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Math Help - integrals and implicit - and fast

  1. #1
    Member cassiopeia1289's Avatar
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    integrals and implicit - and fast

    ok help: we're given the following:
    \int_{0}^{5}{f(x)dx = 8}
    f(x) is even function
    \int_{0}^{5}{g(x)dx = 4}
    g(x) is odd function

    find the following (if possible):
    \int_{0}^{5}\frac{f(x)}{g(x)}dx


    next:
    f(x) = \int_{1}^{2x}\frac{dt}{(t^2+1)^\frac{1}{3}}
    find f'(1)
    (I feel as though \frac{1}{2^\frac{1}{3}} would be too easy)

    next:
    let {x^2 - 4xy + y^2 = 3}, find expression for \frac{dy}{dx}
    when I do it I get \frac{2y-x}{y-2x}
    then it goes on asking "are there any points on curve where tangent is horizontal or vertical?"
    Last edited by cassiopeia1289; April 24th 2008 at 02:22 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cassiopeia1289 View Post
    ok help: we're given the following:
    integral from 0 to 5 of f(x) = 8
    f(x) is even function
    integral from 0 to 5 of g(x) = 4
    g(x) is odd function

    find the following (if possible):
    integral from 0 to 5 of [f(x)/g(x)]dx


    next:
    f(x) = integral from [1 to 2x] of [dt/(t^2+1)^(1/3)]
    find f'(1)
    (I feel as though 1/(2^(1/3)) would be too easy)

    next:
    let x^2 - 4xy + y^2 = 3, find expression for dy/dx
    when I do it I get (2y-x)/(y-2x) where (y-x) cancel out and I just get 1 - which doesnt even make sense

    then it goes on asking "are there any points on curve where tangent is horizontal or vertical?" well, obviously its a constant then ...
    I have to have done something wrong


    (AS WE SPEAK - WORKING ON TYPING THESE IN USING ACTUAL MATH FORM - JUST BE PATIENT, LOL)
    The second one you have \int_a^{2x}f(t)dt the derivative would be 2f(2x)
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  3. #3
    Moo
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    Hello,

    For the second one, let G(t) be an antiderivative for g(t)=\frac{1}{(t^2+1)^{1/3}}

    According to the definition of an integral, f(x)=G(2x)-G(1)

    Hence, f'(x)=2g(2x)

    So f'(1)=2g(2)=2 \frac{1}{5^{1/3}}
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  4. #4
    Moo
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    let x^2 - 4xy + y^2 = 3, find expression for dy/dx
    when I do it I get (2y-x)/(y-2x) where (y-x) cancel out and I just get 1 - which doesnt even make sense
    Well...

    \frac{2y-x}{y-2x}=\frac 12 \times \frac{y-\frac x2}{y-2x}

    I don't find that it cancels out
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  5. #5
    Member cassiopeia1289's Avatar
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    ok - yeah it doesnt cancel - but I finally figured out that horiz tangent at x^2 = -4
    but that doesn't work, so there is no horizontal tangent?
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  6. #6
    Moo
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    I had the expression for f'(x) so I just replaced x by 1..
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  7. #7
    Member cassiopeia1289's Avatar
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    (unrelated - but how do you write definite integrals with the numbers there?)
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  8. #8
    Moo
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    \int_{bbbbbb}^{aaaaaa}
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  9. #9
    Member cassiopeia1289's Avatar
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    can you help with the first and last as well?
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cassiopeia1289 View Post
    ok help: we're given the following:
    \int_{0}^{5}{f(x)dx = 8}
    f(x) is even function
    \int_{0}^{5}{g(x)dx = 4}
    g(x) is odd function

    find the following (if possible):
    \int_{0}^{5}\frac{f(x)}{g(x)}dx


    next:
    f(x) = \int_{1}^{2x}\frac{dt}{(t^2+1)^\frac{1}{3}}
    find f'(1)
    (I feel as though \frac{1}{2^\frac{1}{3}} would be too easy)

    next:
    let {x^2 - 4xy + y^2 = 3}, find expression for \frac{dy}{dx}
    when I do it I get \frac{2y-x}{y-2x}
    then it goes on asking "are there any points on curve where tangent is horizontal or vertical?"
    2x-4[xy'+y]+2yy'=0

    solve for y'
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  11. #11
    Moo
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    Tangent is horizontal when y'=0 and vertical when y' tends to infinity.

    So isolate y' and find for which values of x the conditions are verified.



    For the first one, I think it's not possible with such elements
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Tangent is horizontal when y'=0 and vertical when y' tends to infinity.

    So isolate y' and find for which values of x the conditions are verified.



    For the first one, I think it's not possible with such elements
    I agree with Moo...the first one is impossible unless it wanted \frac{\int_0^{5}f(x)dx}{\int_0^{5}g(x)dx}

    But if it was I am sure you could figure it out
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  13. #13
    Member cassiopeia1289's Avatar
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    ok, good to know about the first one

    the last one - I solved and found that I would have to take the square root of a negative, since no imaginary numbers would work for vertical or horizontal tangents, I'm kinda stuck - so are there just none?
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cassiopeia1289 View Post
    ^ yeah, and when I did I got what I put below in the original post
    what I don't know is how to find the horizontal and vertical tangent point/line things
    Horizontal tangent lines occur when the first derivative is 0
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  15. #15
    Member cassiopeia1289's Avatar
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    ^ yes, I realize this - take the derivative yourself and you'll see my problem

    \frac{dy}{dx} = \frac{2y-x}{y-2x}
    now you get 0 = \frac{2y-x}{y-2x}
    I solved for y and got y = \frac{x}{2}
    I plugged that into the original problem so as to find the x solution and therefore find the horizontal line
    so: x^2 - 4x(\frac{x}{2}) + (\frac{x}{2})^2 = 3
    solve and you get x^2 = -4
    this only works if you use imaginary numbers, so is there just not a horizontal tangent line?
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