Author Topic: VCE Specialist 3/4 Question Thread! (Read 2547171 times) Tweet Share

Homer · « **Reply #1095 on:** January 07, 2013, 03:05:29 pm »

The books answer :/

Planck's constant · « **Reply #1096 on:** January 07, 2013, 03:10:24 pm »

Quote from: Battman on January 07, 2013, 02:51:41 pm

The answer is equivalent to what the calculator gives, its just in a different form, so you haven't made a mistake, just not got it in the same form. So it's correct, I'll post up in a couple of minutes to show that they are the same.

$\begin{alignedat}{1}\frac{\frac{15}{4}\sin^{2}\left(2x\right)x^{\frac{1}{2}}+4x^{\frac{5}{2}}\sin^{2}\left(2x\right)-10\sin\left(2x\right)\cos\left(2x\right)x^{\frac{3}{2}}+8x^{\frac{5}{2}}\cos^{2}\left(2x\right)}{\sin^{3}\left(2x\right)} & =\frac{\frac{15}{4}\left(1-\cos^{2}\left(2x\right)\right)x^{\frac{1}{2}}+4x^{\frac{5}{2}}\left(1-\cos^{2}\left(2x\right)\right)-10\sin\left(2x\right)\cos\left(2x\right)x^{\frac{3}{2}}+8x^{\frac{5}{2}}\cos^{2}\left(2x\right)}{\sin^{3}\left(2x\right)}\:\:\:\left(\mathrm{using\:}\sin^{2}\left(2x\right)+\cos^{2}\left(2x\right)=1\right) \\ & =\frac{x^{\frac{1}{2}}\left(15-15\cos^{2}\left(2x\right)+16x^{2}-16x^{2}\cos^{2}\left(2x\right)-40x\sin\left(2x\right)\cos\left(2x\right)+32x^{2}\cos\left(2x\right)\right)}{4\sin^{3}\left(2x\right)} \\ & =\frac{x^{\frac{1}{2}}\left(\left(32x^{2}-16x^{2}-15\right)\cos^{2}\left(2x\right)-40x\sin\left(2x\right)\cos\left(2x\right)+16x^{2}+15\right)}{} \\ & =\frac{x^{\frac{1}{2}}\left(\left(16x^{2}-15\right)\cos^{2}\left(2x\right)-40x\sin\left(2x\right)\cos\left(2x\right)+16x^{2}+15\right)}{4\sin^{3}\left(2x\right)} \end{alignedat} $

nice one, battman
i must say though that this question borders on cruelty

b^3 · « **Reply #1097 on:** January 07, 2013, 03:13:07 pm »

Checking on the calc the book's answer isn't equivalent to the actual answer, so looks like the book's answer is wrong, unless I typed something in wrong.

pi · « **Reply #1098 on:** January 07, 2013, 03:15:00 pm »

I'd use the product rule after doing a quick derivative of csc(x) via quotient rule. The answer will look a lot nicer.

Stick · « **Reply #1099 on:** January 07, 2013, 07:53:57 pm »

Ugh. Questions like these make me hate calculus.

BubbleWrapMan · « **Reply #1100 on:** January 08, 2013, 06:46:49 am »

They don't really have anything that tedious in spesh exams

Jaswinder · « **Reply #1101 on:** January 09, 2013, 11:30:02 am »

haha thank god!

lol btw nice handwriting homer

Homer · « **Reply #1102 on:** January 09, 2013, 06:16:42 pm »

hey guys just a question regarding a worked example I was looking at. When differentiating sin-1(6x) why doesn't square root of 1/36 become a plus or minus 1/6?

Hancock · « **Reply #1103 on:** January 09, 2013, 06:21:20 pm »

If you look at a graph of $y = sin^{-1}(x)$ , you can see that it always has a positive gradient. As such, it's derivative function must always be positive.

TrueTears · « **Reply #1104 on:** January 09, 2013, 06:43:40 pm »

Quote from: Homer on January 09, 2013, 06:16:42 pm

hey guys just a question regarding a worked example I was looking at. When differentiating sin-1(6x) why doesn't square root of 1/36 become a plus or minus 1/6?

$\sqrt{x^2} = |x|$

$\pm \sqrt{x^2} = \pm |x|$

ie, $\sqrt{9} = 3$ but $\pm \sqrt{9} = \pm 3$

wdf is up with vn's latex, so non-standard

Lasercookie · « **Reply #1105 on:** January 09, 2013, 07:00:13 pm »

Quote from: TrueTears on January 09, 2013, 06:43:40 pm

$\sqrt{x^2} = |x|$

$\pm\sqrt{x^2} = \pm|x|$

ie, $\sqrt{9} = 3 \text{ but } \pm\sqrt{9} = \pm{3}$

Does this still give the errors? I have no idea why it's doing it, even though I made minimal changes to your latex. Might be a problem on the end of the latex server than AN is using :/

b^3 · « **Reply #1106 on:** January 09, 2013, 08:21:39 pm »

AN's $\LaTeX$ doesn't seem to like two slash commands in a row without spaces, i.e. the \pm\sqrt to become what laseredd has, \pm \sqrt

Lasercookie · « **Reply #1107 on:** January 09, 2013, 08:42:16 pm »

Quote from: Battman on January 09, 2013, 08:21:39 pm

AN's $\LaTeX$ doesn't seem to like two slash commands in a row without spaces, i.e. the \pm\sqrt to become what laseredd has, \pm \sqrt

Removing the spaces from \pm \sqrt to become \pm\sqrt was the change I made though.

Homer · « **Reply #1108 on:** January 10, 2013, 01:33:17 pm »

yep, I get it thankss guys

Homer · « **Reply #1109 on:** January 10, 2013, 01:59:49 pm »

when anti differentiating -3/square root(9-x^2)

3cos-1(x/3) and -3sin(x/3) be right? the book only has 3cos-1(x/3)

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2547171 times) Tweet Share

Homer

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Planck's constant

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b^3

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pi

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Stick

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BubbleWrapMan

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Jaswinder

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Homer

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Hancock

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TrueTears

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Lasercookie

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b^3

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Lasercookie

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Homer

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Homer

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