Thread: A study of a 2 variables function

1. A study of a 2 variables function

Let $f(x,y)=\frac{xy}{x^2-y^2}$.
1)Determine the domain of $f$.
2)Tell whether $\lim _{(x,y) \to (0,0)} f(x,y)$ exists or not.
3)Calculate via the definition $\frac{\partial f}{\partial x}(0,1)$.
4)Determine the equation of the tangent plane of $g(x,y)=f(x,y)+2y$ in the point $(0,1,g(0,1))$.
5)Determine the equation of the tangent line to the level curve $g(x,y)=2$ in $(0,1)$.
6)Calculate the directional derivative of $g$ in the point $(0,1)$ in the direction $\left( \frac{\sqrt 3}{2},{\frac{1}{2}} \right)$.
My attempt :

1) $dom f =\{ (x,y)\in \mathbb{R}^2 : x\neq \pm y \}$.

2) $\lim _{(x,y) \to (0,0)} f(x,y)=\lim _{x \to 0} f(x,0)=0$.
$\lim _{(x,y) \to (0,0)} f(x,y)=\lim _{x \to 0} f(x,x^2)=-\infty$, thus the limit does not exist.

3) $\frac{\partial f}{\partial x}(0,1)= \lim _{h \to 0} \frac{f(x+h,y)-f(x,y)}{h}$ evaluated in $(0,1)$. This equals $\lim _{h \to 0} \left ( \frac{1}{h} \right ) \left( \frac{h}{h^2-1} \right )=-1$.

4) $g(x,y)=\frac{xy}{x^2-y^2}+2y$.
$(0,1,g(0,1))=(0,1,2)$.
$\nabla g(x,y,z)= \left ( \frac{y-2x^2y}{x^2-y^2} \right ) \hat i+ \left ( \frac{x(x^2-y^2)-(xy)(-2y)}{(x^2-y^2)^2}+2 \right) \hat j -1 \hat k$ $\Rightarrow \nabla g(0,1,2)=-\hat i - \hat k$.
Hence the equation of the tangent plane is $-x-(z-2)=0\Leftrightarrow x+z-2=0$.

5) I don't know how to start it.
6) Idem.

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I'd like to check my results and for part 5) and 6) a little push in the right direction.

2. 5. g(x,y) = f(x,y) + 2y

Let d denote the partial derivative

Since you are on a level curve dg = dg/dx dx + dg/dy dy = 0

You obtain then:

dy/dx = -dg/dx/[dg/dy] at the point (0,1)

Now you have essentially a calculus 1 problem of finding the tangent
line to an implicitly defined curve at the point (0,1).

y = mx +1 where m = dy/dx

6. Like any directional derivative use grad(g(0,1))*

You already have dg/dy and dg/dx from part 5.

3. Originally Posted by arbolis
Let $f(x,y)=\frac{xy}{x^2-y^2}$.
1)Determine the domain of $f$.
2)Tell whether $\lim _{(x,y) \to (0,0)} f(x,y)$ exists or not.
[snip]
Note that if you take the limit along the line $y = mx ~ (m \neq \pm 1)$ the result is $\frac{m}{1 - m^2}$ ....

4. Originally Posted by mr fantastic
Note that if you take the limit along the line $y = mx ~ (m \neq \pm 1)$ the result is $\frac{m}{1 - m^2}$ ....
Yes ahah, as in this thread : http://www.mathhelpforum.com/math-he...-function.html.
That makes the limit not to exist.