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Math Help - Primitive function and optimize problems

  1. #1
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    Primitive function and optimize problems

    Question 1

    Here I will simply find a primitive function

    int 1 / (x ^ 2-4x +4) (x ^ 2-4x +5) dx

    Trying to simplify and go with the first brackets, get it to (x-2) ^ 2 Thus, I write it as

    int 1 / (x-2) ^ 2 (x ^ 2-4x +5) dx

    Then I used partialbråksuppdelning, this makes me a bit confused about whether I was right or if it is particularly clever, given the results further ahead.

    A / (x-2) + B / (x-2) ^ 2 + (CX + D) / (x ^ 2-4x +5)

    Then you should get all the same denominator and then multiplying the terms. But since I only had a 1 from the beginning in the numeratorthen there isare no CX , thenC = 0

    (A (x ^ 2-4x +5) (x 2) + B (x ^ 2-4x +5) + (CX + D) (x-2) ^ 2) / ((x-2) ^ 2 * (x ^ 2-4x + 5))

    Multiplies simply set my A, B, CX, but not sure if I right, or what the problem actually becomes

    x ^ 3 (A + C) = 0
    x ^ 2 (-6A + B-4C + D) = 0
    X (13A-4B +4 C-4D) = 0
    1 (-10A + 5B +4 D) = 1

    Then I'll do the Gaussian elimination and concludes
    C = 0
    B = -1
    D = - 1 / 12
    A = - 1 / 3

    I put in these values in the original approach with partialbråksuppdelning so it will hardly be of any use?

    Only the first term is easy to do primitive functions to, the other two are just as difficult as before ..

    Question 2

    A barrel emits water with 0.52te ^ (-0.89t) cm ^ 3/day when it emits at most per day and how much is it?

    I perceive it as a derivative, it is a rate because it is cm ^ 3/day and want to look up function to find out how much it is. int = integral


    int 0.52te ^ (-089t) dt

    Do partial integration chooses t g, e, f and move out beyond 0.52 int

    0.52 int t * e ^ (-0.89t) dt = [(e ^-0.89t) * t / (-0.89) * t] - int 1 * e ^-0.89t

    What goes up must come up with intergralen to the last and it is simple it is (e ^ -0.89 t) / (-0.89)

    Thus, the function will

    f (t) = 0:52 * ((e ^ -0.89 t) * t / (-0.89) e ^ -0.89 t / (-0.89))
    Finds zero, put the = 0 and you get a zero, which is
    t = 1

    Then do I simply set the value of the primitive function .. but it feels wrong? How am I going to anyway?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    2. You find the derivative, not the integral.

    You were given the rate at which water is emitted, you have to find the highest rate of emission and the time. So, you are looking for a maximum, which is found when the derivative is equal to 0.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    2. You find the derivative, not the integral.

    You were given the rate at which water is emitted, you have to find the highest rate of emission and the time. So, you are looking for a maximum, which is found when the derivative is equal to 0.
    Wasn't i given the derivate from the beginning? Becuase it says cm^3/day it's a speed? So if i do the intergration i get the derivitave?

    Lets get it clear, everything i did gave me the derivate ? but why?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Okay, you can consider that to be the speed. Now, you are asked to find the maximum speed. This occurs when the acceleration is previously ascending and reaches 0.

    You got the integral... which is the amount of water (distance) then set distance to 0...
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  5. #5
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    Yes but i want to know both when the speed is fastest and the amount of water
    But the integral i'm doing is not getting right :/
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Ok, if you now know the shape of the graph that was given to you, you'll notice that you don't have a 'maximum amount of water'.

    What you need to find, is the derivative, then solve to t at the maximum point.

    Then, you will need the 'magnitude of this speed' by substituting the t you just found in the original formula given to you.

    1. Now, I believe you did a mistake in the Gaussian Elimination. By inspection, I find find that B = 1. Then, having 4 equations with 3 unknowns, I get A = 0, C = 0 and D becomes -1

    A + C = 0
    -6A + B - 4C + D = 0
    13A- 4B +4C - 4D = 0
    -10A + 5B + 4D = 1

    These become:

    A + C = 0
    -6A - 4C + D + 1= 0
    13A +4C - 4D - 4 = 0
    -5A + 2D + 2 = 0

    Taking the first equation, C = -A
    Taking the last equation, D = (5A - 2)/2

    Taking the second equation and the two equations we just got, we get:

    -6A -4(-A) + (5A - 2)/2 + 1 = 0

    Multiply throughout by 2, then simplify to get A = 0
    Hence C = 0

    And -6(0) - 4(0) + D + 1= 0

    D = -1

    Substitute that in the original integral to get:

    \displaystyle \int \dfrac{1}{(x-2)^2(x^2 - 4x + 5)}\ dx = \int \dfrac{1}{(x-2)^2} - \dfrac{1}{x^2 - 4x + 5}\ dx

    Now, this becomes:

    \displaystyle \int \dfrac{1}{(x-2)^2} - \dfrac{1}{x^2 - 4x + 5}\ dx = \int (x-2)^{-2}\ dx- \int \dfrac{1}{(x-2)^2+1}\ dx

    \displaystyle \int \dfrac{1}{(x-2)^2} - \dfrac{1}{x^2 - 4x + 5}\ dx =  \dfrac{(x-2)^{-1}}{-1} - \int \dfrac{1}{(x-2)^2+1}\ dx

    \displaystyle \int \dfrac{1}{(x-2)^2} - \dfrac{1}{x^2 - 4x + 5}\ dx =  \dfrac{1}{2-x} - \int \dfrac{1}{(x-2)^2+1}\ dx

    Substitute \tan\theta = x - 2, \sec^2\theta\ d\theta = 1\ dx

    \displaystyle \dfrac{1}{2-x} - \int \dfrac{1}{\tan^2\theta+1}(\sec^2\theta)\ d\theta = \dfrac{1}{2-x} - \int \ d\theta

    \displaystyle \dfrac{1}{2-x} - \int\ d\theta = \dfrac{1}{2-x} - \theta + C

    \displaystyle \dfrac{1}{2-x} - \theta  + C= \dfrac{1}{2-x} - \tan^{-1}(x-2) + C
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