assignment 7

course mth 272

If your solution to a 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.

007. `query 7

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Question: `q5.2.2. (previously 5.2.36 (was 5.2.34) ) integral of x^2 (1-x^3)^2 by formal substitution.

What is the integral of the given function?

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Your solution:

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

let u = 1-x^3

let u’ = -3x^2

-1/3 (1/3) (1-x^3)^3 = -1/9(1-x^3)^3

confidence rating #$&*:3

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Given Solution:

`a If we let u = 1 - x^3 then u ' = - 3 x^2 and the x^2 in our integrand is - u ' / 3.

(1-x^3)^2 is u^2, so the integrand is - u ' / 3 * u^2 = -1/3 u^3 u ' .

So the integral is you have -1/3 u^2 du. The integral of u^2 u ' is 1/3 u^3.

Thus the integral of -1/3 u^2 u ' is -1/3 of 1/3 u^3, or -1/9 u^3.

So your integral should be -1/9 u^3 = -1/9 (1-x^3)^3.

The general antiderivative is -1/9 ( 1 - x^3)^3 + c. **

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Self-critique (if necessary):

Its hard to type everything you write…

I know. That's why I only ask you to submit only a few selected assigned problems.

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Self-critique rating #$&*:3

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Question: `qWhat is the derivative of your result?

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Your solution:

-1/9(1-x^3)^3

Taking the derivative

(3)(-3x^2)(1/9)(1-x^3)^2 = x^2 (1-x^3)^2

confidence rating #$&*:3

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Given Solution:

`a The derivative of -1/9 (1-x^3)^3, using the Chain Rule, is the product of -1/9, 3(1-x^3)^2, and the derivative -3x^2 of the 'inner function' (1-x^3). Multiplying these factors we get -1/9 (-3x^2) * 3(1-x^3)^2 = x^2 (1-x^3)^2. **

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Self-critique (if necessary):

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Question: `q 5.2.4 (previously 5.2.54 (was 5.2.52)) find x | dx/dp = -400/(.02p-1)^3, x=10000 when p=100

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Your solution:

dp*-400/(.02p-1)^3

u=.02p-1

du= .02 = 50 du

-20000 u^-3 du

10,000 (.02p-1)^-2 + c

10,000=10,000 (.02p-1)^-2 + c

C=0

10,000 (.02p-1)^-2

confidence rating #$&*:

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Given Solution:

`a The equation rearranges to dx = -400 * dp * (.02 p - 1)^-3. An antiderivative of the left-hand side could be just x.

An antiderivative of dp * (.02 p - 1)^-3 is found using u = .02 p - 1, so du = .02 dp and dp = du / .02 = 50 du. Thus the right-hand side becomes -400 * 50 u^-3 du = -20000 u^-3 du, with antiderivative 20000 / 2 * u^-2 + c = 10,000 u^-2 + c.

So we have x = 10,000 * u^-2 + c = 10,000 * (.02 p - 1)^-2 + c.

Note that dx / dp is therefore 10,000 * -2 * .02 (p-1)^-3 = -400 (.02 p - 1)^-3, consistent with the original equation.

Since x = 10,000 * (.02 p - 1)^-2 + c and x = 10,000 when p = 100 we have

10,000 = 10,000 * (.02 * 100 - 1)^2 + c

10,000 = 10,000 / 1^2 + c

10,000 = 10,000 + c so

c = 0.

The solution is therefore

x = 10,000 * (.02 p - 1)^-2 + 0 or just

x = 10,000 * (.02 p - 1)^-2.

**

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Self-critique (if necessary):

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Question: `q5.3.1 (previously 5.3.04 (was 5.3.04)) integral of e^(-.25 x) by Exponential Rule

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Your solution:

e^(-.25 x) * du

u = -.25x = du = -.25

so you will need -4 * -.25 to equal the derivative

so the answer = -4 e^(-.25 x)

confidence rating #$&*:3

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Given Solution:

`a Simple substitution u = -.25 x gives us du/dx = -.25 so that du = -.25 dx and dx = du / (-.25) = -4 du.

Our original integrand e^(-.25 x) dx therefore becomes e^u * (-4 du) = -4 e^u du. Our general antiderivative will be -4 e^u + c, meaning -4 e^(-.25 x) + c.

The derivative of -4 e^(-.25 x) + c is -4 ( -.25 e^-.25 x) = e^-.25 x, verifying our result.

The General Exponential Rule is equivalent to this:

u = -.25 x so du/dx = -.25. Thus the integral is of e^u / (du/dx) = e^(-.25 x) / (-1/4) = -4 e^(-.25 x). *&*&

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Self-critique (if necessary):

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Question: `q 5.3.2 (previously 5.3.10 (was 5.3.10)) integral of 3(x-4)e^(x^2-8x) by Exponential Rule

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Your solution:

3(x-4)e^(x^2-8x)

u = x^2-8x = du = 2x – 8

(x-4) = 2x -8

X= ˝

3/2e^(x^2-8x) + c

confidence rating #$&*:3

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Given Solution:

`a if u=x^2 - 8x then du / dx = 2x - 8

x-4 = 1/2(2x-8) so 3(x-4) = 3/2 du/dx.

Thus 3(x-4)e^(x^2-8x) is 3/2 e^u du/dx.

The general antiderivative of e^u du/dx is e^u + c, so the integral of 3/2 e^u du/dx is 3/2 e^u.

Substituting x^2 - 8x for u we have 3/2 e^(x^2-8x) + c. **

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Self-critique (if necessary):

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Question: `qproblem 5.3.3 (previously 5.3.16) integral of 1/(6x-5) by Log Rule

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Your solution:

1/(6x-5)

u = 6x – 5 = du = 6

1/6 ln (6x-5) +c

confidence rating #$&*:3

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Given Solution:

`a du/dx is the derivative of 6x-5, so du/dx = 6

If we let u = 6x - 5 then du = 6 dx so dx = 1/6 du and the integral becomes that of 1/2 * ln(u) * du/6 = 1/3 ln(u) du

The integral of 1/3 ln(u) du is 1/3 * 1 / u = 1/3 * 1 / (6x-5) = 1 / [ 3(6x-5) ]. **

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Self-critique (if necessary):

I don’t understand where the ˝ came into the problem.

1/2 shouldn't have been there. Not sure where it originated. 1/2 of 1/6 is 1/12 anyway.

Your solution is correct.

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Self-critique rating #$&*:2

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Question: `q 5.3.5 (previously problem 5.3.22 (was 5.3.20)) integral of x/(x^2+4) by Log Rule

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Your solution:

x/(x^2+4)

u = x^2+4 = du = 2x

(1/2) ln (x^2+4) +c

confidence rating #$&*:3

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Given Solution:

`a If we let u = x^2 + 4 we get du/dx = 2x so that the x in the numerator is 1/2 du/dx.

The integral of x / (x^2 + 4) is the integral of 1/2 * ( 2x / (x^2+4) ) = 1/2 (1/u du/dx).

The general antiderivative is therefore 1/2 ln(u) + c = 1/2 ln |x^2+4| + c. **

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Self-critique (if necessary):

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Question: `qWhat is the derivative of your result?

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Your solution:

x^2 +4 = 2x

so the derivative of (1/2) ln (x^2+4) +c

is ˝ * 2x * 1 / (x^2 + 4)

confidence rating #$&*:3

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Given Solution:

`a The derivative of ln(x^2+4) * (1/2) is 1/2 * 2x * 1 / (x^2 + 4) or x / (x^2 + 4). This confirms that ln(x^2+4) * (1/2) is a solution to the equation.

The general antiderivative is of course ln(x^2+4) * (1/2) + c. **

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Self-critique (if necessary):

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Question: `q 5.3.7 (previously 5.3.28 (was 5.3.24) (was 5.3.24) ) integral of e^x/(1+e^x) by Log Rule

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Your solution:

u=1+e^x = du = e^x

ln(1+e^x) + c

confidence rating #$&*:

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Given Solution:

`a let u = 1 + e^x. Then du/dx = e^x.

We are therefore integrating 1 / (1 + e^x) * e^x, which is 1/u du/dx.

The antiderivative is ln |u| + c = ln | 1 + e^x | + c. **

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Self-critique (if necessary):

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Question: `q 5.3.9 (previously 5.3.46 (was 5.3.34) (was 5.3.34) ) integral of (6x + e^x) `sqrt( 3x^2 + e^x)

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Your solution:

U = 3x^2+e^x = du =6x + e^x

U^.5 = u^1.5

2/3 (6x + e^x)^(3/2)

confidence rating #$&*:3

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Given Solution:

`a Here are two detailed solutions:

(6x + e^x) `sqrt( 3x^2 + e^x) = `sqrt(u) * du/dx = u^(1/2) du/dx.

The antiderivative is thus

2/3 u^(3/2) = 2/3 (3x^2 + e^x)^(3/2).

Alternatively

If u = 3x^2 + e^x then du = 6x + e^x and we have the integral of `sqrt(u) du, which is just

2/3 u^(3/2) + c = 2/3 (3x^2 + e^x)^(3/2) + c. **

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Self-critique (if necessary):

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Question: `q 5.3.11 (previously 5.3.58 (was 5.3.54) (was 5.3.52) ) dP/dt = -125 e^(-t/20), t=0, P=2500 and interpretation.

Give your complete solution.

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Your solution:

U=-t/20 so you will multiply -20 by -125 to get 2500

2500 e^(-t/20)

2500= 2500 e^(0/20) +C

0 =C

2500 e^(t/20)

confidence rating #$&*:3

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Given Solution:

`a If dP/dt = -125 e^(-t/20) then dp = -125 e^(-t/20) dt. Integrating both sides we get

p = 2500 e^(-t/20) + c ( to integrate the right-hand side start with u = -t / 20, etc.

If p = 2500 when t = 0 we have

2500 = 2500 e^(-0/20) + c so

2500 = 2500 + c and c = 0.

The final solution is thus

p = 2500 e^(-t/20)

After 15 days the population is p(15) = 2500 e^(-15/20) = 1000, give or take a couple hundred (you can evaluate the expression).

All the trout are considered dead when the population is below 1/2. So you need to solve 1/2 = 2500 e^(-t/20) for t.

Dividing both sides of this equation by 2500 then taking the natural log of both sides you get

-t/20 = ln( 1/2500 ) so

t = -20 * ln (1/2500) = -11 or -12 or so.

Thus t is about 200 days, give or take a little.

Alternative reasoning of the particular solution:

If u = -t/20 then e^u du/dt = e^(-t/20) * -1/20. -125 e^(-t/20) is 2500 * ( -1/20 e^(-t/20) ) = 2500 e^u du/dx.

The integral is 2500 e^u + c = 2500 e^(-t/20) + c.

If t = 0, P=2500 then 2500 = 2500 e^0 + c = 2500 + c, so c = 0. Thus the particular solution is

P = 2500 e^(-t/20).

Alternative solution for the time when all trout are dead:

2500 e^(-t/20) < .5 means

e^(-t/20) < .0002 so -t/20 < ln(.0002) so

-t < ln(.0002) * 20 so

-t < -170.34 and

t > 170.34.

The probability is that all trout are dead by day 171.

STUDENT QUESTION: I couldn't figure out the time for all the trout to die because the ln 0 is undefined

** When the population falls below 1/2 of a fish it rounds off to 0 and you assume that all the trout are dead.

You can think of this in terms of probability. The function doesn't really tell us the precise number but the probable number. When the probability is againt that last fish being alive we figure that it's most likely dead. **

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Self-critique (if necessary):

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Self-critique rating #$&*:3

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