course mth272 ?????l?U????|??€assignment #005
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15:05:34 5.1.12 integrate 3 t^4 dt and check by differentiation
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RESPONSE --> 3t^(5)/5 differentiate 15t(4)/5 3t^4 confidence assessment: 3
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15:05:43 An antiderivative of the power function t^4 is one power higher so it will be a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative of t^4 is be t^5 / 5. By the constant rule the antiderivative of 3 t^4 is therefore 3 * t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative3 t^5 / 5 + c. The derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4), verifying our antiderivative. **
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RESPONSE --> self critique assessment: 3
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15:08:59 5.1.20 (was 5.1.18) integrate v^-.5 dv and check by differentiation
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RESPONSE --> v^(1/2)/(1/2) + C 2v^(1/2) + C differentiate 2v^(1/2) + C v^(-1/2) confidence assessment: 3
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15:09:07 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 --> self critique assessment: 3
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15:09:27 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Its not that hard....yet. confidence assessment: 3
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v??M??c??????assignment #006 006. `query 6 Applied Calculus II 04-08-2009
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15:12:48 5.1.40 (was 5.1.30)(was 5.1.34 int of 1/(4x^2)
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RESPONSE --> (1/4) (x^-2) (1/4)(x^-1/-1) (1/4)(-1/x) =(-1/4x) confidence assessment: 3
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15:13:04 *& 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 --> always forget to add my constant...
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15:19:52 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 --> (1/5)(x^2/2) - 2x f(x)= x^2/10 - 2x + C f(10)= 100/10 - 20 + C= -10 C= 0 f(x)= x^2/10 - 2x confidence assessment: 3
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15:20:29 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 --> self critique assessment: 3
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15:23:38 Is the derivative of your particular solution equal to 1/5 * x - 2? Why should it be?
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RESPONSE --> f(x)= x^2 - 2x = 20x/100 - 2 = 1/5*x - 2 Because the particular solution is the exact formula that we differentiate to find the integrand confidence assessment: 3
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15:23:50 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 --> self critique assessment: 3
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15:33:11 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 --> f ' (x)= x^3/3 + C 6= x^3/3 + C C= 6 f(x) x^3/3 + 6 = 1/3 (x^3) + 6 = 1/3 ( x^4/4) + 6x + C = x^4/12 + 6x + C 3= 0 + 0 + C C=3 f(x)= x^4/12 + 6x + 3 confidence assessment: 3
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15:34:04 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 --> self critique assessment: 3
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15:36:00 Is the second derivative of your particular solution equal to x^2? Why should it be?
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RESPONSE --> Yes, because the particular solution we integrated out of the given information is the orginial function. We just went the opposite way confidence assessment: 3
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15:36:09 *& 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 --> self critique assessment: 3
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15:46:03 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 --> =500(t^2.06/2.06) + C 50,000= 500(0/0) + C C= 50,000 f(10)= 500(10^2.06/2.06) + 50,000 = 77,867.81 confidence assessment: 2
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15:47:43 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 --> self critique assessment: 3
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15:53:23 5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2
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RESPONSE --> f(3-x^3)^ 1/2 (3x^2) ={(3-x^3)^3/2}/3/2 = (2/3) (3-x^3)^3/2 confidence assessment: 2
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16:00:52 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 --> I see how the 2/3 becomes negative but I still dont quite understand the whole process...I'll need to do more of these self critique assessment: 2
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16:12:55 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 * (3)(1/3) 1/3f 3x^2(x^3-1)^-2 (1/3)(x^3-1)^-1/-1 1/3*1/(-x^3+1) 1/(-3x^3+3) confidence assessment: 3
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16:13:21 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 --> self critique assessment: 3
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16:39:00 5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3)
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RESPONSE --> x^2(1-x^3)^1/2 (1/3)(3) 1/3f 3x^2(1-x^3) 1/3{(1-x^3)^3/2/(3/2) (1/3)(2/3)(1-x^3)^3/2 2/9(1-x^3)^3/2 confidence assessment: 2
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16:41:08 *& 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 --> definitely need more work with subsititution self critique assessment: 2
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