3
$\begingroup$

I really don't have idea if I am correct, but I would like to check if what is making sense to me, is indeed correct.

If I have a surface in $R^3$ that corresponds to the equation below: $$ z = 6 - x - x² - 2y² $$ enter image description here When I'm asked to implicit differentiate that, for example, in terms of $\frac{\partial z}{\partial y}$, why should I treat $x$ as a constant?!?!

The reason for why I should do that, in my head, is because we just want to check what a tiny change in $z$ is going to change our $y$, so it's just like taking a derivative of a curve in a plane that intersects my surface, and that plane is described by a fixed $x$ value.

enter image description here Because we're fixing $x$ to get that intersection, we treat it as a constant?! Am I right?

Since: $$ \frac{\partial z}{\partial y} = -4y $$ Can assume that $-4y$ is going to be the slope of the tangent line for any point of that intersection?!

For example, the point $\alpha = (0,1,4)$ is on the plane $x = 0$, and satisfies the equation for the surface. Now, if want to get the tangent line on that point, I just need to get another point that is going to follow the slope of $-4$ because $-4y : y = 1 \Rightarrow -4$, and is on the plane. That point can be, for example $\gamma = (0,2,0)$. Now I can get a director vetor $\vec{t}$, and get a parametric equation for that tangent line: $$ \vec{t} = \alpha - \gamma = (0,-1,4) $$

$$ x = 0 + \beta \cdot 0\\ y = 1 + \beta \cdot -1\\ z = 4 + \beta \cdot 4\\ $$

enter image description here

Is everything correct? If it isn't, please try to explain what have I done wrong. Thanks!

$\endgroup$
4
  • $\begingroup$ What graphing software are you using? $\endgroup$
    – GFauxPas
    Sep 23, 2016 at 19:37
  • $\begingroup$ I'm using geogebra! why? $\endgroup$
    – Bruno Reis
    Sep 23, 2016 at 19:40
  • $\begingroup$ just to see if there's something better than Geogebra I should know about :P $\endgroup$
    – GFauxPas
    Sep 23, 2016 at 19:41
  • $\begingroup$ Geogebra is sick hahaha! $\endgroup$
    – Bruno Reis
    Sep 23, 2016 at 19:42

1 Answer 1

0
$\begingroup$

By definition;

$$\frac{\partial f}{\partial x}(a,b) = \lim_{h \to 0}\ \frac{f(a+h,b) - f(a,b)}{h}$$

Therefore if you wish to compute $z_x$ then $z = z(x,y)$ and so;

$$\frac{\partial z}{\partial x}(a,b) = \frac{d}{dx} z(x,b) \Bigr|_{x=a}$$

Hence; the partial derivative in the direction of $y$ at $(a,b)$ is the same as differentiating $z(x,b)$ i.e holding $y$ as a constant.

$\endgroup$
2
  • $\begingroup$ Thanks for answering. So I am correct? $\endgroup$
    – Bruno Reis
    Sep 23, 2016 at 21:12
  • $\begingroup$ This was only to answer your question about why can you hold the variable as a constant. I haven't read the other stuff, but does this make sense now. $\endgroup$ Sep 23, 2016 at 21:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .