# Approximate a function at given points

I have forgot the exact method how to approximate a function at given points and I need your help to recall it (sure I will show how I think about it, because my question would not be just only problem posting, I will show my effort).

Suppose that our function is given by following two variable form, $f(x,y)=\sqrt{x^3+y^3}$, and we want to approximate the value of this functon at the point $(2.02,1.97)$. As I remember I should use the total differential method (generally tangent line's geometrical interpretation is as best approximation of function at given points), so I used

$$Df(x,y) = \frac{3x^2}{\sqrt{x^3+y^3}}dx + \frac{3y^2}{\sqrt{x^3+y^3}}dy.$$

I took the exact value of $(x,y)$ as $x = 2$ and $y = 1$ and consequently $dx = .02$ and $dy = .97$. So if I put this variable I would get the required answer, right? Or is there another method?

Thanks a lot.

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Looks fine, but I'll choose $y = 2$ and $dy = -.03$ –  martini May 2 '12 at 20:16
i see thanks @martini –  dato datuashvili May 2 '12 at 20:20
then post it as answer if you would like,i will accept it –  dato datuashvili May 2 '12 at 20:30
There is a "2" missing in the denominators, as $\sqrt{x}' = \frac 1{2\sqrt x}$ –  martini May 2 '12 at 20:51
aaaaa i guessed,thanks a lot of –  dato datuashvili May 2 '12 at 21:02

As dato writes in his question we have $$df(x,y) = \frac{3x^2}{2\sqrt{x^3+y^3}}\,dx + \frac{3y^2}{2\sqrt{x^3+y^3}}\,dy$$ at $(x,y) = (2,2)$ we have $f(2,2) = \sqrt{8+8} = \sqrt{16}=4$ and $$df(2,2) = \frac{12}{8}\,dx + \frac{12}{8}\,dy = \frac 32\cdot (dx + dy)$$ So we have $$f(2.02, 1.97) \approx f(2,2) + df(2,2)[.02, -.03] = 4 + \frac 32 \cdot \left(-\frac 1{100}\right) = \frac{797}{200}.$$
$797/200=3.985$ compares well with the exact value $3.9859479424599\cdots$. –  lhf May 2 '12 at 22:15