Estimating (0.97)^3(2.02) using a linear approximation

[_N3WTON_]

Homework Statement

Use an appropriate linear approximation to estimate the value (0.97)^{3}(2.02). Write your answer as a simplified decimal.

Homework Equations

Linearization:
L(x,y) = f(a,b) + \frac{\partial z}{\partial x}(a,b)(x-a) + \frac{\partial z}{\partial y}(a,b)(y-b)

The Attempt at a Solution

I am a little confused on this one about how to get going, but once I get started this shouldn't pose to much of a problem. I was thinking (I came here to confirm this thought) that I could define a function:
z = x^{3}y
Then I would take:
a = 1 \hspace{2 mm} b = 2
Then I would go from there and formulate the linearization. I just wanted to see if this was the right way to proceed or not..

 

[Mark44]

Homework Statement

Use an appropriate linear approximation to estimate the value (0.97)^{3}(2.02). Write your answer as a simplified decimal.

Homework Equations

Linearization:
L(x,y) = f(a,b) + \frac{\partial x}{\partial z}(a,b)(x-a) + \frac{\partial y}{\partial z}(a,b)(y-b)

The Attempt at a Solution

I am a little confused on this one about how to get going, but once I get started this shouldn't pose to much of a problem. I was thinking (I came here to confirm this thought) that I could define a function:
z = x^{3}y
Then I would take:
a = 1 \hspace{2 mm} b = 2
Then I would go from there and formulate the linearization. I just wanted to see if this was the right way to proceed or not..

Looks like you're mostly on the right track, but your partials aren't right.

Since z = f(x, y), the partials you want are $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. It might be that you wrote them upside down inadvertently.

 

[Ray Vickson]

Homework Statement

Use an appropriate linear approximation to estimate the value (0.97)^{3}(2.02). Write your answer as a simplified decimal.

Homework Equations

Linearization:
L(x,y) = f(a,b) + \frac{\partial x}{\partial z}(a,b)(x-a) + \frac{\partial y}{\partial z}(a,b)(y-b)

The Attempt at a Solution

I am a little confused on this one about how to get going, but once I get started this shouldn't pose to much of a problem. I was thinking (I came here to confirm this thought) that I could define a function:
z = x^{3}y
Then I would take:
a = 1 \hspace{2 mm} b = 2
Then I would go from there and formulate the linearization. I just wanted to see if this was the right way to proceed or not..

Best policy: just go ahead and do it.

 

[_N3WTON_]

Looks like you're mostly on the right track, but your partials aren't right.

Since z = f(x, y), the partials you want are $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. It might be that you wrote them upside down inadvertently.

Ah, thank you...I just wrote it down wrong by accident :)