Tangent line and partial derivatives

Question

Consider the funcion $f(x, y)=\displaystyle\frac{2xy^2}{x^2+y^4}$ for $(x, y)\neq(0,0)$ and $f(0, 0)=0$ and the curve $\gamma(t) =(t, t, z(t))$, $t\in\mathbb{R}$, whose image is contained in the graph of $f$. Prove that the Tangent line $T$ in $\gamma(0)$ is not contained in the plan given by $$z-f(0,0)=\frac{\partial f}{\partial x}(0,0)(x-0)+ \frac{\partial f}{\partial y}(0,0)(y-0)$$

In this case I have

$$\frac{\partial f}{\partial x}(0,0)=\displaystyle\lim_{x\rightarrow 0}= \frac{\dfrac{2x0^2}{x^2+0^4}-0}{x-0}=0$$

and

$$\frac{\partial f}{\partial y}(0,0)=\displaystyle\lim_{y\rightarrow 0}= \frac{\dfrac{2\cdot 0y^2}{0^2+y^4}-0}{y-0}=0$$

Moreover $\gamma(0)$ belongs to the plane and

What to do next?

Could anyone please help me

Can u please write the question as it is in the book? Bcoz right now, it is not clear what $\gamma$ is. — Koro, Jun 22 '20 at 19:52
I have updated my answer incorporating calculation of $\gamma'(0)$ — Koro, Jun 22 '20 at 20:55
It's wrong. Please note that you can't replace $z(t) $ (in the definition of $\gamma(t) $) by $2t/(1+t^2)$ as that holds only when $t\ne 0$. Also, once your question has been answered, ideally you should not change your question. However, if you do want to change, write edit (or update) in the end of your post and then put those changes there. — Koro, Jun 22 '20 at 21:18

score 0 · Answer 1 · answered Jun 22 '20 at 19:03

0

In case you take the partial derivative of $(x,y,f(x,y))$ in respect to $x$ you'll get $(1,0,f_x(x,y))$, taking the derivative in respect to $y$ you'll get $(0,1,f_y(x,y))$. The (negative) cross product of these vectors is $(f_x(x,y),f_y(x,y),-1)$, which is normal to the surface, hence a normal of the tangent plane.

answered Jun 22 '20 at 19:03

Michael Hoppe

18,103
3
32
49

Ok, but related to the main question (the line is not contained in the plane), I can't write $(1,1,0)\cdot (0,0,-1)=0$. Because this would mean that $T$ belongs to the plan. Where am I failing? – Marcos Paulo Jun 22 '20 at 19:20

Koro · Accepted Answer · 2020-06-22T20:54:14.183

$f(x, y)=\begin{cases}\displaystyle\frac{2xy^2}{x^2+y^4}; (x,y)\ne (0,0)\\ 0; (x,y)=(0,0) \end {cases}$,

is actually the level surface of $g(x,y,z)=f(x,y)-z $.
Now partial derivatives, $\frac{\partial f (0,0)}{\partial x}=0=\frac{\partial f (0,0)}{\partial y}$ as you have rightly shown.
Tangent plane to $g(x,y,z)=0$ at $(0,0,f(0,0))$ is given by:
$\nabla g(0,0,0).(x-0,y-0,z-f(0,0))=0$, where $.$ is dot product and $\nabla $ is gradient of $g$ , which is equal to $(f_x (0,0), f_y(0,0), -1)$ at $(0,0,f(0,0))$
Simplifying the above to get: $z-f(0,0)=\frac{\partial f}{\partial x}(0,0)(x-0)+ \frac{\partial f}{\partial y}(0,0)(y-0)$
$\implies z=0 \tag{1}$
If tangent to $\gamma (0)$ is contained in $(1)$, then we must have $\gamma '(0)\perp \text{Normal to plane in (1)}$, that is we must have $\gamma'(0). (0,0,1) =0\tag {2}$
Now we have, $\gamma'(0)=(1,1,2)$ and clearly this doesn't satisfy $(2)$ and hence tangent to $\gamma (0)$ can't be contained in $(1)$

Response to comment: Why is $\gamma'(0)=(1,1,2)$?
$\gamma (t)=(t,t,z(t))\implies \gamma'(t)=(1,1,z'(t))$
$f(t,t)=\frac{2t}{1+t^2}=z(t)$, where $t\ne 0$ and $z(0)=0$ and hence $z'(0)=\lim_{t\to 0}\frac{z(t)-z(0)}{t}=\lim_{t\to 0} \frac{2}{1+t^2}=2$

Tangent line and partial derivatives

2 Answers2