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:
27.0052/3 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=x2/3 ).


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 Dx= [blank]

as y=x2/3, Þ y=272/3+Dy

using Kerwin's approximation, (presumably valid only for small Dx as derivative must be evaluated at x) we have
Dy= dy
dx
 X

, where
dy
dx
 = d
dx
 (x2/3)

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 = 272/3 + 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