Author Topic: what am i doing wrong? (Read 3017 times) Tweet Share

nels · « **on:** October 15, 2009, 06:45:37 pm »

this is a question off checkpoints (q138), which is the first short answer question in 2004 exam 1.

show that, for 0< x <0.5, $\frac{\mathrm{d} (\sin^{-1}(\sqrt{2x}))}{\mathrm{d} x}= \frac{1}{\sqrt{2x(1-2x)}}$

so i figured that the chain rule should be used.

$u=\sqrt{2x}= (2x)^{0.5}$

$\therefore \frac{\mathrm{d} u}{\mathrm{d} x}= \frac{1}{\sqrt{x}}$

$y=\sin^{-1} (u)$

$\therefore \frac{\mathrm{d} y}{\mathrm{d} u}= \frac{1}{\sqrt{1-u^{2}}}$

then i multiplied the derivatives together and ended up with $\frac{1}{\sqrt{x(1-2x)}}$

was there something i carelessly left behind? i looked at the solutions and they multiplied the whole chain rule by 2. why?

please help.

TrueTears · « **Reply #1 on:** October 15, 2009, 06:50:31 pm »

nels · « **Reply #2 on:** October 15, 2009, 06:56:03 pm »

wow, im really kicking myself right now. i spent an hour trying to figure it out. :tickedoff:

thank you!

nels · « **Reply #3 on:** October 15, 2009, 08:56:27 pm »

another quick question:

can someone please tell me why $\frac{1}{2}\int_{0}^{1}e^{2x}.dx$ does NOT simplify to $\frac{1}{4}(e^{2}-1)$ ?

man i feel so dumb.

/0 · « **Reply #4 on:** October 15, 2009, 09:04:17 pm »

actually I think it does...

you can quickly check your integrals with your calculator

nels · « **Reply #5 on:** October 15, 2009, 09:14:40 pm »

yeah i thought so as well but...

the question is from a vcaa exam (question 16MC exam 1 '05) and the answer is $\frac{1}{2}(e^{2}-1)$

heres the question:
$\int_{0}^{\frac{\pi }{8}}\sec^{2}(2x)e^{2tan(2x)}$ is equal to? $:-\$

TrueTears · « **Reply #6 on:** October 15, 2009, 09:19:23 pm »

Let $u = tan(2x)$

$\frac{du}{dx}$

Rest is trivial.

nels · « **Reply #7 on:** October 15, 2009, 09:22:58 pm »

hmmm...did that but still ended up with $\frac{1}{4}(e^{2}-1)$

kamil9876 · « **Reply #8 on:** October 15, 2009, 09:25:50 pm »

Quote from: nels on October 15, 2009, 09:22:58 pm

hmmm...did that but still ended up with $\frac{1}{4}(e^{2}-1)$

$\int_0^{tan\frac{\pi}{4}} \frac{1}{4}(e^{2u})du$ to be precise.

Hooligan · « **Reply #9 on:** October 15, 2009, 09:34:47 pm »

Quote from: kamil9876 on October 15, 2009, 09:25:50 pm

Quote from: nels on October 15, 2009, 09:22:58 pm
hmmm...did that but still ended up with $\frac{1}{4}(e^{2}-1)$

$\int_0^{tan\frac{\pi}{4}} \frac{1}{4}(e^{2u})du$ to be precise.

Shouldn't it be: $\int_0^{tan\frac{\pi}{4}} (e^{2u})du$ ?

TrueTears · « **Reply #10 on:** October 15, 2009, 09:36:08 pm »

Quote from: Hooligan on October 15, 2009, 09:34:47 pm

Quote from: kamil9876 on October 15, 2009, 09:25:50 pm
Quote from: nels on October 15, 2009, 09:22:58 pm
hmmm...did that but still ended up with $\frac{1}{4}(e^{2}-1)$

$\int_0^{tan\frac{\pi}{4}} \frac{1}{4}(e^{2u})du$ to be precise.

Shouldn't it be: $\int_0^{tan\frac{\pi}{4}} (e^{2u})du$ ?

Yeah, it is. Dono what kamil did =S

Hooligan · « **Reply #11 on:** October 15, 2009, 09:39:19 pm »

Quote from: TrueTears on October 15, 2009, 09:36:08 pm

Quote from: Hooligan on October 15, 2009, 09:34:47 pm
Quote from: kamil9876 on October 15, 2009, 09:25:50 pm
Quote from: nels on October 15, 2009, 09:22:58 pm
hmmm...did that but still ended up with $\frac{1}{4}(e^{2}-1)$

$\int_0^{tan\frac{\pi}{4}} \frac{1}{4}(e^{2u})du$ to be precise.

Shouldn't it be: $\int_0^{tan\frac{\pi}{4}} (e^{2u})du$ ?

Yeah, it is. Dono what kamil did =S

Yeah, thought so, 'cause if you do $\int_0^{tan\frac{\pi}{4}} \frac{1}{4}(e^{2u})du$ you get $1/8(e^2-1)$

nels · « **Reply #12 on:** October 15, 2009, 09:40:37 pm »

Quote from: Hooligan on October 15, 2009, 09:34:47 pm

Shouldn't it be: $\int_0^{tan\frac{\pi}{4}} (e^{2u})du$ ?

what happened to the 1/2 that was supposed to be in front of the integral?

TrueTears · « **Reply #13 on:** October 15, 2009, 09:41:06 pm »

There was none, was there? =S

Hooligan · « **Reply #14 on:** October 15, 2009, 09:42:08 pm »

Quote from: nels on October 15, 2009, 09:40:37 pm

Quote from: Hooligan on October 15, 2009, 09:34:47 pm

Shouldn't it be: $\int_0^{tan\frac{\pi}{4}} (e^{2u})du$ ?

what happened to the 1/2 that was supposed to be in front of the integral?

The 1/2 doesn't come in, until AFTER you integrate...

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