find cube root 0.9

s_ingram · #1 $Old$ June 29th, 2009, 05:57 PM

Hi again guys,

Two posts from me. But they aren't too hard. For you at least!

Given that $y = x^{-\frac{1}{3}}$ use the calculus to determine an approximate value for $\frac{1}{\sqrt [3]{0.9}}$ Normally with these there is a descernable x and $\delta x$ but I can't see a way to giggle 0.9 into anything useful. Can you?

Random Variable · #2 $Old$ June 29th, 2009, 06:25 PM

I would find the first few terms of the Taylor series of $x^{-1/3}$ centered at a=1.

mr fantastic · #3 $Old$ June 29th, 2009, 07:38 PM

Quote:

Originally Posted by s_ingram $View Post$

Hi again guys,

Two posts from me. But they aren't too hard. For you at least!

Given that $y = x^{-\frac{1}{3}}$ use the calculus to determine an approximate value for $\frac{1}{\sqrt [3]{0.9}}$ Normally with these there is a descernable x and $\delta x$ but I can't see a way to giggle 0.9 into anything useful. Can you?

Use the linear approximation: $f(x + \delta x) \approx f(x) + \delta f'(x)$ .

Take $f(x) = x^{-1/3}$ , $x = 1$ and $\delta = -0.1$ .

Soroban · #4 $Old$ June 29th, 2009, 09:36 PM

Hello, s_ingram!

Quote:

Given that: $y \:=\: x^{-\frac{1}{3}}$ use the calculus to determine an approximate value for $\frac{1}{\sqrt [3]{0.9}}$

We're expected to use: . $x = 1\:\text{ and }\:dx = \text{-}0.1$

We have: . $dy \:=\:-\dfrac{1}{3}x^{-\frac{4}{3}}dx \:=\:-\frac{dx}{3\sqrt[3]{x^4}}$

If $x = 1,\:dx = \text{-}0.1$ . we have: . $y + dy \:=\:1^{-\frac{1}{3}} - \frac{\text{-}0.1}{3\sqrt[3]{1^4}} \:=\:1 + \frac{1}{30} \:=\:1.0333\hdots$

Therefore: . $\frac{1}{\sqrt[3]{0.9}} \;\approx\;1.033$

s_ingram · #5 $Old$ June 30th, 2009, 04:30 AM

AH!!!! So easy! I was sure that being a cube root I had to pull a 2 out, so I was playing with 2 x 9/80 = 2. 0.1125 and I wasn't getting very far! Thanks guys - you are both Fantastic!