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Math Help - derivative

  1. #1
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    Red face derivative

    If y = e^t cost, x=e^t sint, and y''(x+y)^2 = K(xy'-y)
    Find K
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  2. #2
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    y = e^t \cos{t}, x=e^t \sin{t}

    y''(x+y)^2 = K(xy'-y)

    Well, y = e^t \cos{t}, so y' = e^t \cos{t} - e^t \sin{t},
    so y'' = e^t \cos{t} - e^t \sin{t} - e^t \sin{t} - e^t \cos{t} = -2e^t \sin{t}

    (x + y)^2 = e^{2t} (\cos^2{t} + \sin^2{t}) + 2e^t \sin{t} \cos{t} = e^{2t} + 2e^t \sin{t} \cos{t}

    xy' = e^t( \cos{t} - \sin{t}) \times e^t \sin{t} = e^{2t} \sin{t} (\cos{t} - \sin{t})

    y''(x+y)^2 = K(xy'-y)

    \Rightarrow K = \frac{y''(x + y)^2}{xy' - y}

    \Rightarrow K = \frac{-2e^t \sin{t} \times (e^{2t} + 2e^t \sin{t} \cos{t})}{e^{2t} \sin{t} (\cos{t} - \sin{t}) - e^t \cos{t}}

    \Rightarrow K = \frac{-2e^t \sin{t} \times (e^t + 2 \sin{t} \cos{t})}{e^t \sin{t} (\cos{t} - \sin{t}) - \cos{t}}

    Hmmm I can't see this going where we want.....
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  3. #3
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    Quote Originally Posted by nzmathman View Post
    y = e^t \cos{t}, x=e^t \sin{t}

    y''(x+y)^2 = K(xy'-y)

    Well, y = e^t \cos{t}, so y' = e^t \cos{t} - e^t \sin{t},
    so y'' = e^t \cos{t} - e^t \sin{t} - e^t \sin{t} - e^t \cos{t} = -2e^t \sin{t}

    (x + y)^2 = e^{2t} (\cos^2{t} + \sin^2{t}) + 2e^t \sin{t} \cos{t} = e^{2t} + 2e^t \sin{t} \cos{t}

    xy' = e^t( \cos{t} - \sin{t}) \times e^t \sin{t} = e^{2t} \sin{t} (\cos{t} - \sin{t})

    y''(x+y)^2 = K(xy'-y)

    \Rightarrow K = \frac{y''(x + y)^2}{xy' - y}

    \Rightarrow K = \frac{-2e^t \sin{t} \times (e^{2t} + 2e^t \sin{t} \cos{t})}{e^{2t} \sin{t} (\cos{t} - \sin{t}) - e^t \cos{t}}

    \Rightarrow K = \frac{-2e^t \sin{t} \times (e^t + 2 \sin{t} \cos{t})}{e^t \sin{t} (\cos{t} - \sin{t}) - \cos{t}}

    Hmmm I can't see this going where we want.....
    That's because in this case, y isn't a function of x, it's a function of t.

    Remember that y' = \frac{dy}{dx}.


    You need to use the chain rule y' = \frac{dy}{dt}\times\frac{dt}{dx}.

    Use implicit differentiation to find y''.
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  4. #4
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    Ah yes!

    I would have another go, but it is getting late and I'm getting tired...
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  5. #5
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    Quote Originally Posted by varunnayudu View Post
    If y=e^t cost, x=e^t sint, and y''(x+y)^2 = K(xy' - y).Find K
    how about posting your initial calculations?
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  6. #6
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    Sure....

    i have done this .....

    y' = \frac{dy}{dt} = e^t (-sint)

    therefore

    y'' = \frac{d^2 y }{dt} = e^t (-cost)
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  7. #7
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    Quote Originally Posted by varunnayudu View Post
    i have done this .....

    y' = \frac{dy}{dt} = e^t (-sint)

    therefore

    y'' = \frac{d^2 y }{dt} = e^t (-cost)
    First you should carefully review the product rule for differentiating a product of two functions. Then read post #3 more carefully.
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  8. #8
    MHF Contributor kalagota's Avatar
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    well, take a look on what the previous posters replied and also the comment.. you should be able to get the answer..
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