# Linear Approximation

1. Oct 13, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
Find the linear approximation of (xy)/z at the point (-3,2,1)

3. The attempt at a solution
So the example my book gives has 2 variables so i'm struggling a bit with this, But I started off by taking the partial derivative with respect to each variable and solving for it.

d/dx = (y/z) = 2 = A
d/dy = (x/z) = -3 = B
d/dz = -(xy)/z^2 =6 = C

now... find the equation of a plane passing through that point maybe?

A(x-xo)+B(y-yo)+C(z-zo) = 0
2(x+3) - 3(y-2) + 6(z-1) = 0
2x+6 - 3y+6 +6z-6 = 0
2x-3y+6z+6=0

am I on the right track here..?

2. Oct 13, 2013

### arildno

Good try, but not quite. Here, you are NOT asked about the equation for some tangent plane, but how the linearized function Lf looks like, with f=xy/z

That is essentially to determine the first 4 terms of the Taylor series about (-3,2,1), that is:
$$Lf=f(-3,2,1)+\frac{\partial{f}}{\partial{x}}(x-(-3))+ \frac{\partial{f}}{\partial{y}}(y-2) + \frac{\partial{f}}{\partial{z}}(z-3)$$
where the partial derivatives are evaluated at (-3,2,1)

3. Oct 13, 2013

### D H

Staff Emeritus
arildno: That last term should be $\frac{\partial f}{\partial z}(z-1)$. (You had (z-3) instead of (z-1)).

4. Oct 13, 2013

### D H

Staff Emeritus
Close, but no.

Look at the simpler problem finding a linear approximation to f(x) at some point x0. What you did is the equivalent of saying A=df/dx at x=x0, and thus A(x-x0) = 0. That's obviously *not* a linear approximation of f(x). What one needs to do is to replace that 0 on the right hand side with fapprox(x)-f(x0), yielding fapprox(x)=f(x0)+A(x-x0).

You need to do the same with your function of three variables.

5. Oct 13, 2013

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