Author Topic: Increase linear approx (Read 2001 times) Tweet Share

fredrick · « **on:** November 06, 2008, 01:34:54 pm »

given $y=10-\frac{5}{x}$ and that the y-value increases from 5 by a small amount $\frac{p}{10}$ , find in terms of p the approx change in x.

bucket · « **Reply #1 on:** November 06, 2008, 01:42:04 pm »

fredrick · « **Reply #2 on:** November 06, 2008, 01:46:40 pm »

ohhhhhhhh shit change in x (as in dy/dx) not change in x as i thought x final subtract x initial. :idiot2:

shinny · « **Reply #3 on:** November 06, 2008, 01:48:10 pm »

Seems like bucket's going to finish off the working out so I'll just chuck up my answer to cross-check:

$\Delta x = \frac{p}{50}$

fredrick · « **Reply #4 on:** November 06, 2008, 01:53:19 pm »

yep thanks got the answer

riadnicolas · « **Reply #5 on:** November 06, 2008, 01:55:23 pm »

can anyone do the working out?

bucket · « **Reply #6 on:** November 06, 2008, 01:58:02 pm »

Uhm the answer I got is different to shinjitsuzx's... I'll let him do it

;
I got $\frac{p}{2}$

shinny · « **Reply #7 on:** November 06, 2008, 02:02:03 pm »

$y=10-\frac{5}{x}$

$\mbox{Let y=f(x)}$

$f'(x)=\frac{5}{x^{2}}$

At y=5
$5=10-\frac{5}{x}$

$\frac{5}{x}=5$

$x=1$

Using $\frac{\Delta y}{\Delta x} \approx f'(x)$

$\Rightarrow \Delta x \approx \frac{\frac{p}{10}}{5}$

$\Rightarrow \Delta x \approx \frac{p}{50}$

Hope that's right o.O Haven't done this kinda question before.

unknown id · « **Reply #8 on:** November 06, 2008, 02:03:09 pm »

i got $\Delta x \approx \frac{p}{50}$ as well

fredrick · « **Reply #9 on:** November 06, 2008, 02:06:48 pm »

$\frac{dx}{dy}=\frac{\Delta x}{\Delta y}$
$\Delta x=\frac{dx}{dy}\Delta y$
and we know $\frac{dy}{dx}=\frac{5}{x^2}$
therefore $\Delta x=\frac{x^2}{5}*\frac{p}{10}$ , we know x=1 at y=5
so $\frac{p}{50}$

edit: beaten to it

bucket · « **Reply #10 on:** November 06, 2008, 02:10:14 pm »

Oh, I made a stupid error :s I thought $\frac{\frac{p}{10}}{5} \rightarrow \frac{5p}{10}$ =_=

shinny · « **Reply #11 on:** November 06, 2008, 02:15:33 pm »

lol hate when I do that too when my fractions aren't obvious enough on paper so $\frac{\frac{p}{10}}{5}$ looks like $\frac{p}{\frac{10}{5}}$

fredrick · « **Reply #12 on:** November 06, 2008, 02:22:32 pm »

Heres another: a 2% error is made in measuring the radius of a sphere. Find the % change in the surface area.

onlyfknhuman · « **Reply #13 on:** November 06, 2008, 02:24:59 pm »

Quote from: fredrick on November 06, 2008, 02:06:48 pm

$\frac{dx}{dy}=\frac{\Delta x}{\Delta y}$
$\Delta x=\frac{dx}{dy}\Delta y$
and we know $\frac{dy}{dx}=\frac{5}{x^2}$
therefore $\Delta x=\frac{x^2}{5}*\frac{p}{10}$ , we know x=1 at y=5
so $\frac{p}{50}$

edit: beaten to it

Lol that is much clearer.

unknown id · « **Reply #14 on:** November 06, 2008, 02:40:38 pm »

Quote from: fredrick on November 06, 2008, 02:22:32 pm

Heres another: a 2% error is made in measuring the radius of a sphere. Find the % change in the surface area.

$\Delta x = 0.02r$

$A_{sphere} = 4\pi r^2$

$\frac{dA}{dr} = 8\pi r$

$\therefore \%$ change in $f(x)$

$= \frac{\Delta x f'(x)}{f(x)} \times 100$

$= \frac{0.02r \times 8\pi r}{4\pi r^2} \times 100$

$= \frac{8 \times 0.02 \times 100}{4}$

$= 4 \%$

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Author Topic: Increase linear approx (Read 2001 times) Tweet Share

fredrick

Increase linear approx

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