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Linear approximation and rational numbers

  • Thread starter camino
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1. Homework Statement

Use linear approximation to approximate sqrt((3.2)^2 + 2(2.1) + (4.3)^2)

with a rational number (a ratio of integers).

2. Homework Equations

f(x,y) = sqrt(x^2 + 2y + z^2)
f(x,y) = (x^2 + 2y + z^2)^1/2

3. The Attempt at a Solution

x = 3 ∆x = 2/10
y = 2 ∆y = 1/10
z = 4 ∆z = 3/10

≈ sqrt(3^2 + 2(2) + 4^2) + [3/sqrt(3^2 + 2(2) + 4^2)] (2/10) + [2/sqrt(3^2 + 2(2) + 4^2)] (1/10) + [4/sqrt(3^2 + 2(2) + 4^2)] (3/10)

≈ [sqrt(29)] + [6/sqrt(29)(10)] + [2/sqrt(29)(10)] + [12/sqrt(29)(10)]


I'm not sure where to go from here to simplify it down to a single ratio. Am I on the right track?
 
32,575
4,305
1. Homework Statement

Use linear approximation to approximate sqrt((3.2)^2 + 2(2.1) + (4.3)^2)

with a rational number (a ratio of integers).
The estimate should probably be to 1 decimal place, since that's the precision that your numbers seem to be given to. The estimate will be between 5 and 6, based on a quick mental calculation. If you end up with, say, 5.6, that is a rational number. Any decimal fraction that terminates is a rational number.
2. Homework Equations

f(x,y) = sqrt(x^2 + 2y + z^2)
f(x,y) = (x^2 + 2y + z^2)^1/2
Your function has three variables, so it should be f(x, y, z) = ...
3. The Attempt at a Solution

x = 3 ∆x = 2/10
y = 2 ∆y = 1/10
z = 4 ∆z = 3/10

≈ sqrt(3^2 + 2(2) + 4^2) + [3/sqrt(3^2 + 2(2) + 4^2)] (2/10) + [2/sqrt(3^2 + 2(2) + 4^2)] (1/10) + [4/sqrt(3^2 + 2(2) + 4^2)] (3/10)
You have omitted the crucial step of finding ∆f, which involves the first partials with respect to x, y, and z, and the increments, ∆x, ∆y, and ∆z.
≈ [sqrt(29)] + [6/sqrt(29)(10)] + [2/sqrt(29)(10)] + [12/sqrt(29)(10)]


I'm not sure where to go from here to simplify it down to a single ratio. Am I on the right track?
 
42
0
Mark, I have been going through this problem as in my notes and I'm not quite sure what I omitted. Could you expand on that? Or show the step I omitted?
 
32,575
4,305
Sure. You want to approximate f(x + ∆x, y + ∆y, z + ∆y), which is approximately equal to
f(x, y, z) + fx(x, y, z) ∆x + fy(x, y, z) ∆y + fz(x, y, z) ∆z

In this formula, x, y, and z are the values 3, 2, and 4, ∆x, ∆y, and ∆z are .2, .1, and .3 (it's silly to write them as 2/10, 1/10, and 3/10), and fx(x, y, z) is the partial of f, with respect to x, evaluated at x, y, z. Similar for the other two partials.

This estimate is the three-dimension counter part to linear approximations for functions of one variable, f(x). For such a function, f(x0 + ∆x) [itex]\approx[/itex] f(xx0) + f'(x0)∆x. Here x0 is the base x-value at which the function and its derivative are easy to evaluate. I used x0 here but not in what I showed for the function of your problem. There is no difference intended between x0 here and plain old x in the earlier bit I wrote.

Clear?
 

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