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Just wanted to ask a simple question I've forgotten how to solve (lost my mind completely).

Find the equation of the tangent plane to the surface $z=x^2 + y^2$ at $(1,2)$.

The fact it has $3$ variables is what is putting me off. I don't know whether to differentiate with respect to $x$ or $y$.

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Do you know how to take partial derivatives? – Argon Jun 10 '12 at 16:15
@Argon, I'd know how to do partial derivatives if it was f(x,y) = x$^2$+y$^2$ but that would give f xy = 0 – Ricky Rozay Jun 10 '12 at 16:23
@TundeBaba It means you have not accepted any answers before. By the way, shouldn't the point given (i.e. $(1, 2)$) have three numbers, e.g. $(1, 2, 3)$? – Argon Jun 10 '12 at 16:26
@TundeBaba: Please consider accepting some of the answers to your previous question if you haven't done it yet. – Gigili Jun 10 '12 at 16:27
@Gigili i have done now – Ricky Rozay Jun 10 '12 at 16:42
up vote 1 down vote accepted

This plane has normal $(-\dfrac{\partial z}{\partial x}(1,2),-\dfrac{\partial z}{\partial y}(1,2),1) = (-2,-4,1)$.

Then the plane has the form \begin{equation} -2x -4y + z + d = 0. \end{equation} And the plane pass by (1,2,5), hence the plane is \begin{equation} -2x -4y + z + 1 = 0. \end{equation}

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Your normal made me check my tangents and revealed an error. Thanks. – user20266 Jun 10 '12 at 16:36
the answer should be z=-5+2x+4y which isn't the same as yours – Ricky Rozay Jun 10 '12 at 16:45
This plane do not contain the point $(1,2,1^{2} + 2^{2})$. How can it be the tangent at (1,2)? – user29999 Jun 10 '12 at 17:21
You got your $d$ wrong, though. – user20266 Jun 11 '12 at 19:32

Let $M$ denote the surface described by your equation. One way of writing down that plane is to find a point in the plane (you can take the one you are given, $p=(1,2, 5)$) and calculate two tangent vector $v, w$ and then let the tangent plane to the surface M in $p$ $$T_pM = \{q\in \mathbb{R}^3: q = p + tv +sw; s, t \in \mathbb{R}\}$$ It remains to find $v, w$. These you can take as the derivative in $t=0$ of a curve in $M$ through $p$ like, in your case, $(c(t), 2, c^2(t)+4)^T$ and $(1, \bar{c}(t), \bar{c}^2(t)+1)^T$ such that $c(0)=1, \bar{c(0)}= 2$. You may take $c(t) = 1 +t $ and $\bar{c}(t)=2+t$. This results in, e.g., $v=(1, 0, 1)^T$ and $w=(0,1,4)^T$

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Let $z:\mathbb{R}^2\rightarrow \mathbb{R}$ be differentiable at $(x_{0},y_{0})$. The plane in $\mathbb{R}^3$ defined by the equation $$T_{p}=z(x_{0},y_{0})+\left[\frac{\partial z}{\partial x}(x_{0},y_{0})\right](x-x_{0})+\left[\frac{\partial z}{\partial y}(x_{0},y_{0})\right](y-y_{0})$$ is called the tangent plane of the graph of $z$ at the point $(x_{0}, y_{0})$.

This is the definition, now we have to find the tangent plane for $z=x^2+y^2$ at the $(1,2)$.

$$z(1,2)=5;$$ $$\left[\frac{\partial z}{\partial x}(1,2)\right](x-1)=2(x-1)=2x-2 ;$$ $$\left[\frac{\partial z}{\partial y}(1,2)\right](y-2)=4(y-2)=4y-8 .$$

So, the equation for tangent plan at point $(1,2)$ is :

$$T_{p}=5+2x-2+4y-8=2x+4y-5. $$

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