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Math Help - find dy/dx

  1. #1
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    find dy/dx

    consider the cardioid r=1+sin(t) remember that x=rcos(t) and y=rsin(t)

    which means x = (1+sin(t))cos(t)
    and y = (1+sin(t))sin(t)

    now dy/dx : [ (dy/dt) // (dx/dt) ]

    should i use chain rule on the x and y, or would i just distrube the cos and the sin? thanks
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I think that it's easier to use the chain rule; less work, faster result.
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  3. #3
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    I got the dy/dx to be : (cos(t)*(2sin(t)+1)) / - (2sin(t) - 1)*(sin(t)+1) can anyone confirm this for me? TY
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Hmm, okay...

    I'm applying the chain rule.

    \frac{dx}{dt} = (1+sin(t))(-sin(t)) + cos(t).cos(t)

    \frac{dy}{dt} = (1+sin(t))(cos(t)) + sin(t).cos(t)

    Then,

    \frac{dy}{dx} = \frac{(1+sin(t))(cos(t)) + sin(t).cos(t)}{(1+sin(t))(-sin(t)) + cos(t).cos(t)}

     = \frac{cos(t)(1 + 2sin(t))}{-(sin(t) + 1)(2sin(t)-1)}

    Yes, it's correct.
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  5. #5
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    Hello, action812!

    \text}Consider the cardioid: }\;r\:=\:1+\sin\theta

    \text{Find }\:\dfrac{dy}{dx}

    I would derive the formula first . . .


    \begin{array}{ccccccccc}x &=& r\cos\theta & \Rightarrow & \dfrac{dx}{d\theta} &=& \text{-}r\sin\theta + r'\cos\theta \\ \\[-4mm]<br />
y &=& r\sin\theta & \Rightarrow & \dfrac{dy}{d\theta} &=& r\cos\theta + r'\sin\theta \end{array}

    \text{Hence: }\;\dfrac{dy}{dx} \;=\;\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}  } \;=\;\dfrac{r\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta} . [1]


    We have: . \begin{Bmatrix}r \:=\:1+\sin\theta \\ r' \:=\:\cos\theta \end{Bmatrix}


    Substitute into [1]:

    . . \dfrac{dy}{dx} \;=\;\dfrac{(1+\sin\theta)\cos\theta + \cos\theta\sin\theta}{\text{-}(1+\sin\theta)\sin\theta + \cos\theta\cos\theta}

    . . . . . =\;\dfrac{\cos\theta + \sin\theta\cos\theta + \sin\theta\cos\theta}{\text{-}\sin\theta - \sin^2\!\theta + \cos^2\!\theta}

    . . . . . =\;\dfrac{2\sin\theta\cos\theta + \cos\theta}{\cos^2\!\theta - \sin^2\!\theta - \sin\theta}

    . . . . . =\;\dfrac{\sin2\theta + \cos\theta}{\cos2\theta - \sin\theta}
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  6. #6
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    thanks guys!!
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