Query 1

#$&*

course Mth 174

I'm still lost on a couple things, do you have any tutoring hours available?

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Question: Problem 4 section 6.3.

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:

Antiderivative of 100 - 32t is:

100t - 16t^2 + C

if s= 50 then:

50= 100(0) - 16(0)^2 + C

C= 50

s(t)= 100t - 16t^2 + 50

confidence rating #$&*:

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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: ok

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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:

S(t) = 6

-16t^2 + 40t +30 = 6

-16t^2 + 40t + 24 = 0

2t^2 - 5t - 3 = 0

by factoring we get:

(2x+1)(x-3) or x= 3 and -.5

to get a negative number we use x= 3

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

56 ft/s downward

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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12:13:49

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Self-critique (if necessary):

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Self-critique Rating:

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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:

Im not sure what two clock times this question is referring to. In one solution I got t= 3.1 and in another I got 3.

3.1 - 3 = .1 change in time

0 - 6 = -6 change in position

-6/.1 = -60 ft/s

confidence rating #$&*: 1

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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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12:15:31

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Self-critique (if necessary):

I’m still not sure where these times came from exactly.

@& The given times came from the statement of the problem in the textbook.*@

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Self-critique Rating: ok

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Question: What function describes the position of the balloon as a function of time? How can this function be used to answer the various questions posed in this problem?

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Your solution:

I’m not really sure what to do here. I would think that you would be able to substitute t from the equation that has been used for the two balloon problems. Since the original equation was for velocity, which is distance over time, the derivative should concern distance with respect to time. From here I would assume that one would be able to substitute a time into this equation and get a distance that the balloon has reached at that time.

@& Your instincts are pretty good here, but note that velocity is the derviative of position, not vice versa. So you would have to integrate the velocity function.*@

confidence rating #$&*: 1

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Given Solution:

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

30 = -16 * 0^2 + 40 * 0 + c or

30 = c.

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

s(t) =- 16 t^2 + 40 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 + 40 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 + 40 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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12:16:06

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Self-critique (if necessary):

I’m still not sure what the equation would be based off of this answer

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

@& In particular a good self-critique would consider at least the first few lines of the solution, which read

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

30 = -16 * 0^2 + 40 * 0 + c or

30 = c.

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

s(t) =- 16 t^2 + 40 t + 30.'

Do you recognize this as the answer to your question about the equation?

The remainder of the given solution uses this equation to answer the original question.*@

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Self-critique Rating:

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Question: Problem 3 Section 6.4

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:

Very very confused on what the t,x, 1 portion of this problem means

confidence rating #$&*:

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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 Second 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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12:23:36

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Self-critique (if necessary):

I shall attempt to ask the math teacher at emory about this but I am still unsure about it

&#When you do not answer a question you need to continue in your self-critique with a phrase-by-phrase analysis of the solution, detailing everything you do and do not understand. Specific questions are encouraged if you do not understand everything.

You should of course always attempt a solution and detail your thinking about what the problem means and how you might solve it. Even an incorrect attempt forms a basis for correction and subsequent understanding.

&#

@& To start with:

The second line reads

'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.'

What do you and do you not understand about this line?

Do you understand the first statement

'The 2d Fundamental Theorem says that d/dx [ int(f(t)), t, c, x ] = f(x). ' ?

If so, how do you understand it? What meaning do you draw from it?

Do you then understand the statement

'When applying this Theorem you don't find an antiderivative.'

If so, what meaning to you attribute to it?*@

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Self-critique Rating: ok

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

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Your solution:

See above

confidence rating #$&*:1

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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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12:24:24

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: Problem 4 Section 6.4.

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:

Unsure as to why there is a t, and a 0 present

confidence rating #$&*:1

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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 ).

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Self-critique (if necessary):

Confused as to why the derivative of this wouldn’t involve finding the derivative of t^2 and multiplying it at any point

@& t is the variable of integration. As you know, once you find the antiderivative you substitute the limits on the integral for the variable of integration. So there is no t^2 in the completed definite integral.

The Second Fundamental Theorem cuts through all that.*@

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Self-critique Rating: ok

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Question: Find the derivative with respect to x of the integral of e^(t^2) between the limits cos(x) and 3.

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Your solution:

confidence rating #$&*:

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Given Solution:

If we were finding (int(e^(t^2),t, x, 3) 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),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 (if necessary):

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Self-critique Rating:

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Question: Section 6.3 Problem 5

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

What is the general solution to this differential equation?

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11:18:57

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Your solution:

X^2 + x + c

C can be any constant as taking the derivative will eliminate it

confidence rating #$&*:

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Given Solution:

student answer: x^2 / 2+x

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

So is x^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 + x + c . **

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11:18:58

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Self-critique (if necessary): ok

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Self-critique Rating: ok

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

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Your solution:

confidence rating #$&*:

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Given Solution:

** 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 + x + 12. **

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Self-critique (if necessary):

&#You did not answer the given question. You need to always at least explain what you do and do not understand about the question. A phrase-by-phrase analysis is generally required when you cannot otherwise answer a question.

&#

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Self-critique Rating:

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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:

Since the only part of the graph you are changing is c, you end up with a stack of parabolas that vary in respect to the local minimum

@& Good.*@

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 + x is a parabola; you can find its zeros and its vertex using the quadratic formula.

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

So you get a 'stack' of parabolas.

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

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11:20:51

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Self-critique (if necessary): ok

@& I have inserted some notes, many related to the process of self-critique. In general you aren't telling me the specifics of what you do and do not understand. If you do I'm sure I can help.

Not that you're doing badly here. The Second Fundamental Theorem, for example, can be elusive.

I suggest you add detailed self-critiques on these questions and resubmit so I can focus my responses on your specific situation.*@