Author Topic: Specialist Problem (Read 11301 times) Tweet Share

ell · « **Reply #60 on:** May 25, 2008, 09:05:03 pm »

Probably makes it a bit easier if we draw the pyramid.

a) Position vectors of A,B,C and D:
So $\hat{\vec{AB}}=i$ $\hat{\vec{BC}}=j$ and $\hat{\vec{OV}}=k$ , and the length of one side is 4 cm.
From diagram ,
$\vec{OA}=-2i-2j$
$\vec{OB}=2i-2j$
$\vec{OC}=2i+2j$
$\vec{OD}=-2i+2j$

b)
We have to do a bit of travelling to get from P to M.

So $\vec{PM}=\vec{PB}+\vec{BC}+\vec{CM}$

We know $\vec{PB}$ and $\vec{BC}$ :
$\vec{PM}=2i+4j+\vec{CM}$

Now $\vec{CV}=\vec{CQ}+\vec{QO}+\vec{OV}$

$\vec{CV}=-2i-2j+2hk$

$\vec{CM}=\frac{1}{2}\vec{CV}$

Therfore $\vec{PM}=2i+4j+\frac{1}{2}(-2i-2j+2hk)$

$\vec{PM}=2i+4j-i-j+hk$

$\vec{PM}=i+3j+hk$

Similar process for finding $\vec{QN}$ -
$\vec{QN}=\vec{QB}+\vec{BA}+\vec{AN}$

$\vec{QN}=-2j-4i+\vec{AN}$

$\vec{AV}=2i+2j+2hk$

$\vec{AN}=\frac{1}{2}\vec{AV}$

$\vec{QN}=-2j-4i+\frac{1}{2}(2i+2j+2hk)$

$\vec{QN}=-3i-j+hk$

Tea.bag · « **Reply #61 on:** May 26, 2008, 04:07:40 pm »

Solve:

$sin(4x)=cos(2x)$ where $x\in{[0,\pi]}$

Mao · « **Reply #62 on:** May 26, 2008, 04:16:22 pm »

$\begin{align*} \sin(4x)&=\cos(2x) \\ 2\sin(2x)\cos(2x) &= \cos(2x)\\ \end{align*}$

dividing by cos(2x) on both sides, we must apply the constraint $\cos(2x)\neq 0$
hence possible solutions are:
$\begin{align*} \cos(2x)=0 & \mbox{ and } \sin(2x) = \frac{1}{2}\\ x=\frac{\pi}{4} & \mbox{ and } 2x = \left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}\\ \end{align*}$

$x = \left\{\frac{\pi}{12},\frac{\pi}{4},\frac{5\pi}{12}\right\}$

AppleXY · « **Reply #63 on:** May 26, 2008, 04:17:55 pm »

yeah, was gonna post but Mao did it lol.

Remember to exhaust all of your trig ids

Tea.bag · « **Reply #64 on:** May 26, 2008, 04:18:54 pm »

ahhhh.. so simple :uglystupid2:
thnx

humph · « **Reply #65 on:** May 27, 2008, 03:05:54 am »

Quote from: Mao on May 26, 2008, 04:16:22 pm

$\begin{align*} \sin(4x)&=\cos(2x) \\ 2\sin(2x)\cos(2x) &= \cos(2x)\\ \sin(2x) &= \frac{1}{2}\\ 2x &= \left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}\\ x &= \left\{\frac{\pi}{12},\frac{5\pi}{12}\right\}\\ \end{align*}$

you can't just divide through by $\cos(2x)$ as it might be zero, and so you lose possible solutions that way. in fact, the answer should be $x = \left\{\frac{\pi}{12},\frac{\pi}{4},\frac{5\pi}{12}\right\}$

ell · « **Reply #66 on:** May 27, 2008, 07:41:47 am »

You can move the $\cos{(2x)}$ to the LHS to avoid missing out on an answer.

$ \begin{align*} 2\sin{(2x)}\cos{(2x)}-\cos{(2x)}&=0\\ \cos{(2x)}(2\sin{(2x)}-1)&=0\\ \end{align*} $

Mao · « **Reply #67 on:** May 27, 2008, 05:15:18 pm »

thats a very good point~

I forgot about the constraints

Tea.bag · « **Reply #68 on:** June 29, 2008, 05:15:35 pm »

can u tell me way i should tackle these kind of questions?

Find the area between the curves $y=\sqrt{9-x}$ and $y=x+3$ from $x=-1$ and $x=2$

Mao · « **Reply #69 on:** June 29, 2008, 05:26:25 pm »

Quote from: Tea.bag on June 29, 2008, 05:15:35 pm

can u tell me way i should tackle these kind of questions?

Find the area between the curves $y=\sqrt{9-x}$ and $y=x+3$ from $x=-1$ and $x=2$

$\int_a^b f(x)-g(x)\; dx$

polky · « **Reply #70 on:** June 29, 2008, 05:52:04 pm »

Quote from: Tea.bag on June 29, 2008, 05:15:35 pm

can u tell me way i should tackle these kind of questions?

Find the area between the curves $y=\sqrt{9-x}$ and $y=x+3$ from $x=-1$ and $x=2$

Sketch both graphs to figure out which one lies on top and which one is at the bottom of the region you want to integrate. Then you can set about integrating it!

Tea.bag · « **Reply #71 on:** June 30, 2008, 09:39:28 pm »

Find the antiderivative:

1. $\int\frac{x^2}{x^2+4}$

Mao · « **Reply #72 on:** June 30, 2008, 09:45:47 pm »

Quote from: Tea.bag on June 30, 2008, 09:39:28 pm

Find the antiderivative:

1. $\int\frac{x^2}{x^2+4}$

$\implies \int \frac{x^2+4-4}{x^2+4}\; dx=\int 1-2\frac{2}{x^2+4}\; dx = x-2\tan^{-1} \frac{x}{2}+c$

Tea.bag · « **Reply #73 on:** June 30, 2008, 09:47:42 pm »

thnx. i forgot to that i could divide the numerator by denominator

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