Use of Linear approximation..

[kenhung123]

I'm not sure if I am even clear about the idea of it.
Is it used to predict the y value as x changes provided we know the derivative?

I am not sure what's the point of linear approx? Essentials seem to have some emphasis on it in particular the applications of diff exponentials and circular functions.

[the.watchman]

Ok, let's say you wake up from a dream thinking that: I must find

Obviously, you can't get an exact value for this, but using the linear approx. formula, you can get an approx. value for it

So let

Then you can use the formula to find

[kenhung123]

Hmm, I'm not really sure what you mean you can't find exact value for that?

[the.watchman]

Hmm, I'm not really sure what you mean you can't find exact value for that?

I mean you can't come up with an accurate answer to five or six decimal places of that in your head (unless you are a super genius)
So a (quick) linear approximation can find an approximation for these sorts of expressions

[kenhung123]

Oh ok (unless your name is watchman more like it )
Anyway, would it work also if I split it up into f(7.9+0.05)=0.05*f'(x)+f(7.9)?

[the.watchman]

It would, but finding also requires super genius abilities
Therefore, using is much easier

[kenhung123]

Hold on, does linear approximation just mean you kind of round a value (+h which is small rounding) in order to get a simple whole number answer (in this case a perfect cube root)?

[the.watchman]

Hold on, does linear approximation just mean you kind of round a value (+h which is small rounding) in order to get a simple whole number answer (in this case a perfect cube root)?

Sort of, but it's more than just rounding
The error is approximated by the term in the formula which, if you draw a diagram, makes sense

[kenhung123]

Oh lol I actually get it now
The formula is basically y=mx+c really..
where h=x f'(x)=m and f(x)=c
Basically, you need to use a close and easy point (like perfect cube root) and find the gradient at that point and its y value
To predict the y value of a point very close to that 'easy' point, you multiply the small change in x by the gradient + y value at the 'easy' point

[m@tty]

Yes, it is putting a tangent from a known point and finding the coordinate at some other point. Hence it approximates. You needed to understand this for exam 1 last year.

[kenhung123]

Hey is 'h' always a positive number or it can be negative also?

[tcg93]

Hey is 'h' always a positive number or it can be negative also?

Can be negative too

[Martoman]

I had a similar problem last year. Understand what you are doing graphically. At any give point on a curve you draw in a tangent line. Notice how close to the point of contact how the tangent line and he curve are very similar? This gives you the intuition of why small h values lead to higher accuracy of estimation. ALL linear approximation is doing is saying create a tangent line on a point that is easy. This tangent line is of the form y = mx+c. Then you are simply finding the y value of this tangent line, provided x is a small variation from your *easy* spot of contact.  :smitten: Doing this give you a good *approximation* to the actual y value of the curve as the tangent line and curve are similar for small h (i feel like im repeating myself a tad) 

[tram]

lol, thinking of the last question of EXAM 1 are we matoman?

but yea....i hate linear approximation, waste of time

[TrueTears]

try "3D" linear approximation, aka planar approximation with partial derivatives pretty sik stuff