Assignment 6

course Mth 272

1-23-10 6:52pm

006. `query 6*********************************************

Question: `q5.1.6 (previously 5.1.40 (was 5.1.30)(was 5.1.34) ) integral of 1/(4x^2)

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Your solution: 1/(4x^-1/-1)+c

1/(4x) +c

confidence rating #$&*:3

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Given Solution:

`a*& An antiderivative of 1 / (4 x^2) is found by first factoring out the 1/4 to get 1/4 ( x^-2).

An antiderivative of x^-2 is -1 x^-1.

So an antiderivative of 1/4 (x^-2) is 1/4 (-x^-1) = 1 / 4 * (-1/x) = -1/(4x).

The general antiderivative is -1 / (4x) + c.

STUDENT QUESTION: I know I haven't got the right answer, but here are my steps

int 1/4 x^-2 dx

1/4 (x^-1 / -1) + C

-1/ 4x + C

INSTRUCTOR ANSWER: This appears correct to me, except that you didn't group your denominator (e.g. 1 / (4x) instead of 1 / 4x, which really means 1 / 4 * x = x / 4), but it's pretty clear what you meant. The correct expression should be written -1/ (4x) + C.

To verify you should always take the derivative of your result.

The derivative of -1/(4x) is -1/4 * derivative of 1/x. The derivative of 1/x = x^-1 is -1 x^-2, so the derivative of your expression is -1/4 * -1 x^-2, which is 1/4 x^-2 = 1 / (4x^2).

STUDENT ERROR:

The derivative

By rewriting the equation to (4x^2)^-1 I could then take the integral using the chain rule.

** it's not clear how you used the Chain Rule here. You can get this result by writing the function as 1/4 x^-2 and use the Power Function Rule (antiderivative of x^n is 1/(n+1) x^(n+1)), but this doesn't involve the Chain Rule, which says that the derivative of f(g(x)) is g'(x) * f'(g(x)).

The Chain Rule could be used in reverse (which is the process of substitution, which is coming up very shortly) but would be fairly complicated for this problem and so wouldn't be used. **

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Self-critique (if necessary):ok

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Self-critique rating #$&*:3

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Question: `q5.1.5 (previously 5.1.50 (was 5.1.46)(was 5.1.44) ) particular solution of f ' (x) = 1/5 * x - 2, f(10)=-10.

What is your particular solution?

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Your solution: (x^2)/10 - 2x + c is the antidirevative

10^2 /10 - 20 +c= -10

-10 + c = -10

c=0

(x^2)/10-2x + 0

confidence rating #$&*:3

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Given Solution:

`a An antiderivative of x is 1/2 x^2 and an antiderivative of -2 is -2x, so the general antiderivative of 1/5 x - 2 is 1/5 * (1/2 x^2) - 2 x + c = x^2 / 10 - 2x + c.

The particular solution will be f(x) = x^2 / 10 - 2x + c, for that value of c such that f(10) = -10.

So we have -10 = 10^2 / 10 - 2 * 10 + c, or -10 = -10 + c, so c = 0.

The particular solution is therefore f(x) = x^2 / 10 - 2 x + 0 or just x^2 / 10 - 2 x. **

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Self-critique (if necessary):ok

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Self-critique rating #$&*:3

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Question: `qIs the derivative of your particular solution equal to 1/5 * x - 2? Why should it be?

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Your solution: yes because the direvative is equal to it

confidence rating #$&*:3

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Given Solution:

`a The derivative of the particular solution f(x) = x^2 / 10 - 2 x is f ' (x) = (x^2) ' / 10 - 2 ( x ) '. Since (x^2) ' = 2 x and (x) ' = 1 we get

f ' (x) = 2 x / 10 - 2 * 1 = x / 5 - 2, which is 1/5 * x - 2.

The derivative needs to be equal to this expression because the original problem was to find f(x) such that f ' (x) = 1/5 * x - 2. *&*&

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Self-critique (if necessary):ok

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Self-critique rating #$&*:3

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Question: `q5.1.7 (previously 5.1.60 (was 5.1.56)(was 5.1.54) )

f''(x)=x^2, f(0)=3, f'(0)=6.

What is your particular solution?

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Your solution: f'(x)= (1/3) x^3+c c=6

f(x)= (4/3)x^4 + c c=3

f'(x) = (1/3)x^3 + 6

f(x) = (4/3)x^4 + 3

confidence rating #$&*:3

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Given Solution:

`a Since you have the formula for f ''(x), which is the second derivative of f(x), you need to take two successive antiderivatives to get the formula for f(x).

The general antiderivative of f''(x) = x^2 is f'(x) = x^3/3 + C. If f'(0) = 6 then 0^3/3 + C = 6 so C = 6. This gives you the particular solution f'(x) = x^3 / 3 + 6.

The general antiderivative of f'(x) = x^3 / 3 + 6 is f(x) = (x^4 / 4) / 3 + 6x + C = x^4 / 12 + 6 x + C.

If f(0) = 3 then 0^4/12 + 6*0 + C = 3 and therefore C = 3. Thus f(x) = x^4 / 12 + 6x + 3. **

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Self-critique (if necessary):Ok I must have messed up on the f(x) antidirvative I think

going from

f'(x) = (1/3)x^3 + 6 to f(x), should have been 1/3 * (1/4 x^4) + 6 x + c, etc.. Don't want to forget to integrate the 6 to get the 6x.

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Self-critique rating #$&*:2

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Question: `qIs the second derivative of your particular solution equal to x^2? Why should it be?

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Your solution: the dirvative of x^4 /12 + 6x + 3 is x^3/3 +6 and that dirvative is x^2

confidence rating #$&*:3

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Given Solution:

`a*& The particular solution is f(x) = x^4 / 12 + 6 x + 3. The derivative of this expression is

f ' (x) = (4 x^3) / 12 + 6 = x^3 / 3 + 6. The derivative of this expression is

f ''(x) = (3 x^2) / 3 = x^2.

Thus f '' ( x ) matches the original condition of the problem, as it must.

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Self-critique (if necessary):ok

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Self-critique rating #$&*:3

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Question: `q5.1.10 (previously 5.1.76 (was 5.1.70) ) dP/dt = 500 t^1.06, current P=50K, P in 10 yrs

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Your solution: had to look

confidence rating #$&*:1

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Given Solution:

`a You are given dP/dt. P is an antiderivative of dP/dt. To find P you have to integrate dP/dt.

dP/dt = 500t^1.06 means the P is an antiderivative of 500 t^1.06. The general antiderivative is

P = 500t^2.06/2.06 + c

Knowing that P = 50,000 when t = 0 we write

50,000 = 500 * 0^2.06 / 2.06 + c so that

c = 50,000.

Now our population function is

P = 500 t^2.06 / 2.06 + 50,000.

So if t = 10 we get

P = 500 * 10^2.06 / 2.06 + 50,000 = 27,900 + 50,000= 77,900. **

DER

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Self-critique (if necessary):ok I believe I understand just trying to figure out the antidirvatives sometime.

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Question: `q5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2

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Your solution: (3-x^3)^(1/2) * 3x^2

(2/3)(3-x^3)^(3/2) * x^3 + c

confidence rating #$&*:3

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Given Solution:

`a You want to integrate `sqrt(3-x^3) * 3 x^2 with respect to x.

If u = 3-x^3 then u' = -3x^2.

So `sqrt(3-x^3) * 3x^2 can be written as -`sqrt(u) du/dx, or -u^(1/2) du/dx.

The General Power Rule tells you that this integral of -u^(1/2) du/dx with respect to x is the same as the integral of -u^(1/2) with respect to u.

The integral of u^n with respect to u is 1/(n+1) u^(n+1).

We translate this back to the x variable and note that n = 1/2, getting -1 / (1/2+1) * (3 - x^3)^(1/2 + 1) = -2/3 (3 - x^3)^(3/2).

The general antiderivative is -2/3 (3 - x^3)^(3/2) + c. **

DER

COMMON ERROR: The solution is 2/3 (3 - x^3)^(3/2) + c.

The Chain Rule tell syou that the derivative of 2/3 (3 - x^3)^(3/2), which is a composite of g(x) = 3 - x^3 with f(z) = 2/3 z^(3/2), is g'(x) * f'(g(x)) = -3x^3 * `sqrt (3 - x^3). You missed the - sign. **

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Self-critique (if necessary):what happened to the dirvative of 3x^2

You're doing an antiderivative.

If the 3 x^2 wasn't there, you'd be up a creek. The derivative of 2/3 * (3 - x^2)^(3/2) isn't sqrt( 3 - x^3). The derivative of 2/3 * (3 - x^3)^(3/2) is (3 - x^3) * (-3 x^2); the -3 x^2 comes from the chain rule.

This is the tricky thing about integration at this point. It can be tricky to 'reverse' the chain rule, but it's gotta be done in one way or the other. Those tricks are basically what Chapter 5 is about.

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Self-critique rating #$&*:1

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Question: `q5.2. 18 (was 5.2.16 integral of x^2/(x^3-1)^2

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Your solution: 1/(3(1-x^3)) + c

confidence rating #$&*:2

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Given Solution:

`a Let u = x^3 - 1, so that du/dx = 3 x^2 and x^2 = 1/3 du/dx.

In terms of u we therefore have the integral of 1/3 u^-2 du/dx. By the General Power Rule our antiderivative is

1/3 (-u^-1) + c, or

-1/3 (x^3 - 1)^-1 + c = -1 / (3 ( x^3 - 1) ) + c.

This can also be written as

1 / (3 ( 1 - x^3) ) + c. **

DER

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Self-critique (if necessary):ok

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Self-critique rating #$&*:3

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Question: `q5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3)

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Your solution: x^2((1-x^3)^-1/2

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

2/3x^3(1-x^3)^1/2+ c

confidence rating #$&*:3

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Given Solution:

`a*& If we let u = 1 - x^3 we get du/dx = -3 x^2 so that x^2 = -1/3 du/dx.

So the exression x^2 / sqrt(1-x^3) is -1/3 / sqrt(u) * du/dx

By the general power rule an antiderivative of 1/sqrt(u) du/dx = u^(-1/2) du/dx will be (-1 / (-1/2) ) * u^(1/2) = 2 sqrt(u).

So the general antiderivative of x^2 / (sqrt(1-x^3)) is -1/3 ( 2 sqrt(u) ) + c = -2/3 sqrt(1-x^3) + c.

*&*&

DER

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Self-critique (if necessary):how close am I. I noticed I forgot the negative sign again but what about the x^3 where did that go

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Self-critique rating #$&*:2"