Numerical Methods: Numerical Differentiation

**zeugma** · February 25th 2009, 02:20 AM

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). $D(t)=-70+7t+70e^(-t/10)=-70+7(10.0)+70e^\frac{-10.0}{10}=25.752$ .
$V(10)=\frac{D(10.0)}{10.0}=\frac{25.752}{10.0}=2.5 752$

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