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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