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Math Help - Find y in terms of x

  1. #1
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    Find y in terms of x

    Hey,

    Could anyone help me with this calculus question?

    Find y in terms of x

    dy/dx = √x+5 / √x
    Last edited by Oasis1993; September 13th 2009 at 06:27 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Oasis1993 View Post
    Hey,

    Could anyone help me with this calculus question?

    Find y in terms of x

    dy/dx = √x+5/ √x


    \frac{dy}{dx} = \sqrt{x} + \frac{5}{\sqrt{x}}

    dy = \left( \sqrt{x} + \frac{5}{\sqrt{x}}\right) dx

    \int y = \int \left( \sqrt{x} + \frac{5}{\sqrt{x}}\right) dx

    y = \frac{2}{3}(x)^{\frac{3}{2}} + 2(5 \sqrt{x})
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  3. #3
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    Quote Originally Posted by Amer View Post
    \frac{dy}{dx} = \sqrt{x} + \frac{5}{\sqrt{x}}

    dy = \left( \sqrt{x} + \frac{5}{\sqrt{x}}\right) dx

    \int y = \int \left( \sqrt{x} + \frac{5}{\sqrt{x}}\right) dx

    y = \frac{2}{3}(x)^{\frac{3}{2}} + 2(5 \sqrt{x})
    Thank you very much for your reply Amer,
    But my answer sheet says the answer is y= x+10√x+K
    can you relate?

    Thank you.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Oasis1993 View Post
    Thank you very much for your reply Amer,
    But my answer sheet says the answer is y= x+10√x+K
    can you relate?

    Thank you.
    yeah I forgot to add the constant but

    your question is

    \frac{dy}{dx} = \frac{\sqrt{x}+5}{\sqrt{x}} not

    \frac{dy}{dx} = \sqrt{x} + \frac{5}{\sqrt{x}}

    I solved the second one the first is the same , you should just integrate  1 + \frac{5}{\sqrt{x}} instead of \sqrt{x} + \frac{5}{\sqrt{x}}

    that's your fault you should put bracket like this dy/dx = (sqrt(x) + 5)/sqrt(x)
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  5. #5
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    Quote Originally Posted by Amer View Post
    yeah I forgot to add the constant but

    your question is

    \frac{dy}{dx} = \frac{\sqrt{x}+5}{\sqrt{x}} not

    \frac{dy}{dx} = \sqrt{x} + \frac{5}{\sqrt{x}}

    I solved the second one the first is the same , you should just integrate  1 + \frac{5}{\sqrt{x}} instead of \sqrt{x} + \frac{5}{\sqrt{x}}

    that's your fault you should put bracket like this dy/dx = (sqrt(x) + 5)/sqrt(x)
    Sorry I am new to this forum. I cant use the tools properly.
    Could you integrate that for me?
    Thank you very much for your help.
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Oasis1993 View Post
    Sorry I am new to this forum. I cant use the tools properly.
    Could you integrate that for me?
    Thank you very much for your help.

    you should know how to integrate it , I integrate the hard part \frac{5}{\sqrt{x}} you just have to integrate 1 if you can't integrate this it is a big problem you should learn it .
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  7. #7
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    \frac{dy}{dx} = \frac{\sqrt{x}+5}{\sqrt{x}}

    \int dy= \int \left( \frac{\sqrt{x}+5}{\sqrt{x}}\right) dx

    y = \int \left(1 + \frac{5}{\sqrt{x}}\right) dx

    y = x + \frac{5\sqrt{x}}{\frac{1}{2}} + c

    y = x + 10\sqrt{x} + c
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