Question

Let f(x,y)= 5/ (x^2 + y^2). Find an equation of a linear approximation to f(x,y) at (-1,2,1) and use it to estimate f(-1.05,2.1)

Answer 1

wouldnt a linear approx just be a plane?

Answer 2

in fact, the plane is worked up similar to that of pont slope formula; except its point slope slope formula :)

Answer 3

\[z-z_o=f_x(x-x_o)+f_y(y-y_o)\]

Answer 4

not sure @amistre64!

Answer 5

which one's x and whichone's x not etc.

Answer 6

\[f(x,y)=\frac{5}{x^2+y^2}\]
\[f_x=\frac{-5(2x)}{(x^2+y^2)^2}\]
\[f_y=\frac{-5(2y)}{(x^2+y^2)^2}\]

Answer 7

i'm sorry! can you possibly explain how you got fx and fy??

Answer 8

ok, let me post this and ill try

given f( x=-1.05,y=2.1) that gives us:

\[f_x=\frac{10(1.05)}{(1.05)^2+(2.1)^2}\]

\[f_y=\frac{-10(2.1)}{(1.05)^2+(2.1)^2}\]

Answer 9

do you know partial derivatives?

Answer 10

yes

Answer 11

fx means we treat y as a constant
\[\frac{5}{x^2+y^2}=5(x^2+c)^{-1}\]
and derive it

y is the same except for make x^2 = c

Answer 12

oh ok!

Answer 13

is that an OHHH ok? or one of those, i spose ... oks ?

Answer 14

in essense we ignore the variable that is not important in the partial and treat it as any other constant

Answer 15

they give us x0 and y0; to find z0 we solve f(x,y) for the given point and call f(x,y) = z

Answer 16

dat wuz an i kinda get it ok

Answer 17

if your confident with your usual derivative stuff; its just a little extra thinking to ignore that variable

Answer 18

f(x,y) = xy^2

fx = y^2 ; since Cx derives to C and C=y^2

Fy = 2xy since Cy^2 = 2Cy and C = x in that case

Answer 19

i get it!

Answer 20

lol, good :)

Answer 21

i knew this looked familiar

Answer 22

@amistre64 LOL Yup! i came in not that long ago, and saw the bump button available, so i pushed it! LOL

Answer 23

good thing it wasnt shiny and red :)

Answer 24

LOL:)