# Linear Approximation function

1. Oct 29, 2009

### Iconate

1. The problem statement, all variables and given/known data
Use the linear approximation to approximate a suitable function f(x,y) and thereby estimate the following

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

2. Relevant equations

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

3. 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

2. Oct 29, 2009

### Staff: Mentor

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 = fx(x0, y0, z0)*dx + fy(x0, y0, z0)*dy + fz(x0, y0, z0)*dz.

The notation fx(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

3. Oct 29, 2009

### Iconate

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 dont think is right >.<

Last edited: Oct 29, 2009
4. Oct 29, 2009

### Staff: Mentor

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

5. Oct 29, 2009

### Iconate

Ohh I see perfect, thank you very much