6364QA

course MTH 174

09:33 p.m. 6/8/10. Please see my notes about the omissions and errors in this document. These problems have made it more difficult to work with this document. Also, I had worked this out yesterday but forgot to submit it and I know it was due yesterday. I'll be sure to get everything in on time from now on.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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Question: 6.3.8 (previously 6.3 #14, ds / dt = -32 t + 100, s = 50 when t = 0). Find the function s(t).

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Your solution: s(t)= -16t^2 + 100t + 50.

confidence rating #$&* 3

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Given Solution:

s' = 100 - 32t.

Integrating with respect to t we obtain

• s= 100t - 16t^2 + C.

Since s = 50 when t = 0 we have

• 50 = 100(0) - 16(0)^2 + C,

which we easily solve to obtain

• 50 =C.

this into the expression for s(t) we have

• s(t) = 100(t) - 16t^2 + 50

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Self-critique (if necessary): OK

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Self-critique rating #$&* 3

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Question: problem 6.3.6 (previously 6.3 #16) water balloon from 30 ft, v(t) = -32t+40

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Your solution:

I had to look at the given solution to know what this question means and from that I am assuming that this question is supposed to be the same as the homework question that it refers to which is: 6) 6.3.16 A water balloon is launched from the roof of a building at time t = 0 and velocity v(t) = -32t + 40ft/s at time t. v > 0 corresponds to vertical motion. If the roof of the building is 30 feet above the ground, find an expression for the height of the water balloon above the ground at time t. What is the average velocity of the balloon between t = 1.5 and t = 3 seconds? A 6-foot person is standing on the ground below, how fast is the balloon when it hits them in the head.

Since velocity is defined as rate of change of position, ds/dt= v(t), we can integrate the velocity function to find the position function. s(t)=Int(v(t))=Int(-32t+40)= -16t^2 + 40t + c. c is in the initial height and is given to be 30 ft. (this can be verified by using s(0)=30). The time it takes for the balloon to hit the ground is found by setting s=0. s(t)=0= -16t^2 + 40t + 30. Solving here gives one root as being ≈3.104. The velocity as it hits the ground is found with v(3.104)≈ 60 ft/s.

confidence rating #$&* 3

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Given Solution:

Since v(t) = s ' (t), it follows that the antiderivative of the v(t) function is the s(t) function so we haves

• s(t) = -16 t^2 + 40 t + c.

Since the building is 30 ft high we know that s(0) = 30.

Following the same method used in the preceding problem we get

• s(t) = - 16 t^2 + 40 t + 30.

The water balloon strikes the ground when s(t) = 0. This occurs when

-16 t^2 + 40 t + 30 = 0. Dividing by 2 we have

-8 t^2 + 20 t + 15 = 0. The quadratic formula gives us

t = [ -20 +- `sqrt( 400 + 480) ] / ( 2 * -8) or

t = 1.25 +- sqrt(880) / 16 or

t = 1.25 +- 29.7 / 16, approx. or

t = 1.25 +- 1.87 or

t = 3.12 or -.62.

The solution t = 3.12 gives us v(t) = v(3.12) = -32 * 3.12 + 40 = -60, approx.

The interpretation of this result is that when it strikes the ground the balloon is moving downward at 60 feet / second.**

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Self-critique (if necessary): Here our roots differ. I didn't round my sort is and that made the 0.06 difference, still approximately the same though.

Self-critique rating #$&* 3

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Question: How fast is the water balloon moving when it strikes the person's head?

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Your solution: We need to find out how much time it will take for the balloon to reach 6 ft. s(t)= 6= -16t^2 + 40t + 30. Rearranging gives the quadratic equation: -16t^2 + 40t + 24. The quadratic equation gives t= -0.5, 3. So at t = 3 the balloon will strike, now the velocity. v(3)= -32*3 + 40= -56, or 52 ft/s (here the negative indicates direction not quantity).

confidence rating #$&* 3

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Given Solution:

** You have to determine the clock time t when the balloon's altitude is 6 feet. You have the s(t) function. So at what time is the altitude 6 ft?
To answer the question we solve the equation
s(t) = 6, i.e.
-16t^2+40t+30=6. This is a quadratic equation. We rearrange to get
-16t^2+40t+24=0; dividing by -16 we have
t^2-5/2t-3/2=0.
We can solve using the quadratic formula or by factorization, obtaining
t=3 or -.5
When t=3, we have
v(3)=-32*3+40=-56.
Thus the velocity at the 6 ft height is 56 ft/sec downward.
**

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Self-critique (if necessary): OK

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Self-critique rating #$&* 3

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Question: What is the average velocity of the balloon between the two given clock times?

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Your solution: Using the mean value theorem: 1/(3-1.5) Int(v(t), 1.5, 3)= 2/3(-16t^2 + 40t, 1.3, 3) = 2/3(-24-24)= 32 ft/s.

confidence rating #$&*

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Given Solution:

Average Velocity=-32 m/s

average velocity =

change in position / change in clock time =

(s(3) – s(1.5) ) / (3 sec – 1.5 sec) =

(6 ft – 54 ft) / (1.5 sec) =

-32 ft / sec.

Alternatively since the velocity function is linear, the average velocity is the average of the velocities at the two given clock times:

• vAve = (v(1.5) + v(3) ) / 2 = (-56 ft / sec + (-8 ft / sec) ) / 2 = -32 ft / sec.

This method of averaging only works because the velocity function is linear.

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Self-critique (if necessary): I used the mean value theorem and thus didn't get a direction. I see though that it was unnecessary because the velocity function is linear.

Self-critique rating #$&* 3

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Question: What function describes the velocity of the balloon as a function of time?

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Your solution: This is given in the question v(t)= -16t + 40. Where -16 is velocity due to gravity and 40 is the initial velocity. I don't understand if thats what the question is really supposed to be asking though.

Confidence Assessment: 1

Given Solution:

** On this problem you are given s(0) = 30. So we have

30 = -16 * 0^2 + 100 * 0 + c or

30 = c.

Thus c = 30 and the solution satisfying the initial condition is

s(t) =- 16 t^2 + 100 t + 30.

To find the clock time when the object strikes the ground, note that at the instant of striking the ground s(t) = 0. So we solve

- 16 t^2 + 100 t + 30 = 0,

obtaining solutions t = -.60 sec and t = 3.10 sec.

The latter solution corresponds to our ‘real-world’ solution, in which the object strikes the ground after being released. The first solution, in which t is negative, corresponds to a projectile which ‘peaks’ at height 30 ft at clock time t = 0, and which was at ground level .60 seconds before reaching this peak.

To find when the height is 6 ft, we solve

- 16 t^2 + 100 t + 30 = 6,

obtaining t = -.5 sec and t = 3.0 sec. We accept the t = 3.0 sec solution.

At t = 3.0 sec and t = 3.10 sec the velocities are respectively

v(3.104) = -32 * 3.10 + 40 = -59.2 and

v(3.0) = -32 * 3 + 40 = -56,

indicating velocities of -59.2 ft/s and -56 ft/s at ground level and at the 6 ft height, respectively.

From the fact that it takes .104 sec to travel the last 6 ft we conclude that the average velocity during this interval is

-6 ft / (.104 sec) = -57.7 ft / sec.

This is how we find average velocity. That is, ave velocity is displacement / time interval, vAve = `ds / `dt.

Since velocity is a linear function of clock time, a graph of v vs. t will be linear and the average value of v over an interval will therefore occur at the midpoint clock time, and will be equal to the average of the initial and final velocities over that interval.

In this case the average of the initial and final velocities over the interval during which altitude decreased from 6 ft to 0 is

vAve = (vf + v0) / 2 = (-59.2 + (-56) ) / 2 ft / sec = -57.6 ft / sec.

This agrees with the -57.7 ft / sec average velocity, which was calculated on the basis of the .104 sec interval, which was rounded in the third significant figure. Had the quadratic equation been solved exactly and the exact value ( sqrt(55) + 5 – 3) / 2 of the time interval been used, and the exact corresponding initial and final velocities, the agreement would have been exact.

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Self-critique (if necessary): All of this seems to just restate what we have already answered in the previous questions. Did the question mean to ask for average velocity function? I'm not sure whats going on here.

Careless editing on my part. The question should have asked for the position function:

'What function describes the position of the balloon as a function of time?'

This is a 'followup question', which many students fail to answer in their original solutions.

When you encounter such a question, if your solution already contains the answer or the equivalent, you don't need to elaborate further.

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Self-critique rating #$&* 1

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Question: 6.4.3 (previously 6.4 #12) derivative of (int(ln(t)), t, x, 1). What is the derivative of this function?

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Your solution: Evaluating this expression using second fundamental theorem we find f(x)= ln 0 - ln x= -ln x.

confidence rating #$&* 3

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Given Solution:

In the following we'll use the format [int(f(t), t, c, x] to stand for 'the integral of f(t) with respect to t with lower limit c and upper limit x'.
The 2d Fundamental Theorem says that d/dx [ int(f(t)), t, c, x ] = f(x). When applying this Theorem you don't find an antiderivative.
The integral we are given has limits x (lower) and 1 (upper), and is therefore equal to -int(ln(t), t, 1, x). This expression is in the form of the Fundamental Theorem, with c = 1, and its derivative with respect to x is therefore - ln(x).
Note that this Theorem is simply saying that the derivative of an antiderivative is equal to the original function, just like the derivative of an antiderivative of a rate-of-depth-change function is the same rate function we start with.

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Self-critique (if necessary): Not as much explanation in my solution.

My solution is intended as an explanation for someone who doesn't understand the problem very well. Your solutions are expected to demonstrate your understand, but certainly need not be as explanatory or detailed.

Self-critique rating #$&* 3

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Question: Why do we use something besides x for the integrand?

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Your solution: The x variable used as a variable for a limit of the integration allows us to substitute in x to find f(x) without actually integrating f(t) where t is used almost as a dummy variable used just to find f(x).

confidence rating #$&* 2

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Given Solution:

** In the statement of the Theorem the derivative is taken with respect to the variable upper limit of the interval of integration, which is different from the variable of integration.

The upper limit and the variable of integration are two different variables, and hence require two different names. **

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Self-critique rating #$&* 2

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Question: 6.4.4 (previously Section 6.4 #18) derivative of (int(e^-(t^2),t, 0,x^3)

What is the desired derivative?

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Your solution: Using the second fundamental theorem we get e^(-(x^3)2) - e^(-0)^2= e^(-x^6).

Confidence Assessment: 2

Given Solution:

** If we were finding (int(e^-(t^2),t, 0,x) the answer would just be e^-(x^2) by the Second Fundamental Theorem.

However the upper limit on the integral is x^3.

• This makes the expression int(e^(-t^2),t,0,x^3) a composite of f(z) = (int(e^-(t^2),t, 0,z) and g(x) = x^3.

• Be sure you understand how the composite f(g(x)) = f(x^3) would be the original expression int(e^(-t^2),t,0,x^3).

Now we apply the chain rule:

g'(x) = 3x^2, and by the Second Fundamental Theorem f'(z) = e^-(z^2). Thus the derivative is

g'(x) f'(g(x))= 3x^2 * e^-( (x^3)^2 ).

Self-critique (if necessary): I didn't explain the whole thing as formally as the given solution. I knew what was going on but didn't show everything.

Self-critique rating #$&* 3

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Question: Find the derivative with respect to x of the integral of e^(t^2) between the limits t and cos(x).

Your solution: This question seems incomplete and I had to find the question in the book to see what was missing and it seems what it is supposed to say is: derivative of int(e^(t^2),t,cos(x), 3) with respect to x. This is found by composition, that is f(g(x))= f(cos x) used with the second fundamental theorem. Evaluation gives: e^(3)^2- (-sin x)*e^((cos x)^2).

the problem should have read 'between the limits 3 and cos(x). The text also changed the order of integration in one edition or the other, so the limits should be cos(x) and 3.

confidence rating #$&* 2

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Given Solution:

If we were finding (int(e^(t^2),t, x, 0) the answer would just be -e^(x^2) by the Second

Fundamental Theorem (along with the reversal of integration limits and therefore sign).

However the lower limit on the integral is cos(x). This makes the expression

int(e^(t^2),t,cos(x), 3) a composite of f(z) = (int(e^(t^2),t, z, 3) and g(x) = cos(x).

Be sure you see that the composite f(g(x)) = f(cos(x)) would be the original expression

int(e^(t^2),t,cos(x),0).

g'(x) = -sin(x), and by the Second Fundamental Theorem f'(z) = -e^(z^2) (again the

negative is because of the reversal of integration limits).

The derivative is therefore

• g'(x) f'(g(x))= -sin(x) * (-e^( (cos(x))^2 ) = sin(x) e^(cos^2(x)).

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Self-critique (if necessary): Okay, the problem is flip-flopping here. First it has: int(e^(t^2),t,cos(x), 3) and then: int(e^(t^2),t,cos(x),0). My answer had the former because that is what the correlating problem in the book used. I could have just as easily have done the latter using the same composition method.

The 0's are the result of careless editing on my part, and I agree that can be confusing.

The limit 3 is fixed, and leads to a constant term when integrated (that term can't be written out in closed form, because the integral of e^(t^2) can't be written in closed form, but that doesn't mean the constant term isn't there). The derivative of that term with respect to x is zero.

The only variation with x comes from the cos(x) limit on the integral.

The last six lines of the given solution are, I believe, correct, with one edit (replacing a 0 by 3):

int(e^(t^2),t,cos(x), 3) a composite of f(z) = (int(e^(t^2),t, z, 3) and g(x) = cos(x).
Be sure you see that the composite f(g(x)) = f(cos(x)) would be the original expression
int(e^(t^2),t,cos(x),3).
g'(x) = -sin(x), and by the Second Fundamental Theorem f'(z) = -e^(z^2) (again the
negative is because of the reversal of integration limits).
The derivative is therefore
• g'(x) f'(g(x))= -sin(x) * (-e^( (cos(x))^2 ) = sin(x) e^(cos^2(x)).

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Self-critique rating #$&* 3

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Question: 6.3.5 (was 6.3 #10) dy/dx = 2x+1

What is the general solution to this differential equation?

……!!!!!!!!……………………There seems to be a lot missing from this section. I'll add in what seems to be the usual sections.

My solution: Integration gives the general solution to be: y= x^2 + x + x.

Confidence: 3

Here I guess would be the given solution all there is is what's below:

student answer: (x^2)/2+x

Instructor response: ** Good. This is an antiderivative of the given function.

So is x^2 / 2 + x + c for any constant number c, because the derivative of a constant is zero.

The general solution is therefore the function y(x) = x^2 / 2 + x + c . **

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Self-critique rating #$&* This solution is wrong, the question has 2x and the integral of that is x^2 and not (x^2)/2.

That's correct. Again careless editing on this end.

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Question: What is the solution satisfying the given initial condition (part c)?

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My Solution: Again there is a lot of missing information. I don't know what part c is referring to, there is a question in the book using this equation and that part c says: y(1)= 5. So we have 5= 1^2 + 1 + c to get c= 3.

Here is where the Given Solution section should be.

** If for example you are given the initial condition y(0) = 12, then since you know that y(x) = x^2 / 2 + x + c, you have y(0) = 0^2 / 2 + 0 + c = 12.

Thus c = 12 and your particular solution is y(x) = x^2 / 2 + x + 12. **

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Self-critique (if necessary): I don't know where the solution comes up with y(0)= 12. I guess it is just a random example.

That initial condition was a 'for example ... ' solution.

Self-critique rating #$&* 1

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Question: What three solutions did you graph, and what does your graph of the three solutions look like?

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Your solution: Here the question must be referring to problem 5 in the 6.3 assignment page (it would be nice to know because sometimes its not as clear as this one was). I used c = 0, 1, and 2 for my three solutions and my graphs were all upward opening parabolas with y-intercepts of 0, 1, and 2 respectively and all the values for y can be found by adding or subtracting the different c values from the c=0 solution.

no new problem is listed, so the question still refers to the same numbered problem

confidence rating #$&* 3

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Given Solution:

** To graph the three solutions you could choose three different values of c.

The graph of x^2 / 2 + x is a parabola; you can find its zeros and its vertex using the quadratic formula.

The graph of x^2 / 2 + x + c just lies c units higher at every point than the graph of x^2 / 2 + x.

So you get a 'stack' of parabolas.

Be sure you work through the details and see the graphs. **

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Self-critique (if necessary): OK

Self-critique rating #$&* 3

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I would like to add that there are a lot of omissions and errors in this document. As you have seen in several parts I had to guess as to what was being asked. I'm not sure if that is just a flaw with this document or if this will continue to be the case. What is your policy on working assignments with these types of difficulties?

Problem statement in these documents are abbreviated, and are more for my reference than yours. It is expected that you have done the problems as assigned, and already know what the problems are asking. In any case, the full statement of every question here should be in the text.

Naturally when there is an error and you catch it, I remember that, and there's a permanent record of it in your posted work. If you are in a situation at the end of the term where your grade is on a borderline, this could definitely make a difference in your favor.

I'll try to proofread upcoming Queries to catch editing errors. I don't think there are many, but obviously I had overlooked some errors that occurred in this document.

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

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