Assignments 5-7

course MTH 272

iƍǯ⋓assignment #005

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ХOz{ų~Qw

Applied Calculus I

06-24-2006

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14:01:03

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14:04:35

query 5.1.12 integrate t^4 dt and check by differentiation

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RESPONSE -->

This problem can be solved by using the simple power rule. The integrand x^n dx = x^(n+1) / (n+1) + C. In this case, the antiderivative would be 1/5 * t^5 + C, or (t^5)/5 + C. You can check this by taking the derivative of (t^5)/5. Using the power rule, you get 5*(1/5) t^4= t^4

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14:04:47

** An antiderivative of this power function is one power higher so you will have a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative will be t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative t^5 / 5 + c.

The derivative of 1/5 t^5 is 1/5 * 5 t^4 = t^4), verifying our antiderivative. **

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RESPONSE -->

ok

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14:06:55

query 5.1.18 integrate v^-.5 dv and check by differentiation

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RESPONSE -->

The antiderivative will have a power of +.5. Multiplying v by the reciprocal of .5, which is 2, you get an antiderivative of 2v^.5.

You can check this by taking the derivative, in which case you get v^(-.5)

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14:08:18

** An antiderivative of this power function is a constant multiple of the power function which is one power higher. The power of the present function is -.5 or -1/2; one power higher is +.5 or 1/2. So you will have a multiple of v^.5. Since the derivative of v^.5 is .5 v^-.5 an antiderivative will be v^.5 / .5 = v^(1/2) / (1/2) = 2 v^(1/2). Adding the arbitrary integration constant we end up with general antiderivative 2 v^(1/2) + c.

The derivative of 2 v^(1/2) is 2 * (1/2) v^(-1/2) = v^(-1/2), verifying our antiderivative. **

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RESPONSE -->

Oops, I think I forgot the ""+C"" This is something I'm really going to have to make an effort to remember. Other than that, my answer was right.

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14:09:24

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I MUST remember to add ""+C"" to each answer when integrating!! It's been a long time since I've done this stuff, but it seems to be coming back to me slowly but surely.

It seems to be coming back very well.

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麞`ίUz

assignment #006

ХOz{ų~Qw

Applied Calculus I

06-24-2006

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14:13:32

5.1.36 (was 5.1.34 int of 1/(4x^2)

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RESPONSE -->

the int of 1/(4x^2) is the same as the int of 4x^(-2). Integrating this gives us -4x^(-1) +C, which can be rewritten as -4/x +C

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14:16:32

*&*& 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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RESPONSE -->

I made a dumb mistake. I rewrote the integrand as 4x^(-2), when in fact, the entire thing should have been raised to the power of -2, which means it should have been: int (4x)^(-2). This was a very careless mistake, which is the worst kind of mistake to make. I did check my answer of -4x^(-1) by taking the derivative, and I did come up with 4x^(-2), so I thought it was correct, but obviously, since the 4x^(-2) was wrong, my answer was wrong also. Had I rewritten it correctly, I could have come up with the correct answer.

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14:22:58

5.1.46 (was 5.1.44 particular soln of f ' (x) = 1/5 * x - 2, f(10)=-10.

What is your particular solution?

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RESPONSE -->

If f'(x) = 1/5*(x) - 2, when can solve for the antiderivative, which would give us f(x). x will be raised to the power of 2, which means we must multiply the constant 1/5 times the reciprocal of 2, which is 1/2. This gives us:

f(x) = (1/5)*(1/2)x -2 + C

= (1/10)x - 2 +C

To solve for the particular solution, we plug in 10 for each value of x, and set the entire equation equation to -10

f(10) = (1/10)*(10) -2 +C = -10

=1-2+C = -10

= -1+C = -10

= C = -9

Plugging in -9 for C gives us a particular solution of:

f(x) = (1/10)x - 2 -9

= (1/10)x - 11

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14:26:54

** 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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RESPONSE -->

Another careless mistake! I was so caught up in solving for the antiderivative of (1/5)x correctly, that I completely neglected to solve for the antiderivative of -2! This would have yielded -2x. ALSO, I didn't even get the antiderivative of (1/5)x correct! I forgot to square the x term. I don't know why I made so many mistakes in this problem. I know exactly how to solve these problems, so I guess that's what made me be sort of careless. I apologize- I usually don't make this many dumb mistakes!

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14:31:58

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

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RESPONSE -->

After making so many mistakes on the last question, I started over and correctly solved the antiderivative, giving me a value of 0 for C, and f(x) = (1/10)(x^2) - 2x. The derivative of f(x) is in fact (1/5)x -2. This should be the case because if you take the derivative of f(x), you get f'(x), which is what we started with, and which is what we were supposed to integrate to begin with. This is why, when integrating, you can take the derivative of the answer you come up with, and see if it matches the original integrand to see if you solved correctly.

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14:32:11

** 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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ok

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14:38:22

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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RESPONSE -->

You start by solving for the antiderivative of f""(x), which gives you f'(x).

f'(x) = 1/3 * x^3 + C

f'(0) = 6

--> f(0) = 1/3*(0)^3 +C =6

=C=6

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

You take the antiderivative of f'(x) to get f(x)

f(x) = (1/3)*(1/4)*x^4 + 6x + C

=(1/12)*x^4 +6x +C

f(0) = 3

--> f(0)= (1/12)(0^4) + 6(0) +C = 3

=C=3

f(x) = (1/12)*x^4 +6x +3

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14:38:44

** 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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ok

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14:40:37

Is the second derivative of your particular solution equal to x^2? Why should it be?

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RESPONSE -->

The second derivative of the particular solution is in fact equal to x^2, as it should be, since that is the value of f''(x) which we started out with.

f(x) = (1/12)x^4 +6x +3

First derivative:

f'(x) = 4*(1/12)x^3 + 6

= (1/3)x^3 +6

Second derivative:

f''(x) = 3*(1/3)x^2

= x^2

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14:40:46

*&*& 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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ok

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14:49:38

5.1.76 (was 5.1.70 dP/dt = 500 t^1.06, current P=50K, P in 10 yrs

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RESPONSE -->

You must first solve for P(t) by integrating dP/dt. This gives you:

P(t) = (1/2.06)*500t^2.06 +C

=(500/2.06)t^2.06 +C

The current P (when t=0) is 50,000. Plugging in zero of t, and setting the equation equal to 50,000 yields the value of C:

P(0) = (500/2.06)*0^2.06 + C = 50,000

=0+C = 50,000

C = 50,000

This gives us the particular solution of:

P(t) = (500/2.06)t^2.06 + 50,000

The get P in 10 years, plug in t=10:

P(10) = (500/2.06)*10^2.06 + 50,000

= 27,867.80635 + 50,000

= 77,867.80635

= 77,867.81

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14:50:38

** 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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RESPONSE -->

77,867.81 is the response I got. I would have gotten 77,900 if I had rounded off.

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14:51:08

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I clicked ""next question/answer"" but nothing appeared, so I'll click it again and hope it doesn't take me straight to the answer.

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14:56:48

5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2

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RESPONSE -->

This can be rewritten as the integral of (3-x^3)^(1/2) * (3x^2)dx. You have to use the chain rule to solve this integral. u^n = (3-x^3)^.5, and du/dx = 3x^2 dx, since du = 3x^2

int u^.5 du/dx = 2/3 u^(3/2) + C

=(2/3) * (3-x^3)^(3/2) +C

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14:57:27

06-24-2006 14:57:27

** 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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NOTES -------> The integral of u^n with respect to u is 1/(n+1) u^(n+1)

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15:03:26

5.2. 18 (was 5.2.16 integral of x^2/(x^3-1)^2

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This can be rewritten as the integral of x^2*(x^3-1)^-2dx.

u=(x^3-1), and du = 3x^2. This means that the first term of the integrand, x^2, must be multiplied by 3 in order to give us the correct du of 3x^2. You multiply this by 3 and also divide by 3, which gives you:

1/3 int ((x^3-1)^(-2) * 3x^2dx. This gives us the correct du/dx, so we can now integrate.

1/3 * (-1)*(x^3 -1)^(-1) + C

= -(1/3)(x^3 - 1)^(-1) +C

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15:03:37

** 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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ok

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15:09:25

5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3)

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This can be rewritten as the integral of (1-x^3)^(-.5) * x^2 dx. Using the chain rule, u = (1-x^3), and du= -3x^2. This means that in order to get to correct du/dx, we must multiply x^2 by -3, and divide everything by -3.

This gives -(1/3) int (u^(-1/2)), and we can now integrate:

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

= -2/3 (1-x^3)^(1/2) +C

or

= (-2 *sqrt(1-x^3)) / 3

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15:09:47

*&*& 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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ok

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ɡ_ʿĩz˕PWv

assignment #007

ХOz{ų~Qw

Applied Calculus I

06-25-2006

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14:38:15

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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RESPONSE -->

u= (1-x^3)

du= -3x^2

Multiply and divide by -3 so that the first term equals -3x^2, and we have du/dx

This leaves us with:

-1/3 int (1-x^3)^2 and we can simply integrate

(-1/3)*(1/3)(1-x^3)^3 +C

= (-1/9)*(1-x^3)^3 +C

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14:38:48

** 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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ok

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14:41:50

What is the derivative of your result?

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RESPONSE -->

The derivative of (-1/9)(1-x^3)^3 is:

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

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

This is what we started out with after multiplying and dividing the original expression by 3 in order to get du/dx. The original expression was (x^2)(1-x^3)^2 dx. I'm not sure how to continue deriving from -1/3 (1-x^3)^2 dx to end up at this original expression, or if -1/3 (1-x^3)^2 dx is as far as you can go.

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14:45:10

** 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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RESPONSE -->

Oh yes, I forgot about the chain rule. I was either overthinking or not thinking enough, because it's pretty obvious that the chain rule is what you have to use. I make way too many careless mistakes and that's going to end up hurting me come test time.

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15:07:06

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

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RESPONSE -->

u= .02p - 1

du = .02

-400 must be multiplied by (.02/-400) in order to result in .02 dx. Therefore, we must also multiply the entire expression by (-400/.02), or -20,000. This leaves us with:

-20,000 int ((.02p-1)^-3), and integrating gives us:

-20,000 * (-1/2) * ((.02p-1)^-2)

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

We know that x=10,000 when p=100, so substituting in these known values results in:

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

10,000 (1) + C = 10,000

C = 0

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

After looking at this problem so more, I realize that maybe you can just pull out the -400 from the beginning since it is a constant. This leaves you with -400 int ((.02p - 1)^(-3)), and integrating would give you:

-400*(-1/2) * (.02p-1)^(-2)

200 * (.02p -1)^(-2) + C

Plugging in x=10,000 and p=100 gives you:

200*(.02(100) - 1)^(-2) +C = 10,000

200 + C = 10,000

C = 9800

200(.02p-1)^(-2) + 9800

I'm not sure which answer is correct.

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15:11:03

** 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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RESPONSE -->

This is the answer I got using the first method, so I learned that you can't just pull out a constant. This will probably prove to be a helpful thing to know. However, one concern I have is that I'm having trouble following the steps to the solution given here, particularly the second paragraph:

""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.""

This isn't how I solved it, but since I came up with the same answer, I'm guessing the way I solved it is correct, just a little different?

Your solution is completely equivalent to this one, with only minor differences in notation. Your solution is fine.

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15:16:23

5.3.04 (was 5.3.04 integral of e^(-.25 x) by Exponential Rule

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RESPONSE -->

u = -.25x

du = -.25 or -1/4

You have to multiply and divide both sides by -1/4, which the same as multiplying the entire expression by -4. This leaves the expression in the form of int e^u du/dx * dx, which equals int e^u du = e^u+C

-4e^(-.25x) + C

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15:17:31

** 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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RESPONSE -->

I don't remember ever having used simple substitution before, so I still don't really understand that, and I especially don't understand how it differs from the General Exponential Rule

Your notation is a little different, but you appear to completely understand not only the process but the reasons for the process.

One comment on your solution:

u = -.25x

du = -.25 or -1/4

-.25 is the derivative of u with respect to x. So du/dx = -.25. You can't just say that du = -.25. In fact, since du/dx = -.25, we have du = -.25 dx.

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15:32:37

5.3.10 (was 5.3.10 integral of 3(x-4)e^(x^2-8x) by Exponential Rule

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RESPONSE -->

u = x^2 - 8x

du = 2x - 8= 2(x-4)

What we have is 3(x-4), so we should multiply and divide both sides by 2/3, which will give us:

3/2 int (e^(x^(2-8x)))

= 3/2 e^(x^(2-8x)) + C

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15:32:54

** 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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RESPONSE -->

ok

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15:39:11

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RESPONSE -->

I accidentally pressed space bar before I started typing in this box, so it took me straight to the answer. I'll try solving it without looking at the answer, then I'll look at the correct response and critique my answer.

u= 6x - 5

du = 6

In the numerator, we have 1, and we need a 6 in order for the integral to be in the form (du/dx)/u * dx = int (1/u)du = ln |u| +C. To get a 6 in the numerator, we multiply by 6, and then divide the entire expression by 6. Now it is the correct form and we can substitute 6x-5 in for u, and we get:

1/6 ln |6x-5| +C

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15:47:05

problem 5.3.20 integral of x/(x^2+4) by Log Rule

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RESPONSE -->

u = x^2 + 4

du = 2x

Right now, we have x in the numerator, but we need 2 x, so we multiply by 2, and then factor out 1/2. This leaves it in the form we need so that the derivative is 1/2 ln |u| +C

= 1/2 ln |x^2 +4| +C

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15:47:12

** 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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RESPONSE -->

ok

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15:50:19

What is the derivative of your result?

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RESPONSE -->

The derivative of 1/2 ln |x^2 +4| +C = x/(x^2+4), which makes sense considering since is the integrand that we started with.

1/2 ln |x^2 +4|

1/2 * 1/(x^2 +4) *2x

1/2 * 2x/ (x^2 + 4)

2x / 2(x^2 +4)

x / (x^2 +4)

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15:50:33

** 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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RESPONSE -->

ok

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15:53:07

5.3.24 (was 5.3.24 integral of e^x/(1+e^x) by Log Rule

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RESPONSE -->

This expression is already in the form int (du/dx)/u *dx, which equals int (1/u)du, which equals ln|u|+C. u= 1+e^x, and du= e^x, which is already in the numerator.

ln|u| +C=

ln |1+e^x| +C

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15:53:19

** 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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RESPONSE -->

ok

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15:56:53

5.3.34 (was 5.3.34 integral of (6x + e^x) `sqrt( 3x^2 + e^x)

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RESPONSE -->

This can be rewritten as the integral of (6x + e^x) * (3x^2 + e^x)^1/2

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

du = ex + e^x

We already have the correct form of du, so no additional multiplication or division is necessary, and we can take the integral of (ex^2 + e^x)^(1/2). If we increase the power of 1/2 by 1, we get 3/2, and the integral ends up being:

2/3 (ex^2 + e^x)^(3/2) + C

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15:57:08

** 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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RESPONSE -->

ok

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16:05:46

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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RESPONSE -->

u= -t/20

du= -1/20= -.05

We need -.05 dt, but we have -125 dt, so we need to multiply and divide by (.05/125). If we divide the entire expression by this, we get everything multiplied by 125/.05, which equals 2500.

2500 * integral of e^(-t/20)

= 2500 e^(-t/20) + C

When t=0 and P=2500:

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

=2500 + C = 2500

C= 0

Solution: 2500e^(-t/20) + C

=2500e^(-t/20) + 0

=2500e^(-t/20)

Interpretation: 2500e^(-t/20) is the rate at which P changes. Since the variable, t, is negative, the value of P must be decreasing.

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16:07:47

** 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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RESPONSE -->

I'm sorry, I didn't see the other parts to the question. The part that I did do, I got right. I've worked through the other parts now though, and I understand how to solve them.

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Your work is very good. I've inserted a couple of notes, but you appear to be in excellent shape. You did make a few errors, but you understand them and are now aware of them, so you're much less likely to make such errors on the test.