Author Topic: TrueTears question thread (Read 66309 times) Tweet Share

dejan91 · « **Reply #570 on:** October 09, 2009, 03:18:11 pm »

Quote from: Damo17 on October 09, 2009, 02:12:16 pm

Quote from: TrueTears on October 09, 2009, 12:17:30 am
What's the systematic way of working out $\tan(\theta) = \sqrt{3}-2$

This requires the use of the trig identity: $tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}$

What identity is this? Is it derived from something (as in manipulated) or are we meant to know it?

TonyHem · « **Reply #571 on:** October 09, 2009, 03:33:53 pm »

Quote from: dejan91 on October 09, 2009, 03:18:11 pm

What identity is this? Is it derived from something (as in manipulated) or are we meant to know it?

$\frac{sin^2(x)}{cos^2(x)}$
$cos(2x) = 1-2sin^2(x)$
$2sin^2(x) = 1-cos(2x)$
$cos(2x) = 2cos^2(x)-1$
$2cos^2(x)= 1+cos(2x)$
$so\; \frac{sin^2(x)}{cos^2(x)} = \frac{1-cos(2x)}{1+cos(2x)}$

TrueTears · « **Reply #572 on:** October 09, 2009, 04:20:58 pm »

Quote from: Damo17 on October 09, 2009, 02:12:16 pm

Quote from: TrueTears on October 09, 2009, 12:17:30 am
What's the systematic way of working out $\tan(\theta) = \sqrt{3}-2$

This requires the use of the trig identity: $tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}$

$tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}= 7-4\sqrt{3}$

$1-cos(2 \theta)=7-4\sqrt{3}+7cos(2 \theta)-4\sqrt{3}cos(2 \theta)$

$4\sqrt{3}-6=cos(2 \theta)(8-4\sqrt{3})$

$\frac{4\sqrt{3}-6}{8-4\sqrt{3}} \times \frac{8+4\sqrt{3}}{8+4\sqrt{3}}=cos(2 \theta)$

$\frac{\sqrt{3}}{2}=cos(2 \theta)$

$\therefore$ $\theta =\pm \frac{\pi}{12}$

But we take the negative as $\sqrt{3}-2$ is negative, so:

$\theta= -\frac{\pi}{12}$

Awesome, thanks, how did you think of $7 - 4 \sqrt{3}$ ?

TonyHem · « **Reply #573 on:** October 09, 2009, 04:23:09 pm »

tan^2(theta) = 3 - 4sqrt{3} + 4

TrueTears · « **Reply #574 on:** October 09, 2009, 04:26:30 pm »

Quote from: TonyHem on October 09, 2009, 04:23:09 pm

tan^2(theta) = 3 - 4sqrt{3} + 4

indeed, thanks.

and yeah thanks again Damo!!!!!!!!!!!!!!!!!!!!!!! pr0!!!!!!!!!!!!!!!

Damo17 · « **Reply #575 on:** October 09, 2009, 04:38:29 pm »

Quote from: TrueTears on October 09, 2009, 04:26:30 pm

and yeah thanks again Damo!!!!!!!!!!!!!!!!!!!!!!! pr0!!!!!!!!!!!!!!!

No problem. It took me a while, but I rather enjoyed it.

TrueTears · « **Reply #576 on:** October 23, 2009, 04:48:38 pm »

A small object is dropped vertically down from the top of a building which is h metres high. It takes 0.4 seconds to travel the last 8 metres before it hits the ground. If the air resistance is negligible, show that the height of the building is : $h = \frac{(100+g)^2}{50g}$

Pretty simple question IF you assume it starts from rest. So can you assume it starts from rest?

kurrymuncher · « **Reply #577 on:** October 23, 2009, 04:56:18 pm »

Quote from: TrueTears on October 23, 2009, 04:48:38 pm

A small object is dropped vertically down from the top of a building which is h metres high. It takes 0.4 seconds to travel the last 8 metres before it hits the ground. If the air resistance is negligible, show that the height of the building is : $h = \frac{(100+g)^2}{50g}$

Pretty simple question IF you assume it starts from rest. So can you assume it starts from rest?

Well, it says "dropped", so Im assuming it starts from rest.

maybe? i dont know lol

TrueTears · « **Reply #578 on:** October 23, 2009, 04:57:48 pm »

Quote from: kurrymuncher on October 23, 2009, 04:56:18 pm

Quote from: TrueTears on October 23, 2009, 04:48:38 pm
A small object is dropped vertically down from the top of a building which is h metres high. It takes 0.4 seconds to travel the last 8 metres before it hits the ground. If the air resistance is negligible, show that the height of the building is : $h = \frac{(100+g)^2}{50g}$

Pretty simple question IF you assume it starts from rest. So can you assume it starts from rest?

Well, it says "dropped", so Im assuming it starts from rest.

maybe? i dont know lol

Yeah, but when something is dropped it would also be dropped with a certain velocity.

ed_saifa · « **Reply #579 on:** October 23, 2009, 05:06:49 pm »

I'm sure that drop implies from rest. You can't really drop something with a certain velocity; that would be called thrown

TrueTears · « **Reply #580 on:** October 23, 2009, 05:14:12 pm »

Ahhh haha, alright thanks you guys!

TrueTears · « **Reply #581 on:** October 23, 2009, 05:58:42 pm »

A particle is moving so that its position at time t seconds is given by $r(t) = 2cos\left(\frac{\pi}{10}t\right) i + 3t j$ . Find the cartesian equation of the particle, state the domain and range.

Let $x(t) = 2cos\left(\frac{\pi}{10}t\right)$ and $y(t) = 3t$

So the cartesian equation is $y = \frac{30}{\pi}\cos^{-1}\left(\frac{x}{2}\right)$ .

So I've always done that the range of $x(t)$ = domain of cartesian and range of $y(t)$ = range of cartesian.

Okay, so the range of $x(t)$ for $t \ge 0$ is $[-2,2]$ which is right domain for cartesian, however range of $y(t)$ is $[0, \infty)$ however the cartesian is not defined for $[0, \infty)$ does that mean we take the highest possible range that is defined for the cartesian? ie, $[0, 30]$

kamil9876 · « **Reply #582 on:** October 23, 2009, 06:17:17 pm »

The crux is in this step:

$x=2cos(\frac{y\pi}{30})$

when "inverting":
$<br />cos^{-1}(\frac{x}{2})=\frac{y\pi}{30} + 2k\pi$ and that other scenario that I cbf working out.

THis is analogous to $y=x^2 \implies x=\sqrt{y}$ not being neccesary.

TrueTears · « **Reply #583 on:** October 23, 2009, 06:19:38 pm »

Quote from: TrueTears on October 23, 2009, 05:58:42 pm

Okay, so the range of $x(t)$ for $t \ge 0$ is $[-2,2]$ which is right domain for cartesian, however range of $y(t)$ is $[0, \infty)$ however the cartesian is not defined for $[0, \infty)$ does that mean we take the highest possible range that is defined for the cartesian? ie, $[0, 30]$

So... what range do I take?

Mao · « **Reply #584 on:** October 23, 2009, 06:25:17 pm »

the cartesian equation is not necessarily a function

The cartesian equation is $x = 2\cos \left( \frac{\pi y}{30}\right)$

Domain is [-2,2], range is [0,infinity)

The graph is a wave travelling vertically upwards.

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