course mth 272 ˆZ¸~l¹€ü¶w˜˜²xúéñ¦êžÙïÎŽassignment #006 006. `query 6 Applied Calculus II 06-29-2008
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08:13:04 5.1.40 (was 5.1.30)(was 5.1.34 int of 1/(4x^2)
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RESPONSE --> = -4 integral sign x^-2 = -4 (X^-1) /(-1) + C =4x + C confidence assessment: 1
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08:19:01 *& 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 --> ok. so you factor out 1/4 then you get (1/4) (-x^-1) = 1/4 * (-1/x) =-1/(4x) + C making sure 4x is grouped by itself in the denominator. self critique assessment: 2
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08:25:09 5.1.50 (was 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 --> i know you first want to integrate to find the general solution. I'm not sure how to integrate this though. Is the 1/5 separate from the rest of the problem or is it going to become (1/5)x - 2. but once you integrate and get whatever the general solution is you use the initial condition of f(10) = -10 to find C. then plug C into the general solution equation to get the particular solution. confidence assessment: 1
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08:25:21 5.1.50 (was 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 --> confidence assessment:
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08:33:18 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 --> ok so antiderivative of 1/5 x - 2 is 1/5 * (1/2 x^2) - 2 x + c = x^2 / 10 - 2x + c. then plug in f(10) =-10 -10 = -10 + C c=0 part. solution = x^2 / 10 - 2 x self critique assessment: 2
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08:34:24 Is the derivative of your particular solution equal to 1/5 * x - 2? Why should it be?
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RESPONSE --> yes It should be because you are finding the anti-derivative. and the anti-derivative and the derivative are inverses of each other. confidence assessment: 3
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08:36:27 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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RESPONSE --> ok, what I stated before actually worked out would be: Derivative f(x) = x^2 / 10 - 2 x is f ' (x) = (x^2) ' / 10 - 2 ( x ) '. Since (x^2) ' = 2 x and (x) ' = 1 you get..... f ' (x) = 2 x / 10 - 2 * 1 = x / 5 - 2, which is 1/5 * x - 2. self critique assessment: 2
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08:41:56 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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RESPONSE --> i think you need to integrate twice because it gives you the 2nd derivative. = integral sign (x^3/3) + C im unsure about what the 2nd integral will be. but then i know you follow the same steps as 2 problems ago plugging in the initial condition and finding C, then pluggin C into the general solution. confidence assessment: 1
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08:50:07 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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RESPONSE --> oh ok so we find the complet particular solution with just the first antiderivative. f(x) = x^3/3 + C. f(0) = 6 0^3/3 + C = 6 so C = 6 part. solution = f(x) =x^3 / 3 + 6 then find antiderivative of this which is f(x) = (x^4 / 4) / 3 + 6x + C = x^4 / 12 + 6 x + C plug in f(0) =3 0^4/12 + 6*0 + C = 3 C= 3 f(x) = x^4 / 12 + 6x + 3 self critique assessment: 2
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08:52:03 Is the second derivative of your particular solution equal to x^2? Why should it be?
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RESPONSE --> yes, because you found the 2nd antiderivative so it makes sense that because the derivative and antiderivative are inverses of each other that the 2nd derivative and 2nd antiderivative would also be inverses. confidence assessment: 2
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08:53:25 *& 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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RESPONSE --> Particular solution is f(x) = x^4 / 12 + 6 x + 3. derivative of this = f ' (x) = (4 x^3) / 12 + 6 = x^3 / 3 + 6. derivative of this = f ''(x) = (3 x^2) / 3 = x^2. self critique assessment: 3
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09:01:28 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 --> find anti-derivative dP/dt = 500 t^1.06 = 500 (t^2.06/2.06) + C confidence assessment: 2
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09:06:50
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RESPONSE --> ok so I found the anti-derivative correctly. then I need to plug in P = 50,000 when t = 0 50,000 = 500 * 0^2.06 / 2.06 + c c = 50,000. pop function then equals P = 500 t^2.06 / 2.06 + 50,000 when t =10 P = 500 * 10^2.06 / 2.06 + 50,000 = 27,900 + 50,000 = 77,900 self critique assessment: 2
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09:17:31 5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2
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RESPONSE --> sqrt(3-x^3) * 3x^2 =integral sign (3-x^3)^1/2 (3x^2)dx =(3-x^3)^3/2 / (3/2) + C 2/3 (3-x^3)^3/2 + C confidence assessment: 2
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09:19:07 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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RESPONSE --> ok so it looks like I missed the - sign. i now understand what you've said about the chain rule. self critique assessment: 3
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09:27:24 5.2. 18 (was 5.2.16 integral of x^2/(x^3-1)^2
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RESPONSE --> x^2/(x^3-1)^2 = integral sign (x^3-1)^2 *(x^2)dx = (x^3-1)^3 / 3 + C =1/3(x^3-1)^3 + C confidence assessment: 2
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09:28:29 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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RESPONSE --> instead of writing (1/3)(x3-1)^3 + C it should have been written out 1 / (3 ( 1 - x^3) ) + C self critique assessment: 3
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09:30:23 5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3)
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RESPONSE --> unsure what to do when sqrt is in denominator confidence assessment: 0
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09:35:14 *& 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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RESPONSE --> let u = 1 - x^3 du/dx = -3 x^2 so that x^2 = -1/3 du/dx. then antiderivative of x^2 / (sqrt(1-x^3)) = -1/3 ( 2 sqrt(u) ) + C = -2/3 sqrt(1-x^3) + C self critique assessment: 2
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