Query 2

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course MTH174

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.

Optional Preliminary Questions:

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Question: `q001. Show that if s(t) = 2 t^3 + 3 t^2 + 5 t + 3 is substituted into the differential equation

ds/dt = 6 t^2 + 6 t + 5

we obtain an identity.

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Your solution:

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Question: `q002. Find a function s(t) that satisfies the differential equation

ds/dt = 3 t^2 + 4 t - 6.

Find also a function s(t) that satisfies the above equation as well as the initial condition s(0) = 4.

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Your solution:

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Self-critique Rating:

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Question: `q003. If the function s(t) represents the position of an object at time t, then ds / dt = s ' (t) represents the instantaneous rate at which the position is changing at clock time t, which is the instantaneous velocity v(t) of the object. s '' (t) represents the instantaneous rate at which s ' (t) is changing, i.e., the instantaneous rate at which the velocity is changing, which is the acceleration a(t) of the object.

Thus

a(t) = dv / dt

and

v(t) = ds / dt.

If a(t) = 3, then dv/dt = 3. What is the general function v(t) that satisfies this equation?

Plug your v(t) function into the equation ds/dt = v(t). What is your equation?

What is the general function s(t) that satisfies your equation?

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Your solution:

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Self-critique Rating:

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Question: `q004. What is the definite integral of f(x) = x^2 with respect to x, from x = 2 to x = t?

What is the derivative of your integral?

Repeat for the function f(x) = cosine(x).

• What is the derivative with respect to t of your first integral?

• What is the derivative with respect to t of your second integral?

• In terms of the meanings of integrals and derivatives, why do the results obtained in this problem make sense?

For a general function f(x), if you were to integrate f(x) from x = -3 to x = t, then take the derivative of the result, what would you expect to get?

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Your solution:

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Self-critique Rating:

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Question: `q005. What is the definite integral of f(x) = x^2 with respect to x, from x = -4 to x = e^(t^2)?

What is the derivative of your integral?

Repeat for the function f(x) = cosine(x).

In terms of the meanings of integrals and derivatives, and the chain rule, why do these results make sense?

For a general function f(x), if you were to integrate f(x) from x = -3 to x = e^(t^2), then take the derivative of the result, what would you expect to get?

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Your solution:

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Self-critique Rating:

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Question: `q006. What would be the derivative with respect to t of the integral of e^(x^2), from x = 0 to x = t?

What would be the derivative with respect to t of the integral of e^(x^2), from x = 0 to x = cos(t)?

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Your solution:

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Self-critique Rating:

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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:

Integrate:

-32t + 100

[-32t^(1+1)]/2 + [100t^(0+1)]/1

(-32t^2)/2 +100t + C

Plug t = 0

C= 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:

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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:

(a)

Integrate:

-32t + 40

-16t^2 + 40t + 30

Solve for t

-40 + sqrt(-40^2 - 4(-16)(30))/2(-16)

T = approx. -40.56

Plug T

-32(-40.56) + 40

Approx.. 1337.88

(b)

[-16(3^2) + 40(3)] - [-16(1.5^2) + 40(1.5)]

-24 - 24 = -48 velocity

(c)

S(6) = -16(6)^2 + 40(6) + 30

-306 velocity

confidence rating #$&*:2

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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): Plugged the quadratic formula wrong

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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:

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): used the s(t) function incorrectly

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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:

See above

confidence rating #$&*: 2

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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):

Used the wrong process to calculate velocity

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Self-critique Rating: 3

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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 s(t) = -16t^2 + 40t + 30

confidence rating #$&*: 3

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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):OK

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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:

confidence rating #$&*: 1

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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):

I understand now.

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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:

X is used as the upper limit in the construction theorem.

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): OK

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Self-critique Rating:

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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:

Integrating e^x gives us e^x.

With an exponent in the exponent we apply the chain rule.

3x^2 (e^(x^3)) - e^(0^2)

3x^2(e^(x^3)) - 1

confidence rating #$&*:: 3

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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):OK

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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:

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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): I understand now

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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:

Integrate 2x + 1

[2x^(1+1)]/2 + [1x^(0+1)]/1

x^2 + x + C

confidence rating #$&*: 3

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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:

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

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Your solution:

X^2 + x + C

1^2 + 1 + C = 5

C = 3

confidence rating #$&*: 3

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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): OK

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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:

C=1;2;3;

They are quadratic graphs with increasing y values.

confidence rating #$&*: 2

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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): Didn’t mention the effect the constant has on the graph.

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Self-critique Rating: 2

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Question: `q007. A ball is tossed straight upward at 20 feet / second from a height of 5 feet. Its acceleration function is

a(t) = -32 feet/second^2 * t.

Its velocity function satisfies the equation

dv / dt = a,

where v and a stand for the velocity and acceleration functions v(t) and a(t). Find the general solution to this equation, then evaluate the integration constant so your velocity function matches the given initial velocity (i.e., so that v = 20 ft / second when t = 0).

The position function satisfies the equation

ds / dt = v

where s and v stand for the position and velocity functions s(t) and v(t). Find the general solution to this equation, then evaluate the integration constant so your position function matches the given initial position (i.e., so that s = 5 feet when t = 0).

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Your solution:

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Self-critique Rating:

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

#*&!

Query 2

&#Your work looks good. Let me know if you have any questions. &#