Linear Approximation of (xy)/z at (-3,2,1)

[PsychonautQQ]

Homework Statement

Find the linear approximation of (xy)/z at the point (-3,2,1)

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..?

 

[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)

 

[D H]

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

 

[D H]

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..?

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 f_approx(x)-f(x₀), yielding f_approx(x)=f(x₀)+A(x-x₀).

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

 

[arildno]

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

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