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Math Help - Differentiable functions

  1. #1
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    Differentiable functions

    Hi there,

    I'm having difficulty with this problem:

    Assume that x and y are differentiable functions of t. Find dy/dt when x^2-xy+y^2=19, dx/dt=3 for x=2, and y>0.

    Anyone have any ideas?

    Thanks in advanced for the help!
    Last edited by juicysharpie; December 7th 2011 at 12:51 AM.
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  2. #2
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    Re: Differentiable functions

    Quote Originally Posted by juicysharpie View Post
    Hi there,

    I'm having difficulty with this problem:

    Assume that x and y are differentiable functions of t. Find d/dt when x^2-xy+y^2=19, dx/dt=3 for x=2, and y>0.

    Anyone have any ideas?

    Thanks in advanced for the help!
    d/dt is NOT a function, it's a differential operator.

    Which function are you hoping to find the derivative of?
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  3. #3
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    Re: Differentiable functions

    Oh sorry, I wrote it wrong. I'm supposed to find dy/dt. I'll correct it in my question.
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  4. #4
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    Re: Differentiable functions

    Quote Originally Posted by juicysharpie View Post
    Hi there,

    I'm having difficulty with this problem:

    Assume that x and y are differentiable functions of t. Find dy/dt when x^2-xy+y^2=19, dx/dt=3 for x=2, and y>0.

    Anyone have any ideas?

    Thanks in advanced for the help!
    \displaystyle \begin{align*} x^2 - x\,y + y^2 &= 19 \\ \frac{d}{dt}\left(x^2 - x\,y + y^2\right) &= \frac{d}{dt}(19) \\ \frac{d}{dt}\left(x^2\right) - \frac{d}{dt}\left(x\,y\right) + \frac{d}{dt}\left(y^2\right) &= 0 \\ \frac{d}{dx}\left(x^2\right)\frac{dx}{dt} - \left(x\,\frac{dy}{dt} + y\,\frac{dx}{dt}\right) + \frac{d}{dy}\left(y^2\right)\frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} - x\,\frac{dy}{dt} - y\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} &= 0 \end{align*}

    Now substitute your values and solve.
    Last edited by Prove It; December 7th 2011 at 01:07 AM.
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  5. #5
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    Re: Differentiable functions

    Thanks for that part. But I wonder if y(dx/dt) is supposed to be negative for 2x(dx/dt) - [x(dy/dt)+y(dx/dt)] + 2y(dy/dt) because of the subtraction sign in front where you used the product rule? Also I have a hard time solving for the y-value. When I do it, there seems to be imaginary numbers. Am I just solve wrong?
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  6. #6
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    Re: Differentiable functions

    Quote Originally Posted by juicysharpie View Post
    Thanks for that part. But I wonder if (dx/dt)y for 2x(dx/dt) - [(dy/dt)x+(dx/dt)y] + 2y(dy/dt) supposed to be a negative too because of the subtraction sign in front where you used the product rule? Also I have a hard time solving for the y-value. When I do it, there seems to be imaginary numbers. Am I just solve wrong?
    Yes it should be negative.

    I doubt that there would be imaginary numbers, considering you are told y > 0, and complex numbers can't be ordered.

    Once you have substituted everything, you'll have a DE to solve for y.
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  7. #7
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    Re: Differentiable functions

    But don't I need y to solve for dy/dt? So, I would need to substitute x, dx/dt, and y. I plugged in x=2 into the original function x^2-xy+y^2=19 in order to find y, but that's when I kept getting imaginary numbers. My result of (2)^2-(2)y+y^2=19 is y^2+2y+15=0, which I can't factor.
    Last edited by juicysharpie; December 7th 2011 at 01:33 AM.
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  8. #8
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    Re: Differentiable functions

    Ok I figured it out. I just made mistakes with my signs. (2)^2-(2)y+y^2=19 is y^2+2y-15=0. So, y=5 or -3. But since the problem says y>0, y must equal 5.

    Thanks for the help!
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