assignment 2 query

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course mth 174

time 822date 9-20-10

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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:

S=50 or C and this makes the equation 16t^2+100(t)+50=s(t)

confidence rating #$&* 2.5

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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):

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Self-critique rating #$&*3

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Question: Problem 6 section 6.3

problem 6.3.17 was 6.3.6 (previously 6.3 #16) water balloon from 30 ft, v(t) = -32t+40

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Your solution:

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

Then we say that the c is 30 due to the height of the building so we have -16t^2+40t+30=0

c = 30 because the height is y = -16t^2+40t+c, and we know that y(0) = -16 * 0^2 + 40 * 0 + c = c, and y(0) = 30.

I also assume that as in the other problem we need to solve using the quadratic and the I found it to be t 3.12 or -.62

good, but you still need to find v

confidence rating #$&* 2.5

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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):

Ok I did not think of how that the information was to be used and I see that this is all related to physics and I can relate all of this material . I see that is was used to find the speed of on object as it strikes the ground.

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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:

It is traveling at a speed of balloon at six feet is calculated by the same method as above and then setting it equal to 6 and then we put this into the quadratic and solve for t then we substitute this into the equation and find that the balloon is traveling at a speed of 56ft/s at the impact onto the six feet tall mans head .

right; you should have actually done that, but I think you could have

confidence rating #$&* 2.5

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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):ok

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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:

The average velocity is calculated by taking delta x/delta T so we find this to be

3-1.5/3-1.5

6-54/1.5

So then we have 32ft/s down so its negative. Because its traveling down.

confidence rating #$&* 3

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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 like problems like this we just went over this in physics does this mean that I am behind here or in physics I am so confused at times.

I don't know what the pace of your physics course is or when calculus, with position, velocity and acceleration functions, is used in that particular course. So I can't tell you anything about that.

But you're connecting the two subjects, which is good.

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Self-critique rating #$&*2.5

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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:

The function of balloon as a function of clock time is 16t^2+40t+C=0 and then this would give us a formula that we can change to find 0 to find anything and c to find any change or variance in what we wish to find.

confidence rating #$&*

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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.

The position function is s(t) =- 16 t^2 + 40 t + 30.

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12:16:06

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Self-critique (if necessary):

Ok I see that you have reiterated about all of the preceeding problems but what was the main function that was requested of this solution?

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Self-critique rating #$&*2

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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:

The derivative of this function is -lnx

confidence rating #$&* 2.5

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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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12:23:36

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Self-critique (if necessary):

Ok I know that I got it right but to add my understanding I think I got this pretty good. It is first finding the integral and then taking that function and finding the derivative so I think that this simply means thatfirst you are finding the area encompassed by a finction and its average value and then you are finding the average of an average over a small portion of that segment.

not a bad statement; I'm glad you're thinking about this

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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:

Because the x is the value as well and it could be easily confused.

confidence rating #$&* 3

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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):

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Self-critique rating #$&*3

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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:

The derivative of this is 3x^2*e^-x^6

confidence rating #$&* 2

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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):

I made a mistake on the last part of the differentiation as you can tell I think It was in my algebra oops. Been working on this for a while.

Your solution was good. (-x^3)^2 = - x^6, so 3 x^2 e^(- (x^3)^2) = 3 x^2 e^(-x^6). Your answer and the given solution agree.

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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 cos(x) and 3.

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Your solution:

sinx*e^cosx^2

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):

Ok I think that I am doing pretty good.

I think I understand this better than what I have in the past.

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Self-critique rating #$&*3

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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:

The general solution to this would be x^2+1x+c=0

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):

Wow that was a lucky guess I thought back to the water balloon problem and I thought that what you wanted for a general form is it always AX^2+bx+c=0 or is this not correct.

x^2 is an antiderivative of 2x and x is an antiderivative of 1. So y = x^2 + x + c is the general antiderivative of y = 2 x + 1.

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Self-critique rating #$&*3

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

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Your solution:

The solution for part c is 12 this is due to the fact that the equation is already in for an anything like y(0)=12 is going to result in a c=12 or any other number or variable.

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):

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Self-critique rating #$&*3

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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:

The three graphs I used the values of c as 12,16,8 and had y values where the graph crossed it at the c values and the graphs were parabolas that were concave up(facing up).

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):

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Self-critique rating #$&*ok

This looks pretty good. See my notes and let me know if you have questions.

See my note on your Mth 173 Blackboard gradebook.

#$&*

assignment 2 query

c = 30 because the height is y = -16t^2+40t+c, and we know that y(0) = -16 * 0^2 + 40 * 0 + c = c, and y(0) = 30.

good, but you still need to find v

right; you should have actually done that, but I think you could have

I don't know what the pace of your physics course is or when calculus, with position, velocity and acceleration functions, is used in that particular course. So I can't tell you anything about that. But you're connecting the two subjects, which is good.

The position function is s(t) =- 16 t^2 + 40 t + 30.

not a bad statement; I'm glad you're thinking about this

Your solution was good. (-x^3)^2 = - x^6, so 3 x^2 e^(- (x^3)^2) = 3 x^2 e^(-x^6). Your answer and the given solution agree.

x^2 is an antiderivative of 2x and x is an antiderivative of 1. So y = x^2 + x + c is the general antiderivative of y = 2 x + 1.

This looks pretty good. See my notes and let me know if you have questions. See my note on your Mth 173 Blackboard gradebook.

I don't know what the pace of your physics course is or when calculus, with position, velocity and acceleration functions, is used in that particular course. So I can't tell you anything about that.

But you're connecting the two subjects, which is good.

This looks pretty good. See my notes and let me know if you have questions.

See my note on your Mth 173 Blackboard gradebook.