course Mth 272 Assignment 5 and 6
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11:52:05 query 5.1.12 integrate t^4 dt and check by differentiation
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RESPONSE --> t^ 5 = 5 t^4 = derivative t^5/5 antiderivative add the Constant hence t^5 / 5 + C so 1/5 t^ 5 * 5t^4 gives you t^4
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11:52:20 ** 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 --> always remember to deal with the constant
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11:57:46 query 5.1.18 integrate v^-.5 dv and check by differentiation
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RESPONSE --> v ^.5 = .5 v ^ -.5 the derivative v^.5/.5 = v^.5 / .5 = 2v^.5 the antiderivative add the constand so 2v^5 + c 2v^.5 = 2*(.5) v^-.5 = v^-.5
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11:57:51 ** 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 --> ok
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11:58:22 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> the issue of the constant is key and checking by seeing the derivative is also key
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??????Q??m??? assignment #006 ?{?h?????????Applied Calculus I 06-10-2006
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12:04:25 5.1.36 (was 5.1.34 int of 1/(4x^2)
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RESPONSE --> factor out 1/4 and do the antiderivative of x^-2 which is -1 x^ -1 so 1/4 (-1x $-1)
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12:05:06 *&*& 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 --> once you get the answer rearrrange the problem if possible. 1 / 4 * (-1/x) = -1/(4x). The general antiderivative is -1 / (4x) + c.
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12:10:24 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 --> .5 x^2 the antiderivative of x -2 is -2x the antiderivative so .5 * (.5 x^2) - 2 x + C = x^2 / 10 - 2x + C the particular solution is when you do -10 = which put into the above ould be -10=10^2 - 2(10)+C in turn -10+c or c is 0
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12:11:02 ** 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 --> so with that you solve the problem with 0 as the C so (x) = x^2 / 10 - 2 x + 0 or just x^2 / 10 - 2 x
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12:11:58 Is the derivative of your particular solution equal to 1/5 * x - 2? Why should it be?
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RESPONSE --> ok
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12:12:32 ** 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 --> find the derivative, and the derivative should be equal to the expression because the problem is to find f(x)
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12:14:03 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 --> if x^2 then the antiderivative is x^3/3 + C put in the value for 6 hence c is 6
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12:16:10 ** 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 --> make sure to write the final function and f(0) = 3 then c= 3 hence the function is x^4 / 12 + 6x + 3
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12:17:47 Is the second derivative of your particular solution equal to x^2? Why should it be?
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RESPONSE --> yes because the function is x^4 / 12 + 6 x + 3 adn the derivative of that is x^3 / 3 + 6 and the derivative of that is x^2
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12:17:50 *&*& 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
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12:18:08 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 --> k
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12:20:37 ** 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 --> To find P you have to integrate dP/dt antiderivative is P = 500t^2.06/2.06 + c and c wil be 50000 solution , p = 500 t^2.06 / 2.06 + 50,000
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12:24:38 5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2
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RESPONSE --> first deriv is -3x^2 ..?
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12:25:25 ** 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 --> use the general power rule so n = 1/2, getting -1 / (1/2+1) * (3 - x^3)^(1/2 + 1) = -2/3 (3 - x^3)^(3/2) antiderivative is -2/3 (3 - x^3)^(3/2) + c
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12:26:36 5.2. 18 (was 5.2.16 integral of x^2/(x^3-1)^2
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RESPONSE --> ok
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12:28:08 ** 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 --> use the general power rule again. 1/3 (-u^-1) + c, or -1/3 (x^3 - 1)^-1 + c = -1 / (3 ( x^3 - 1) ) + c
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12:28:26 5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3)
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RESPONSE --> ok
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12:29:17 *&*& 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 --> use the general power rule. note to self learn the different methods and how to deal with square roots and other complex situations.
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