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Thread: equations help ( differentiation )

  1. #1
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    equations help ( differentiation )

    The curve $\displaystyle y= cx + \frac{d}{x} $ has gradient 6 at the point $\displaystyle ( \frac{1}{2} , 1 ) $ find c and d

    my working,

    sub in the values for x and y and solve simultaneously but i dont get the right answer.....

    first equation = $\displaystyle \frac{1}{2}c + \frac{d}{\frac{1}{2}} $ which I simplfied to $\displaystyle 2c + 2d = 1 $ is this simplification correct?

    $\displaystyle \frac{dy}{dx} = c -dx^{-2}= 6 $

    these are my two equations, and when I try to solve for 'c' and 'd' i dont get the right answer?
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  2. #2
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    Hello,
    Quote Originally Posted by Tweety View Post
    The curve $\displaystyle y= cx + \frac{d}{x} $ has gradient 6 at the point $\displaystyle ( \frac{1}{2} , 1 ) $ find c and d

    my working,

    sub in the values for x and y and solve simultaneously but i dont get the right answer.....

    first equation = $\displaystyle \frac{1}{2}c + \frac{d}{\frac{1}{2}} $ which I simplfied to $\displaystyle 2c + 2d = 1 $ is this simplification correct?
    It's $\displaystyle 1=\frac 12 c +\frac{d}{\frac 12}$
    There's a problem with your simplification :
    $\displaystyle \frac{d}{\frac 12}=2d$
    so you have $\displaystyle 1=\frac 12 c+2d$

    $\displaystyle \frac{dy}{dx} = c -dx^{-2}= 6 $

    these are my two equations, and when I try to solve for 'c' and 'd' i dont get the right answer?
    This one is correct, though it is an equation for x=1/2
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    how do i evaluate $\displaystyle -dx^{-2} $ when x = $\displaystyle \frac{1}{2} $

    is this method correct..

    $\displaystyle \frac{1}{2}^{-2} $ = $\displaystyle -\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4} $

    so the equation is $\displaystyle c -\frac{1}{4}d = 6 $ ?
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    Moo
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    Quote Originally Posted by Tweety View Post
    how do i evaluate $\displaystyle -dx^{-2} $ when x = $\displaystyle \frac{1}{2} $

    is this method correct..

    $\displaystyle \frac{1}{2}^{-2} $ = $\displaystyle -\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4} $

    so the equation is $\displaystyle c -\frac{1}{4}d = 6 $ ?
    wow... I think you've got some problems in algebra :s

    $\displaystyle -x^{-2}=-\frac{1}{x^2}$

    So when x=1/2, we have $\displaystyle x^2=\left(\frac 12\right)^2=\frac{1^2}{2^2}=\frac 14$

    ---> $\displaystyle \frac{1}{x^2}=\frac{1}{\frac 14}=4$

    -----> $\displaystyle -dx^{-2}=-4d$
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  5. #5
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    Quote Originally Posted by Moo View Post
    wow... I think you've got some problems in algebra :s

    $\displaystyle -x^{-2}=-\frac{1}{x^2}$

    So when x=1/2, we have $\displaystyle x^2=\left(\frac 12\right)^2=\frac{1^2}{2^2}=\frac 14$

    ---> $\displaystyle \frac{1}{x^2}=\frac{1}{\frac 14}=4$

    -----> $\displaystyle -dx^{-2}=-4d$
    Yes thats why I ask for help !!

    thanks anyways..........
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