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Math Help - Consider the the cardioid

  1. #1
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    Consider the the cardioid

    Consider the the cardioid r=1+sin(θ)

    find dy/dx. I'm a little unsure where to start here, I'm guessing convert it which, I got x^2 + y^2= (1+sin(θ))^2, am I going in the right direction? Any guidance would be appreciated. Thanks
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  2. #2
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    OK so I did a little more work, and figure I should times both sides by r instead of squaring it.
    which gave me:

    r^2= r + rsin(t)

    x^2+y^2=r+y

    r= (x^2+y^2-y)

    r^2= (x^2+y^2-y)^2

    x^2+y^2= (x^2+y^2-y)^2


    But I'm still stuck on how to find the derivative of this? [dy/dx] = [dy/dt / dx/dt ]
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  3. #3
    MHF Contributor chisigma's Avatar
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    You are going in the right direction!...

    ... because is \theta = \tan^{-1} \frac{y}{x} , y can be expressed as function of x in implicit form as...

    \displaystyle f(x,y)= x^{2} + y^{2} - \{ 1 + \sin (\tan^{-1} \frac{y}{x})\} ^{2} = 0 (1)

    ... so that is...

    \displaystyle \frac{dy}{dx} = - \frac{f_{x} (x,y)}{f_{y} (x,y)} (2)

    Kind regards

    \chi \sigma
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  4. #4
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    in which one was I going in the right direction...lol, my first or second post?
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  5. #5
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    still stuck...any guidance would be appreciated. thanks
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