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Thread: Numerical Methods: Numerical Differentiation

  1. #1
    Jan 2009

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

    $\displaystyle f_x(x,y) =\frac{f(x+h,y) - f(x-h,y)}{2h}+O(h^2)
    $\displaystyle f_y(x,y) =\frac{f(x,y+h) - f(x,y-h)}{2h}+O(h^2)

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

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

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

    Solving for $\displaystyle f_x(x,y)$ using equation (i)
    h_____$\displaystyle f(2+h,3)$_$\displaystyle f(2-h,3)$_____2h______$\displaystyle f_x(2,3)$

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

    Solving for $\displaystyle f_y(x,y)$ using equation (i)
    h_____$\displaystyle f(2,3+h)$_$\displaystyle f(2,3-h)$_____2h______$\displaystyle f_y(2,3)$

    Am I doing this right??

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


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

    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:
    $\displaystyle V(10)=\frac{D(11.0)-D(9.0)}{11.0-9.0}=4.421$

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