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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: bucket on May 28, 2008, 12:15:25 am

Title: linear approximation
Post by: bucket on May 28, 2008, 12:15:25 am: okay, i DONT get this whole concept!!!
what the hell is it showing and how does it work!!

Here is a couple of the questions:

The equation of a curve is $4x^3-8x^2+10$ , find the approximate change in y as x increases from $2$ to $2.02$

and also

Given that $y=x^4-5x^3$ find in terms of $q$ the approximate increase in y as x increases from $2$ to $2+q$ where $q$ is small
Title: Re: linear approximation
Post by: AppleXY on May 28, 2008, 12:44:06 am: $f'(x) = 12x^2 - 16x$

$f(x+h) = f(x) + hf'(x)$

$\mbox{as x = 2 and h = .02}$

$f(2+.02) = 10 + .02(16)$

$f(2.02) = 10 + 0.32$

$\therefore \mbox{the approximate change in y is 0.32}$

$\boxed{= 10.32}$

${\mbox{Question 2}}$

$y=x^4-5x^3$

$\frac{dy}{dx} = 4x^3 - 15x^2$

${\delta y}=\delta \cdot \frac{dy}{dx}$

$= (2+q)\cdot -28$

$\boxed{ = -28(2+q)}$

(fixed for latex screwups)
Title: Re: linear approximation
Post by: bucket on May 28, 2008, 12:53:18 am: the answer for 1 was 0.32, but i understand why! (it increased from 10 by 0.32)
thanks a lot for that explanation!
Title: Re: linear approximation
Post by: AppleXY on May 28, 2008, 01:08:18 am: Yeah. The approximate increase's is just the rearrangement of the 'first principles of differentiation' rule. :)