Author Topic: 4U Maths Question Thread (Read 665304 times) Tweet Share

RuiAce · « **Reply #1155 on:** May 08, 2017, 04:38:46 pm »

Quote from: seventeenboi on May 08, 2017, 04:20:07 pm

HI,
just a conics question:
I can do everything else except q2b(gamma).. helppp!!!
Thanks so much )

Hints: Use PS=e*PM and PS'=ePM' to take care of the LHS.

If you need the full solution please provide the answers to the previous part.

seventeenboi · « **Reply #1156 on:** May 08, 2017, 04:48:21 pm »

Quote from: RuiAce on May 08, 2017, 04:38:46 pm

Hints: Use PS=e*PM and PS'=ePM' to take care of the LHS.

If you need the full solution please provide the answers to the previous part.

OMG THANK YOU SO MUCH
it worked out yayy

kiwiberry · « **Reply #1157 on:** May 09, 2017, 10:28:35 am »

Hello, I'm having trouble starting off this volumes question, any help is appreciated

RuiAce · « **Reply #1158 on:** May 09, 2017, 03:00:03 pm »

Quote from: kiwiberry on May 09, 2017, 10:28:35 am

Hello, I'm having trouble starting off this volumes question, any help is appreciated

$\text{We ask GeoGebra for a bit of help visualising where to get started}$

$\text{In this simulation, the sphere }x^2+y^2+z^2=4\\ \text{was plotted against the cylinder }x^2+y^2=1\\ \text{Note that the radius of the sphere is 2}\\ \text{and the radius of a cross section of the cylinder is 1, so this matches our scenario}$
$\text{What we want to find is the volume of the sphere}\\ \text{LESS the volume of the cylinder specifically INSIDE the sphere}\\ \text{This gives us a ring-shaped thing}$
$\text{Projected onto the }x-y\text{ plane*, a cross section looks like this}$
* - Not fully accurate, but don't worry about that.

$\text{It appears that we are still rotating a circle about the }y\text{-axis}\\ \text{But more accurately, only a cross section of said circle.}$
______________________________________
$\text{We return to our question}\\ \text{Consider an arbitrary sphere with diameter }2a\text{ centred at the origin}\\ \text{It has equation }\boxed{x^2+y^2=a^2}$
$\text{Since we seek to apply the shells method}\\ \text{Because we intend to rotate about the }y\text{-axis, we consider }\boxed{y=\pm \sqrt{a^2-x^2}}$
$\text{The radius of a typical shell will certainly be }x^2\\ \text{The height will be }2\sqrt{a^2-x^2}\text{. You should be able to see why.}\\ \text{Hence a typical shell has volume }\delta V = 2\pi r^2h\delta r=4\pi x^2 \sqrt{a-x^2}\delta x$
$\therefore\boxed{ V=\int_{\frac{a}2}^{a} 4\pi x^2 \sqrt{a^2-x^2}dx}\\ \text{This integral is best tackled with the substitution }x=a\sin \theta, \, -\frac{\pi}2\le \theta \le \frac{\pi}2$

kiwiberry · « **Reply #1159 on:** May 09, 2017, 04:26:17 pm »

Quote from: RuiAce on May 09, 2017, 03:00:03 pm

$\text{We ask GeoGebra for a bit of help visualising where to get started}$ ....

Ahhh thank you so much Rui, that helped a lot

Wales · « **Reply #1160 on:** May 12, 2017, 07:11:49 pm »

Just dropping by with a integration question.

Integral of (1+x)/SQRT(1-x-x^2) dx

From Coroneos' 100 Integrals question 15.

Cheers, Wales

RuiAce · « **Reply #1161 on:** May 12, 2017, 08:03:43 pm »

Quote from: Wales on May 12, 2017, 07:11:49 pm

Just dropping by with a integration question.

Integral of (1+x)/SQRT(1-x-x^2) dx

From Coroneos' 100 Integrals question 15.

Cheers, Wales

$\begin{align*}\int \frac{1+x}{\sqrt{1-x-x^2}}dx&=-\frac12 \int \frac{-2-2x}{\sqrt{1-x-x^2}}dx\\ &= -\frac{1}{2}\int \left(\frac{-2-2x}{\sqrt{1-x-x^2}}-\frac{1}{\sqrt{\frac54-\left(x+\frac12\right)^2}}\right)dx\\ &= - \sqrt{1-x-x^2} + \frac12 \sin^{-1}\left(\frac{2}{\sqrt5}\left(x+\frac12\right)\right)+C\end{align*}$

chelseam · « **Reply #1162 on:** May 12, 2017, 10:41:20 pm »

Hi! I'm having trouble integrating cot^3x, and cot^5x. Thank you

RuiAce · « **Reply #1163 on:** May 12, 2017, 11:03:43 pm »

Quote from: chelseam on May 12, 2017, 10:41:20 pm

Hi! I'm having trouble integrating cot^3x, and cot^5x. Thank you

This integral works the exact same as $\int \tan^3 x\, dx$. You should have a go at that one first.
$\text{Note that the result }\frac{d}{dx}\cot x = -\csc^2x\text{ is standard}\\ \text{and analogous to }\frac{d}{dx}\tan x = \sec^2x\\ \text{which is necessary in computing }\int \tan^3x\,dx$

hanaacdr · « **Reply #1164 on:** May 13, 2017, 09:08:07 am »

Hi
Could i get some help on this question

crazycodpro · « **Reply #1165 on:** May 13, 2017, 06:43:39 pm »

RuiAce · « **Reply #1166 on:** May 13, 2017, 06:48:09 pm »

Quote from: crazycodpro on May 13, 2017, 06:43:39 pm

hi

$\text{This is a standard 3U derivative that you should have no trouble computing with the chain rule}\\ \frac{d}{dx} \arcsin (3x+5) = \frac{1}{\sqrt{1-(3x+5)^2}}\times 3$

RuiAce · « **Reply #1167 on:** May 13, 2017, 07:21:21 pm »

Quote from: hanaacdr on May 13, 2017, 09:08:07 am

Hi
Could i get some help on this question

A bit all over the place. Feel free to point out any areas of confusion

hanaacdr · « **Reply #1168 on:** May 14, 2017, 03:08:47 pm »

Quote from: RuiAce on May 13, 2017, 07:21:21 pm

A bit all over the place. Feel free to point out any areas of confusion
(Image removed from quote.)

WOW!
thank you!

beau77bro · « **Reply #1169 on:** May 19, 2017, 10:11:18 am »

Please help

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