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Thread: solve explicitly for y and diff to get dy/dx in terms of x

  1. #1
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    solve explicitly for y and diff to get dy/dx in terms of x

    Hi. Can you tell me what to do next.
    I got that dy/dx=-72x/50y

    Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

    36x2 + 25y2 = 900

    I think next it would be
    $\displaystyle 25y^2=900-36x^2$
    then square root everything and
    $\displaystyle 5y=300-6x$
    and
    $\displaystyle y=60-6/5x$
    what's next? did i do something wrong?
    thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kimjb View Post
    Hi. Can you tell me what to do next.
    I got that dy/dx=-72x/50y

    Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

    36x2 + 25y2 = 900

    I think next it would be
    $\displaystyle 25y^2=900-36x^2$
    then square root everything and
    $\displaystyle 5y=300-6x$
    and
    $\displaystyle y=60-6/5x$
    what's next? did i do something wrong?
    thanks
    yes, you did something wrong. in general, $\displaystyle \displaystyle \sqrt{a \pm b} \ne \sqrt a \pm \sqrt b$, yet that is what you did with the right side of the equation when you took the square root. Also note that there should be a $\displaystyle \displaystyle \pm$ sign on the right hand side. since if $\displaystyle \displaystyle x^2 = a$, then $\displaystyle \displaystyle x = \pm \sqrt a$


    EDIT: the "first and second quadrants" part of the problem means you should only consider the positive square root.
    Last edited by Jhevon; Oct 9th 2010 at 06:01 PM.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kimjb View Post
    Hi. Can you tell me what to do next.
    I got that dy/dx=-72x/50y

    Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

    36x2 + 25y2 = 900

    I think next it would be
    $\displaystyle 25y^2=900-36x^2$
    then square root everything and
    $\displaystyle 5y=300-6x$
    and
    $\displaystyle y=60-6/5x$
    what's next? did i do something wrong?
    thanks
    $\displaystyle \sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$!!!

    So what you have to do from $\displaystyle 25y^2=900-36x^2$ is take square roots of both sides to get

    $\displaystyle 5y=\sqrt{900-36x^2}\implies 5y=6\sqrt{25-x^2}\implies y=\frac{6}{5}\sqrt{25-x^2}$.

    Thus, your $\displaystyle \dfrac{dy}{dx}=-\dfrac{72x}{50\left(\frac{6}{5}\sqrt{25-x^2}\right)}=-\dfrac{72x}{60\sqrt{25-x^2}}=-\dfrac{6x}{5\sqrt{25-x^2}}$ which should be the same as differentiating $\displaystyle y=\frac{6}{5}\sqrt{25-x^2}$ wrt x.

    Does this make sense?
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  4. #4
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    Quote Originally Posted by kimjb View Post
    Hi. Can you tell me what to do next.
    I got that dy/dx=-72x/50y

    Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

    36x2 + 25y2 = 900

    I think next it would be
    $\displaystyle 25y^2=900-36x^2$
    then square root everything and
    $\displaystyle 5y=300-6x$ *
    and
    $\displaystyle y=60-6/5x$
    what's next? did i do something wrong?
    thanks
    The DE is seperable:

    $\displaystyle \displaystyle 50 \int y \, dy = -72 \int x \, dx \Rightarrow 25 y^2 = -36 x^2 + C$. You obviously know something we don't to get that C = 900.

    By the way, according to the line with * next to it, you seem to think that if $\displaystyle A^2 - B^2 = C^2$ then $\displaystyle A - B = C$. An instructor would expect better from a student studying differential equations ....
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  5. #5
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    Thanks that did make a lot more sense Chris.
    Mr fantastic, I'm actually in a calculus class and I put this in the calculus forum but it got moved, but I do understand about the line with the star. I'm not sure why I put that- that's kind of embarassing.
    Thanks for helping
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