Milka Krasteva

Posted on Sunday, 13 April, 2003 - 01:51 pm:

Hi,

This is a P3 question I am stuck on:
Using differentiation, find approximate values for:
27.005^2/3 and a few other numbers in the same format...I haven't come across this ever before, and would appreciate a hint on how to get started.

Philip Ellison

Posted on Sunday, 13 April, 2003 - 02:14 pm:

I'm not sure exactly what they're after, but I'd imagine that you're probably required to obtain the first few terms in a power series.

Kerwin Hui

Posted on Sunday, 13 April, 2003 - 04:31 pm:

This is an exercise in using
LaTeX Image

(letting y=x^2/3 ).

Kerwin

Milka Krasteva

Posted on Sunday, 13 April, 2003 - 06:49 pm:

Kerwin,
I'm afraid I cannot work out how to apply this to the question...
Could you possibly expand on the method.
Thanks,
Milka

Colin Prue

Posted on Sunday, 13 April, 2003 - 07:07 pm:

i've never seen this done before but here goes (i'll let you fill in the blanks):

let $x = 27$ , therefore $Δ x =$ [blank]

as $y = x^{2 / 3}$ , $\Rightarrow y = 27^{2 / 3} + Δ y$

using Kerwin's approximation, (presumably valid only for small $Δ x$ as derivative must be evaluated at $x$ ) we have $Δ y = \frac{dy}{dx} X$ , where $\frac{dy}{dx} = \frac{d}{dx} (x^{2 / 3})$ evaluated at [blank]

(as a check i got 9.0011)

Milka Krasteva

Posted on Sunday, 13 April, 2003 - 07:45 pm:

Thanx, Colin.
The method is quite clear now, but for some reason I cannot get the correct answer...instead of 9.0011 I get 9.01949 (using Dx=0.005).
Am I doing something wrong?

Chris Tynan

Posted on Sunday, 13 April, 2003 - 08:13 pm:

I get the same as Colin:

y = 27^2/3 + e(y) (e(y) means error of y)
e(y) = 2/3 . 27^-1/3 . e(x) and e(x)=0.005

I'm pretty sure you've placed x=0.005 in the derivative rather than 27, since x=27 and e(x)=0.005.

Chris

Milka Krasteva

Posted on Sunday, 13 April, 2003 - 10:19 pm:

Thank you, Chris. That's exactly what I was doing wrong. Everything clear now.
Thanks to everyone for their help.
Milka