course Mth 272 Ѐ|Vȣcassignment #006
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16:01:50 5.1.40 (was 5.1.30)(was 5.1.34 int of 1/(4x^2)
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RESPONSE --> Antideriv x^-2= -x^-1 Antideriv 1/(4x^2)= 1/4 (-x^-1) Simplified= 1/4(-1/x)= -1/(4x) -1/(4x) + c confidence assessment: 3
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16:02:00 *& 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 self critique assessment: 3
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16:03:00 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 --> Antideriv 1/5x-2= 1/5(1/2x^2) - 2x + c particular solution = x^2/10 -2x+c = -10= -10 +c c=0 sub for c = x^2/ 10 - 2x + 0 confidence assessment: 3
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16:03:24 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 good self critique assessment: 3
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16:06:57 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 --> I accidentally pressed the button for ""next question/answer,"" so it gave me the answer to that supplemental question.I do understand though that the derivative should be equal to 1/5*x-2 because it was direcxted to find f(x) from f'(x)= that expression. self critique assessment: 2
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16:09:12 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 --> 2 antideriv's necessary to get f''(x) antideriv of x^2= x^3/3 + c x= 0 0^3/3 + c= 6 c= 6 = x^3/3 + 6 antideriv x^3/3 + 6= x^4/12+6x+c f(0)=3 c=3 = x^4/12+6x+3 confidence assessment: 2
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16:09:47 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 --> Ok I thought I had more difficulty with the problem than what I did simply because it was a lot of work. self critique assessment: 2
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16:10:27 Is the second derivative of your particular solution equal to x^2? Why should it be?
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RESPONSE --> It should be to match the context of the original problem. confidence assessment: 3
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16:10:33 *& 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 --> ok self critique assessment: 3
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16:11:07 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 --> P= 50,000 T=10 Antideriv P = 500t^2.06/2.06 + c 50,000 = 500*0^2.06/2.06 + c c= 50,000 P= 500t^2.06/2.06 + 50000 confidence assessment:
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16:11:47 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 --> I hit cancel on my confidence assessment and it went to the solution. Did you receive my solution for this problem? self critique assessment: 3
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16:12:41 5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2
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RESPONSE --> Deriv 3-x^3= -3x^2 sqrt(3-x^3)= (3-x^3)^1/2 =[(3-x^3)^1/2]*-3x^2 2/3*(3-x^3)^(3/2) = 2/3*(3-x^3)^(3/2)+c confidence assessment: 3
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16:12:54 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 self critique assessment: 3
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16:13:53 5.2. 18 (was 5.2.16 integral of x^2/(x^3-1)^2
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RESPONSE --> u = x^3 - 1 du/dx= 3x^2 -1/3 (x^3 - 1)^-1 + c Simplified= -1/[3( x^3 - 1)] + c confidence assessment: 3
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16:14:05 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 --> ok looks good self critique assessment: 3
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16:14:48 5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3)
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RESPONSE --> Du/dx of 1-x^3= -3x^2 x^2= -1/3*(-3x^2) = (-1/3)/[(1-x^3)^1/2]* -3x^2 Simplified antiderivative= -2/3*sqrt(1-x^3) + c confidence assessment: 3
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16:15:01 *& 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 --> good ok self critique assessment: 3
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