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Math Help - Numerical Methods: Numerical Differentiation

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    Numerical Methods: Numerical Differentiation

    Hi everyone. I have a homework on numerical differentiation and I just want to know if my answers are correct.

    1.) The partial derivative, f_x(x,y) of f(x,y) with respect to x is obtained by holding y fixed and differentiating with respect to x. Similarly, f_y(x,y) is found by holding x fixed and differentiating with respect to y. The equation below is adapted to partial derivatives:

    f_x(x,y) =\frac{f(x+h,y) - f(x-h,y)}{2h}+O(h^2)<br />
    f_y(x,y) =\frac{f(x,y+h) - f(x,y-h)}{2h}+O(h^2)<br />

    [The two equations above are denoted as equation (i)]

    (a). Let f(x,y)=\frac{xy}{(x+y)}. Calculate the approximations to f_x(2,3) and f_y(2,3) using the formulas in (i) with h = 0.1, 0.01, and 0.001. Compare with the values obtained by differentiating f(x,y).

    MY ANSWERS:
    First, I solved for the derivative of f(x,y) with respect to x. I got:
    f_x(x,y)=\frac{y(x+y)-xy^2}{(x+y)^2}=\frac{3(2+3)-2(3)^2}{(2+3)^2}=-0.12

    Solving for f_x(x,y) using equation (i)
    h_____ f(2+h,3)_ f(2-h,3)_____2h______ f_x(2,3)
    0.1_____1.23529____1.16327____0.20000___0.36014
    0.01____1.20359____1.19639____0.20000___0.36000
    0.001___1.20036____1.19964____0.20000___0.36000



    Solving for f_y(x,y)
    Derivative with respect to y: f_y(x,y)=\frac{x(x+y)-yx^2}{(x+y)^2}=\frac{2(2+3)-3(2)^2}{(2+3)^2}=-0.08

    Solving for f_y(x,y) using equation (i)
    h_____ f(2,3+h)_ f(2,3-h)_____2h______ f_y(2,3)
    0.1_____1.21569____1.18367____0.20000___0.16006
    0.01____1.20160____1.19840____0.20000___0.16000
    0.001___1.20016____1.19984____0.20000___0.16000

    Am I doing this right??


    ================================================== =======
    2.) The distance D = D(t) traveled by an object is given in the table below:

    _t_____D(t)
    8.0___17.453
    9.0___21.460
    10.0__25.752
    11.0__30.301
    12.0__35.084

    (a) Find the velocity V(10) by numerical differentiation
    (b) Compare your answer with D(t)=-70+7t+70e^\frac{-t}{10}.

    My Answers:
    (a). Is it okay if I use central-difference to solve for the velocity? Or should I use forward of backward difference? If so, should I assume different step values (h)? Using central-difference, my answer is:
    V(10)=\frac{D(11.0)-D(9.0)}{11.0-9.0}=4.421

    (b). D(t)=-70+7t+70e^(-t/10)=-70+7(10.0)+70e^\frac{-10.0}{10}=25.752.
    V(10)=\frac{D(10.0)}{10.0}=\frac{25.752}{10.0}=2.5  752
    Last edited by zeugma; February 25th 2009 at 06:00 AM.
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