Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y$.

share|improve this question
    
Do you know how to take partial derivatives? –  Argon Jun 10 '12 at 16:15
4  
That "0% accept rate" is very distracting, I can't help it. –  Gigili Jun 10 '12 at 16:16
    
@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
    
@Gigili what does the 0% accept rate rate mean? sorry i'm not used to this site yet –  Ricky Rozay Jun 10 '12 at 16:24
    
@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

3 Answers 3

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}

share|improve this answer
    
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 $w=(0,1,4)^T$

share|improve this answer

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 :

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.