course MTH 272 7/1 7 pm If your solution to a stated problem does not match the given solution, you should self-critique per instructions at
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think my answer equates to the same thing. Self-critique Rating: ********************************************* 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? 10x or x+10 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a graph with different points plotted with variables and numbers Confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): the solution and explanation helped me figure this type of problem out much better. Self-critique Rating: good ********************************************* Question: `qIs the derivative of your particular solution equal to 1/5 * x - 2? Why should it be? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes it should be. If derived properly, it should come out as that. Confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): got it right Self-critique Rating: good ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Plug in numbers for the corresponding equations and then bring them all together to form one line. F”(x)=x^2, f(0)=3 f’(0)=6 My answer: x^4 / 6 + 7x + 3 Confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): get it better now. These are tough. Self-critique Rating: ********************************************* Question: `qIs the second derivative of your particular solution equal to x^2? Why should it be? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, I got a different answer than that.
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): oh ok got it Self-critique Rating: ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P=50 k 10 years dP/dt=500t^1.06 t=10 P(10)= P=500t^3.06/3.06 + c Now our population function is P = 500 t^2.06 / 2.06 + 5,000. So if t = 10 we get P = 500 * 10^3.06 / 3.06 + 50,000 = 81,600. ** Confidence rating:1.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): understand better now Self-critique Rating: ********************************************* Question: `q5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u' = -3x^2. -u^(1/2) du/dx. 1/(n+1) u^(n+1).= -2/3 (3 - x^3)^(3/2) + C Confidence rating:2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): was correct Self-critique Rating: ********************************************* Question: `q5.2. 18 (was 5.2.16 integral of x^2/(x^3-1)^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Move bottom to top and make exp. Negative. Multiple exp out and divide by new exp. Multiply out to make correct. 1/3 (-u^-1) + 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: move bottom to top and make exp negative. Integrate and multiply out. 2/3(sqrtx^3)-2 Confidence rating:0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: