Author Topic: TrueTears question thread (Read 67600 times) Tweet Share

TrueTears · « **Reply #285 on:** March 09, 2009, 01:40:48 am »

Quote from: kamil9876 on March 08, 2009, 09:58:33 pm

Q1.

Because you are considering the maximum domain and range of the hyperbola, then the only way to yield this is to ensure that the parametric equations have maximum range( I am pretty sure). Hence the $\tan(t-\frac{pi}{3})$ and $\sec(t - \frac{pi}{3})$ bits must hav their maximum ranges. The maximum ranges for $\tan\theta$ and $\sec\theta$ occur when $\theta$ is [0,pi)/{pi/2} for both functions. Now in this case $\theta=t-\frac{pi}{3}$ so t has a domain of $[\pi/3,\pi + \pi/3) without {\pi/2 + \pi/3}$ i.e: a domain translated $\pi/3$ units to the positive side. No duplication arises since $\pi$ is just one period for both of those trig functions and the values don't repeat.

ah yes i fully understand now,

thanks for the yet again awesome explanation kamil !

kamil9876 · « **Reply #286 on:** March 09, 2009, 06:55:25 pm »

Also, make sure (if u havnt already noticed), that the domain of t has one open bracket, and one closed bracket, as tan(pi) and tan(0) are equal to each other and same for sec. Hence if both endpoints of the domain are included then one point is duplicated, if both are not included then one point is missing. So, have one endpoint included, one not.

TrueTears · « **Reply #287 on:** March 09, 2009, 06:56:23 pm »

yeap, noticed that ^^

TrueTears · « **Reply #288 on:** March 09, 2009, 10:39:58 pm »

also this Q7 lol just stumped on it -_-"

Many thanks guys!!

/0 · « **Reply #289 on:** March 09, 2009, 11:25:57 pm »

$|\vec{AB}|=|u|$

$|\vec{AC}|=|u+v|$

If they are equal, $|u|^2=|u+v|^2$

$|u|^2= (u+v)\cdot (u+v)$

$u\cdot u= u \cdot u + 2 u \cdot v + v \cdot v$

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$

$\Righatarrow u \cdot u = 2 u \cdot u + 2u\cdot v$

$2u\cdot v +u\cdot u=0$

$u\cdot (2v + u)=0$

And since, $\vec{AD} = 2v+u$ , we have $AB \perp AD$

TrueTears · « **Reply #290 on:** March 09, 2009, 11:45:20 pm »

ahh nice thanks man

TrueTears · « **Reply #291 on:** March 10, 2009, 12:01:28 am »

and also for Q8, i've worked out $\overrightarrow {BC} = 2i-3j-4k$

so the shortest distance must be a straight line perpendicular to $\overrightarrow {OC}$ , but the question says "...between the POINT with coordinates..." How can you get something perpendicular to a point? Or maybe i'm just misunderstanding the question lol

EDIT: nvm i figured it out

TrueTears · « **Reply #292 on:** March 10, 2009, 04:55:36 pm »

Ok just a simple question which I'm not sure about.

simplify:
$tan(sin^{-1}(\frac{-2}{3}))$

so let $sin^{-1}(\frac{-2}{3}) = x$

$sinx = \frac{-2}{3}$

using the identity $\frac{1}{sin^2x} = 1 + \frac{1}{tan^2x}$

solving for tanx yields $tanx = \frac{2\sqrt{5}}{5}$ or $-\frac{2\sqrt{5}}{5}$

But how do you know which value to take? because sin is negative in 3rd and 4th quadrants, but tan is positive in 3rd and negative in 4th. What value do you take for tanx? the +ve or -ve?
Thanks guys!

GerrySly · « **Reply #293 on:** March 10, 2009, 06:48:10 pm »

Quote from: TrueTears on March 10, 2009, 04:55:36 pm

Ok just a simple question which I'm not sure about.

simplify:
$tan(sin^{-1}(\frac{-2}{3}))$

so let $sin^{-1}(\frac{-2}{3}) = x$

$sinx = \frac{-2}{3}$

using the identity $\frac{1}{sin^2x} = 1 + \frac{1}{tan^2x}$

solving for tanx yields $tanx = \frac{2\sqrt{5}}{5}$ or $-\frac{2\sqrt{5}}{5}$

But how do you know which value to take? because sin is negative in 3rd and 4th quadrants, but tan is positive in 3rd and negative in 4th. What value do you take for tanx? the +ve or -ve?
Thanks guys!

I would take a different approach and think about right angle triangles. sin is $\frac{O}{H}$ , so that means that O = 2, and H = 3, so therefore we can find

$\begin{align*}A &= \sqrt {3^2 - 2^2}\\&= \sqrt{5}\end{align*}$

Now we know that $\tan (\theta) = \frac{O}{H}$ , so therefore, $\tan (sin^{-1} (\frac{-2}{3})) = \frac{2}{\sqrt{5}}$

Just visualise the triangles, then you can derive any trigonometric function from that

TrueTears · « **Reply #294 on:** March 10, 2009, 06:52:20 pm »

yeap, cool thanks gerrysly

TrueTears · « **Reply #295 on:** March 10, 2009, 07:50:11 pm »

Also /0 why is AB = CD?

Quote from: /0 on March 09, 2009, 11:25:57 pm

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$

Mao · « **Reply #296 on:** March 10, 2009, 08:06:49 pm »

Quote from: TrueTears on March 10, 2009, 04:55:36 pm

Ok just a simple question which I'm not sure about.

simplify:
$tan(sin^{-1}(\frac{-2}{3}))$

so let $sin^{-1}(\frac{-2}{3}) = x$

$sinx = \frac{-2}{3}$

using the identity $\frac{1}{sin^2x} = 1 + \frac{1}{tan^2x}$

solving for tanx yields $tanx = \frac{2\sqrt{5}}{5}$ or $-\frac{2\sqrt{5}}{5}$

But how do you know which value to take? because sin is negative in 3rd and 4th quadrants, but tan is positive in 3rd and negative in 4th. What value do you take for tanx? the +ve or -ve?
Thanks guys!

Remembering that $x\in \left[ -\frac{\pi}{2},\frac{\pi}{2}\right]$ for arcsin, since you are taking an arcsin of a negative number, your answer will be in quadrant 4.

Hence, tan of that angle will be negative.

This also applies to GerrySly's response.

TrueTears · « **Reply #297 on:** March 10, 2009, 08:07:51 pm »

yeah thanks mao

Mao · « **Reply #298 on:** March 10, 2009, 08:12:13 pm »

Quote from: TrueTears on March 10, 2009, 07:50:11 pm

Also /0 why is AB = CD?

Quote from: /0 on March 09, 2009, 11:25:57 pm

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$

not sure why he did that but it's incorrect.

$|AC| = |BC|\implies (u + v) \cdot (u + v) = v \cdot v \implies 2 u \cdot v = - u \cdot u$

$DA \cdot AB = (u + 2v) \cdot u = u \cdot u + 2 u \cdot v = u \cdot u - u \cdot u = 0$

$\implies DA \perp AB$

TrueTears · « **Reply #299 on:** March 10, 2009, 08:15:56 pm »

Quote from: Mao on March 10, 2009, 08:12:13 pm

Quote from: TrueTears on March 10, 2009, 07:50:11 pm
Also /0 why is AB = CD?

Quote from: /0 on March 09, 2009, 11:25:57 pm

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$

not sure why he did that but it's incorrect.

$|AC| = |BC|\implies (u + v) \cdot (u + v) = v \cdot v \implies 2 u \cdot v = - u \cdot u$

$DA \cdot AB = (u + 2v) \cdot u = u \cdot u + 2 u \cdot v = u \cdot u - u \cdot u = 0$

$\implies DA \perp AB$

yeah that's what i did XD

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