What is the linear approximation for estimating f(x,y)?

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Homework Statement

Use the linear approximation to approximate a suitable function f(x,y) and thereby estimate the following

f(x,y) = \sqrt{(4.01)^2 + (3.98)^2 + (2.02)^2}

Homework Equations

Not going to type it out, but the formula for f(x,y) http://en.wikipedia.org/wiki/Linear_approximation

The Attempt at a Solution

I just need to find the equation first, I can do the estimation.

My guess is, let x = 4, and y = 2 then

f(x,y) = \sqrt{(x)^2 + (x)^2 + (y)^2}
=(2(x)² + (y)²)^1/2

Is that done correctly? Proceeding this would be to solve the Linear Approximation formula, and check to see that it is infact close to \sqrt{(4.01)^2 + (3.98)^2 + (2.02)^2}

Any help would be great thanks

 

[Mark44]

Homework Statement

Use the linear approximation to approximate a suitable function f(x,y) and thereby estimate the following

f(x,y) = \sqrt{(4.01)^2 + (3.98)^2 + (2.02)^2}

Homework Equations

Not going to type it out, but the formula for f(x,y) http://en.wikipedia.org/wiki/Linear_approximation

The Attempt at a Solution

I just need to find the equation first, I can do the estimation.

My guess is, let x = 4, and y = 2 then

f(x,y) = \sqrt{(x)^2 + (x)^2 + (y)^2}
=(2(x)² + (y)²)^1/2

Is that done correctly? Proceeding this would be to solve the Linear Approximation formula, and check to see that it is infact close to \sqrt{(4.01)^2 + (3.98)^2 + (2.02)^2}

Any help would be great thanks

Your function is actually one with three variables, f(x, y, z), and its formula is sqrt(x^2 + y^2 + z^2)

Here's what you want to use:
f(x₀ + dx, y₀ + dy, z₀ + dy) \approx f(x₀, y₀, z₀) + df
where df = f_x(x₀, y₀, z₀)*dx + f_y(x₀, y₀, z₀)*dy + f_z(x₀, y₀, z₀)*dz.

The notation f_x(x₀, y₀, z₀) means the partial deriviative of f, evaluated at (x₀, y₀, z₀), and so on for the other two partials.

For your problem, x₀ = 4, y₀ = 4, and z₀ = 2
dx = .01, dy = -.02, and dz = .02

 

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Ah, That all makes sense, but when I computed df, I got that equal to 0

df/dx = x/\sqrt{(x)^2 + (z)^2 + (y)^2} = 0.6666...
df/dy = 0.6666...
df/dz = 0.3333...

which means, my approximation is exact, which i don't think is right >.<

 

[Mark44]

It just means that df = 0.
f(x₀ + dx, y₀ + dy, z₀ + dy) \approx
f(x₀, y₀, z₀) + df
The fact that df = 0 doesn't turn the above into an equality.

In this case, the exact value is sqrt(36.009) \approx 6.000075, which is approximately equal to your estimate, sqrt(36) = 6

 

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Ohh I see perfect, thank you very much