1. ## Linear approximation

Last one, guys...

"Given the function f(x,y)=x^2cos(xy)-5xy, estimate the value of f(3,97,0.025)."

My work:

fx(x,y) = 2x(cos(xy))-y(x^2sin(xy)+5)
fy(x,y) = -x(x^2sin(xy)+5)

So, f(x,y) + fx(x,y)(x-a) + fy(x,y)(y-b)

f(3.97,.025) + fx(3.97,.025)(3.97-3.95) + fy(x,y)(.025-.020)

And, after plugging into this equation, my final answer was 15.32.

Did I mess up anywhere? I'm pretty confident about the answer, since (3.97,.025) plugged into the original equation is 15.26, but I just want to make sure I'm not missing anything.

2. You confused me with having 3 coordinates in f(x,y).
I would use a=4 and b=0.

you want $\displaystyle f(x,y)\approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$

3. Originally Posted by matheagle
You confused me with having 3 coordinates in f(x,y).
I would use a=4 and b=0.

you want $\displaystyle f(x,y)\approx f(a,b) + f_x(x,y)(x-a) + f_y(x,y)(y-b)$
Or may be:

$\displaystyle f(x,y)\approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$

(the whole point is to evaluate the function and its partials at easy values)

CB