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Math Help - Help with differential equation problem?

  1. #1
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    Help with differential equation problem?

    The expression dy/dx = x(cubed root(y)) gives the slope at any point on the graph of the function f(x) where f(2) = 8.
    I am supposed to find
    1. The equation of the tangent line to f(x) at point (2, 8).
    2. An expression for f(x) in terms of x.
    and 3.The domain of and minimum value of f(x).

    I know how to separate differential equations and then find C, but I can't figure out how for this equation. Also, I'm not even sure if that's what it's asking! I'm really confused about this. Any help? I'd really appreciate it. Thanks!
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  2. #2
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    Re: Help with differential equation problem?

    Quote Originally Posted by canyouhelp View Post
    The expression dy/dx = x(cubed root(y)) gives the slope at any point on the graph of the function f(x) where f(2) = 8.
    I am supposed to find
    1. The equation of the tangent line to f(x) at point (2, 8).
    2. An expression for f(x) in terms of x.
    and 3.The domain of and minimum value of f(x).

    I know how to separate differential equations and then find C, but I can't figure out how for this equation. Also, I'm not even sure if that's what it's asking! I'm really confused about this. Any help? I'd really appreciate it. Thanks!
    1. You are given the formula for the slope. $\dfrac {dy}{dx}=x\sqrt[3]y$

    At $x=2$ you are told that $f(2)=8$ So plug these points in to $\dfrac {dy}{dx}$ to find the slope.

    Then use point-slope form to get the equation for the line.

    2. You've got a separable diff eq and an initial condition. Solve this just like you've been doing with the rest of the problems.

    3. Examine the function from (2) and find it's domain. Find the minimum as usual by setting the derivative to 0 and solving for x.
    Thanks from canyouhelp and topsquark
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    Re: Help with differential equation problem?

    Quote Originally Posted by romsek View Post
    1. You are given the formula for the slope. $\dfrac {dy}{dx}=x\sqrt[3]y$

    At $x=2$ you are told that $f(2)=8$ So plug these points in to $\dfrac {dy}{dx}$ to find the slope.

    Then use point-slope form to get the equation for the line.

    2. You've got a separable diff eq and an initial condition. Solve this just like you've been doing with the rest of the problems.

    3. Examine the function from (2) and find it's domain. Find the minimum as usual by setting the derivative to 0 and solving for x.
    For part 1 then would the slope be 4? I'm sorry..maybe I didn't do that right? :/

    Part 2.. I'm having trouble separating it with the cubed root :/ Would it be dy/(cubedroot(y))=xdx? I don't think that's right but I don't know.
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    Re: Help with differential equation problem?

    Quote Originally Posted by canyouhelp View Post
    For part 1 then would the slope be 4? I'm sorry..maybe I didn't do that right? :/

    Part 2.. I'm having trouble separating it with the cubed root :/ Would it be dy/(cubedroot(y))=xdx? I don't think that's right but I don't know.
    4 is correct.

    your separation of variables is correct.
    Thanks from canyouhelp
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    Re: Help with differential equation problem?

    Quote Originally Posted by romsek View Post
    4 is correct.

    your separation of variables is correct.
    Okay thanks! How would I turn this into point-slope form (I'm drawing a blank, I used to remember but now I don't..I know the formula is y=mx+b and m=4 but I can't remember how to transfer the rest)
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    Re: Help with differential equation problem?

    Quote Originally Posted by romsek View Post
    4 is correct.

    your separation of variables is correct.
    The problem with my separation though is that when I integrate it I get 3y/(2(cubedroot(y))+c= (x^2/(x))+c With (2,8) plugged in I get 6+c=2+c So I can't find c. Is there something wrong with the separation? Or how I integrated them?
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    Re: Help with differential equation problem?

    Quote Originally Posted by canyouhelp View Post
    Okay thanks! How would I turn this into point-slope form (I'm drawing a blank, I used to remember but now I don't..I know the formula is y=mx+b and m=4 but I can't remember how to transfer the rest)
    look it up
    Thanks from topsquark
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    Re: Help with differential equation problem?

    Quote Originally Posted by canyouhelp View Post
    The problem with my separation though is that when I integrate it I get 3y/(2(cubedroot(y))+c= (x^2/(x))+c With (2,8) plugged in I get 6+c=2+c So I can't find c. Is there something wrong with the separation? Or how I integrated them?
    you integrated incorrectly
    Thanks from canyouhelp
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    Re: Help with differential equation problem?

    Quote Originally Posted by romsek View Post
    look it up
    Is the equation for part 1: y-8=4(x-2)? :/
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    Re: Help with differential equation problem?

    Quote Originally Posted by romsek View Post
    you integrated incorrectly
    Can you help me with what I did wrong? Maybe tell me where I might have messed up?
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    Re: Help with differential equation problem?

    Quote Originally Posted by canyouhelp View Post
    Is the equation for part 1: y-8=4(x-2)? :/
    yes
    Thanks from canyouhelp
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    Re: Help with differential equation problem?

    Quote Originally Posted by canyouhelp View Post
    Can you help me with what I did wrong? Maybe tell me where I might have messed up?
    $\dfrac {dy}{\sqrt[3] y} = y^{-1/3} dy$

    integrating you get

    $\dfrac 3 2 y^{2/3}$

    which isn't what you obtained
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    Re: Help with differential equation problem?

    Quote Originally Posted by romsek View Post
    $\dfrac {dy}{\sqrt[3] y} = y^{-1/3} dy$

    integrating you get

    $\dfrac 3 2 y^{2/3}$

    which isn't what you obtained
    Oh okay. So then I have (2/3)y^(2/3)=(x^2)/x Which is x=(3y^(2/3))/2 right? Would that be it then?..or do I need to plug (2,8) into this?
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    Re: Help with differential equation problem?

    Quote Originally Posted by canyouhelp View Post
    Oh okay. So then I have (2/3)y^(2/3)=(x^2)/x Which is x=(3y^(2/3))/2 right? Would that be it then?..or do I need to plug (2,8) into this?
    Okay so part 2 would be f(x)=(2/3)(sqrt(2/3))(x^(3/2)) right? I think I finally figured that out!
    Last edited by canyouhelp; April 11th 2014 at 04:51 PM.
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