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Math Help - related rates with triangle

  1. #1
    Super Member bigwave's Avatar
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    related rates with triangle

    basically I don't understand the suggestion on how this was set up

    Let \theta (in radians) be an angle in a right triangle, and let x and y, respectively, be the lengths of the sides adjacent to and opposite \theta Suppose that x and y vary with time.

    (a) How are \frac{d\theta}{dt} , \frac{dx}{dt} and \frac{dy}{dt} Related

    the book says that
    \frac{d\theta}{dt}=\frac{cos^2\theta}{x^2}\left( x \frac{dy}{dt}-y\frac{dr}{dt}\right)

    thanks for help

    actually I don't know where the \frac{cos^2\theta}{x^2} comes from I assume the hypotenuse is a constant
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  2. #2
    Super Member TheChaz's Avatar
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    Re: related rates with triangle

    \frac{d\theta}{dt}=\frac{cos^2\theta}{x^2}\left( x \frac{dy}{dt}-y\frac{dr}{dt}\right)

    Start with

    \tan (\theta) = \frac{y}{x}

    Differentiate both sides.

    Quotient rule. Etc.

    (p.s. If x and y are changing, then the hypotenuse is NOT constant!)
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  3. #3
    Super Member bigwave's Avatar
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    Re: related rates with triangle

    i did try that but still couldn't get this... will try again.. guess we don't need to worry about hyp if tan is used

    don't know if you want to show how this was derived.
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  4. #4
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    Re: related rates with triangle

    Quote Originally Posted by bigwave View Post
    basically I don't understand the suggestion on how this was set up

    Let \theta (in radians) be an angle in a right triangle, and let x and y, respectively, be the lengths of the sides adjacent to and opposite \theta Suppose that x and y vary with time.

    (a) How are \frac{d\theta}{dt} , \frac{dx}{dt} and \frac{dy}{dt} Related

    the book says that
    \frac{d\theta}{dt}=\frac{cos^2\theta}{x^2}\left( x \frac{dy}{dt}-y\frac{dr}{dt}\right)

    thanks for help

    actually I don't know where the \frac{cos^2\theta}{x^2} comes from I assume the hypotenuse is a constant
    \frac{d}{dt} \left[\tan{\theta} = \frac{y}{x}\right]

    \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}}{x^2}

    \frac{1}{\cos^2{\theta}} \cdot \frac{d\theta}{dt} = \frac{1}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)

    \frac{d\theta}{dt} = \frac{\cos^2{\theta}}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)
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  5. #5
    Super Member TheChaz's Avatar
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    Re: related rates with triangle

    Quote Originally Posted by skeeter View Post
    \frac{d}{dt} \left[\tan{\theta} = \frac{y}{x}\right]

    \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}}{x^2}

    \frac{1}{\cos^2{\theta}} \cdot \frac{d\theta}{dt} = \frac{1}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)

    \frac{d\theta}{dt} = \frac{\cos^2{\theta}}{x^2} \left(x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}\right)
    Thanks, skeeter! I wasn't about to LaTex all that!
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  6. #6
    Super Member bigwave's Avatar
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    Re: related rates with triangle

    wow, thx, I was close but not close enough

    yep, latex is 80% of the process....
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  7. #7
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    Re: related rates with triangle

    Quote Originally Posted by bigwave View Post
    wow, thx, I was close but not close enough

    yep, latex is 80% of the process....
    90% in this case ...
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