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Thread: A study of a 2 variables function

  1. #1
    MHF Contributor arbolis's Avatar
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    A study of a 2 variables function

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

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

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

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

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

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

    ------------------------------------------------------
    I'd like to check my results and for part 5) and 6) a little push in the right direction.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    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.
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by arbolis View Post
    Let $\displaystyle f(x,y)=\frac{xy}{x^2-y^2}$.
    1)Determine the domain of $\displaystyle f$.
    2)Tell whether $\displaystyle \lim _{(x,y) \to (0,0)} f(x,y)$ exists or not.
    [snip]
    Note that if you take the limit along the line $\displaystyle y = mx ~ (m \neq \pm 1)$ the result is $\displaystyle \frac{m}{1 - m^2}$ ....
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Note that if you take the limit along the line $\displaystyle y = mx ~ (m \neq \pm 1)$ the result is $\displaystyle \frac{m}{1 - m^2}$ ....
    Yes ahah, as in this thread : http://www.mathhelpforum.com/math-he...-function.html.
    That makes the limit not to exist.
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