Linearisation of a Function: How to Compute the Tangent Plane at a Given Point?

[danj303]

Homework Statement

Compute the linearisation of z = x^\alphay^\beta about (1,1) if \alpha & \beta \neq 0.

Homework Equations

The Attempt at a Solution

I can see how it works when \Deltax and/or \Deltay are given but not sure how to do it in this form??

 

[Quinzio]

They ask you a thing like:

\Delta z = k_x \Delta x + k_y \Delta y

What k_x, k_y might then be ?

 

[danj303]

Thanks for the reply but sorry I don't follow you there. That is the entire question as I have written. I can understand how it works when for example you are given a parabola and have to calculate the change in height going a certain distance to one side. But in that case delta x & delta y are given.

Not sure how to go about it in this case.

 

[Quinzio]

It's pretty much the same thing, just that you have 2 variable instead of 1.
How would you proceed with the parabola ?

 

[danj303]

By using a linear approximation to estimate delta z so that

\Deltaz = f(x₀ +delta x , y₀ + delta y) - f(x₀,y₀)

But in this case I don't know delta x and y??

 

[danj303]

Think I got it.

It just goes to [\alpha, \beta]

 

[Quinzio]

Yep.

 

[jambaugh]

Side note, you can also do this implicitly.

Your surface is:
f(x,y,z)= z - x^\alpha y^\beta = 0
Let the vector:
\Delta \vec{r}= \langle \Delta x, \Delta y, \Delta z \rangle
be your local linearized variables:
and the tangent plane in these local variables will be:
\nabla f(1,1,1) \bullet \Delta\vec{r} = 0