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Thread: Differentials...2

  1. #1
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    Differentials...2

    Evaluate f(1, 2) and f(1.05, 2.1) and calculate Δz. Also, use the total differential dz to approximate Δz.

    f(x, y) = x siny

    z_x = siny

    z_x = xcosy

    dz = sinydx + xcosydy

    To find f(1, 2) and f(1.05, 2.1), I evaluate the given function at the two points.

    I am having trouble finding delta z and then using dz to approximate delta z.

    The answer for delta z is given to be -0.00923 and the numerical answer for dz is given to be 0.00385.

    How is this done?
    Last edited by USNAVY; Jan 28th 2017 at 07:16 PM.
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  2. #2
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    Re: Differentials...2

    The value for \Delta z is 1.05\sin{2.1} - 1\sin{2} = -0.00293. This should be approximately z_x\Delta x + z_y\Delta y = (\sin{2})(1.05-1) + (1\cos{2})(2.1-2) = 0.00385.

    This is an unusual example - usually the estimate is pretty good if \Delta x and \Delta y are "small".

    - Hollywood
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  3. #3
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    Re: Differentials...2

    Quote Originally Posted by hollywood View Post
    The value for \Delta z is 1.05\sin{2.1} - 1\sin{2} = -0.00293. This should be approximately z_x\Delta x + z_y\Delta y = (\sin{2})(1.05-1) + (1\cos{2})(2.1-2) = 0.00385.

    This is an unusual example - usually the estimate is pretty good if \Delta x and \Delta y are "small".

    - Hollywood
    Thanks for the input. I solved the problem after several tries.
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  4. #4
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    Re: Differentials...2

    f(x,y) = xsiny

    Let z = f(x,y)

    dz = sinydx + xcosydy

    f(1,2) = 1sin(2) = 0.90930

    f(1.05,2.1) = (1.05)sin(2.1) = 0.906370

    Δx = 0.05

    Δy = 0.1

    Δz = (1.05)*sin(2.1) - sin(2)

    Δz = -0.00293
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