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Math Help - Differentiate and find the square of dy/dx

  1. #1
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    Differentiate and find the square of dy/dx

    Question:

    If x = cos \theta + \theta sin \theta ,  y = sin \theta - \theta cos \theta , then show that

    \frac{dy}{dx} = tan \theta

    Also find  \frac{d^2 y}{dx^2}
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question:

    If x = cos \theta + \theta sin \theta ,  y = sin \theta - \theta cos \theta , then show that

    \frac{dy}{dx} = tan \theta

    Also find  \frac{d^2 y}{dx^2}
    \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}} (which I'm sure will be somewhere in yur class notes or textbook).

    \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{d \theta} \left(\frac{dy}{dx}\right) \cdot \frac{d \theta}{dx}

    If you need more help, please show all your working and state specifically where you are stuck.
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  3. #3
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    "yur"?
    Last edited by mr fantastic; December 5th 2009 at 12:00 PM. Reason: No edit - just commenting (without going off-topic) that I only have a limited number of o's ....
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  4. #4
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    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}} (which I'm sure will be somewhere in yur class notes or textbook).

    \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{d \theta} \left(\frac{dy}{dx}\right) \cdot \frac{d \theta}{dx}

    If you need more help, please show all your working and state specifically where you are stuck.


    I am getting the \frac{d^2 y }{dx^2} = sec^2 \theta .\frac{1}{ \theta cos \theta}


    <br />
\frac{d}{d \theta} \left( \frac{dy}{dx} \right) = \frac{d}{d \theta} (tan \theta) = sec^2 \theta<br />

    <br />
\frac{d \theta}{dx} = \frac{1}{\frac{dx}{d \theta}} = \frac{1}{\theta cos \theta}<br />

    therefore

    <br />
\frac{d^2 y}{dx^2} = sec^2 \theta . \frac{1}{ \theta cos \theta}<br /> <br />

    Is this Right!!!
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    Quote Originally Posted by zorro View Post
    I am getting the \frac{d^2 y }{dx^2} = sec^2 \theta .\frac{1}{ \theta cos \theta}


    <br />
\frac{d}{d \theta} \left( \frac{dy}{dx} \right) = \frac{d}{d \theta} (tan \theta) = sec^2 \theta<br />

    <br />
\frac{d \theta}{dx} = \frac{1}{\frac{dx}{d \theta}} = \frac{1}{\theta cos \theta}<br />

    therefore

    <br />
\frac{d^2 y}{dx^2} = sec^2 \theta . \frac{1}{ \theta cos \theta}<br /> <br />

    Is this Right!!!
    Yes, and if you recall that \frac{1}{\cos \theta} = \sec \theta then you can simplify your answer a little bit.
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  6. #6
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    So after simplifying

    Quote Originally Posted by mr fantastic View Post
    Yes, and if you recall that \frac{1}{\cos \theta} = \sec \theta then you can simplify your answer a little bit.
    = sec^2 \theta . \frac{1}{\theta cos \theta}
    = \frac{sec^3 \theta}{ \theta}

    Is this right now???
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  7. #7
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    Quote Originally Posted by zorro View Post
    = sec^2 \theta . \frac{1}{\theta cos \theta}
    = \frac{sec^3 \theta}{ \theta}

    Is this right now???
    Correct.
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  8. #8
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    By the way, Zorro, \frac{d^2y}{dx^2} is NOT "the square of dy/dx"!
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  9. #9
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    Thanks u every one for helping me

    Thank u Mr fantastic for ur help
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