what values of t does y = cos(5 t) take its maximum value? For what t values does this function take its minimum value?

max = 1, min = -1

@&

cos(5) and cos(-5) are not max and min values of the cosine.

*@

#$&*

What is the maximum possible value of y = 4 cos(83 t)?

-4

@&

4 would be the max; -4 the min

*@

#$&*

Find at least one value of t for which y = cos(2 t + pi) takes the value zero.

t = pi/4 - pi/2

@&

Good.

That would of course simplify to -pi/4, and would give you cos(pi/2), which is zero.

*@

#$&*

Find at least one value of t for which y = 3 cos(2 t + pi/4) takes its maximum value.

3pi/8

@&

Not bad, but that would yield -3, which is a minumum.

*@

#$&*

Find at least one value of t for which the derivative of y = cos(t) is zero.

y' = sin(t) is zero at multiples of pi

@&

Good.

*@

#$&*

Find at least one value of t for which the derivative of y = cos(t) is zero, and the second derivative is negative.

y' = -sin(t), y'' = -cos(t), when t = 0

@&

Good.

*@

#$&*

Find the critical points of y = 4 sin(2 pi t + pi/2), then using a second-derivative test find the values of t at which this function takes its maximum and minimum values.

Feel free to submit a copy of these questions with your answers.

"

@&

I've inserted numerous notes.

You are asking excellent questions and showing very good ability to do this work.

However there are a number of things, mostly at the level of precalculus, that you need to nail down so they don't limit you as you work through this course. Your calculus doesn't look bad, so much of your thinking is at a level higher than that of precalculus. So it should not be difficult to review and if possible learn the necessary precalculus topics.

Check my notes. I'll welcome additional questions.

*@