Author Topic: TrueTears question thread (Read 66210 times) Tweet Share

TrueTears · « **Reply #45 on:** December 28, 2008, 06:07:46 pm »

yeah lol

ed_saifa · « **Reply #46 on:** December 28, 2008, 06:20:58 pm »

Quote from: TrueTears on December 28, 2008, 05:42:31 pm

and...

If $Arg z = \frac{\pi}{4}$ and $Arg(z-3) = \frac{\pi}{2}$ , find $Arg(z-6i})$ .

http://vcenotes.com/forum/index.php/topic,2787.msg34557.html#msg34557

TrueTears · « **Reply #47 on:** December 28, 2008, 07:47:31 pm »

o thanks very much

TrueTears · « **Reply #48 on:** December 28, 2008, 09:32:19 pm »

and just another Q, i got $\sqrt{\frac{1+cos(\alpha)}{1-cos(\alpha)}}$

How can i simplify that further?

My book does...

$\sqrt{(\frac{1+cos(\alpha)}{1-cos(\alpha)}) (\frac{1+cos(\alpha)}{1+cos(\alpha)})}$

why do u multiply by $\frac{1+cos(\alpha)}{1+cos(\alpha)}$ ? its not rationalising or anything is it?

Why can't u multiply it by $\frac{1-cos(\alpha)}{1-cos(\alpha)}$ ?

vce08 · « **Reply #49 on:** December 28, 2008, 09:35:31 pm »

cos the base becomes 1 - (cos(x))^2 which is equal to (sin(x))^2 when u use the book thingy

humph · « **Reply #50 on:** December 28, 2008, 09:59:26 pm »

Quote from: TrueTears on December 28, 2008, 09:32:19 pm

and just another Q, i got $\sqrt{\frac{1+cos(\alpha)}{1-cos(\alpha)}}$

How can i simplify that further?

My book does...

$\sqrt{(\frac{1+cos(\alpha)}{1-cos(\alpha)}) (\frac{1+cos(\alpha)}{1+cos(\alpha)})}$

why do u multiply by $\frac{1+cos(\alpha)}{1+cos(\alpha)}$ ? its not rationalising or anything is it?

Why can't u multiply it by $\frac{1-cos(\alpha)}{1-cos(\alpha)}$ ?

$\sqrt{\frac{1+\cos(\alpha)}{1-\cos(\alpha)}} = \sqrt{\left(\frac{1+\cos(\alpha)}{1-\cos(\alpha)}\right) \left(\frac{1+\cos(\alpha)}{1+\cos(\alpha)}\right)}$
$= \sqrt{\frac{\left(1+\cos(\alpha)\right)^2}{1-\cos^2(\alpha)}}$
$= \sqrt{\frac{\left(1+\cos(\alpha)\right)^2}{\sin^2(\alpha)}}$
$= \frac{1+\cos(\alpha)}{\sin(\alpha)}$

$\sqrt{\frac{1+\cos(\alpha)}{1-\cos(\alpha)}} = \sqrt{\left(\frac{1+\cos(\alpha)}{1-\cos(\alpha)}\right) \left(\frac{1-\cos(\alpha)}{1-\cos(\alpha)}\right)}$
$= \sqrt{\frac{1-\cos^2(\alpha)}{\left(1-\cos(\alpha)\right)^2}}$
$= \sqrt{\frac{\sin^2(\alpha)}{\left(1-\cos(\alpha)\right)^2}}$
$= \frac{\sin(\alpha)}{1-\cos(\alpha)}$

So there are two ways to simplify it. The first way is "nicer" because the denominator is simpler. But it doesn't make that much of a difference...

TrueTears · « **Reply #51 on:** December 28, 2008, 10:09:21 pm »

ahh thanks humph

TrueTears · « **Reply #52 on:** December 29, 2008, 07:04:15 pm »

could someone please list the newton notation way to differentiate $sin^{-1}(x)$ , $cos^{-1}(x)$ and $tan^{-1}(x)$ . Like for product rule in newton notation it's just $f(x)g'(x) + g(x)f'(x)$ . My book doesn't make it very clear how to differentiate the inverse of sin, cos, tan. yeah so pretty much could someone please teach me how to differentiate those 3 inverses. lol

Matt The Rat · « **Reply #53 on:** December 29, 2008, 07:22:19 pm »

There's a couple of different ways to do it.

For $sin^{-1}(x) , cos^{-1}(x) and tan^{-1}(x)$ the derivatives are given on the formulae sheet.

When they become slightly more complex (ie, $sin^{-1}(\frac{3}{x}))$ then either than chain rule or implicit diff could be used

Chain Rule: Make the substitution $u = \frac{3}{x}$ and then $y = sin^{-1}(u)$

Implicit Diff: $y = sin^{-1}(\frac{3}{x}) => sin(y) = \frac{3}{x}$ Use that info to make a right angled triangle (with $x$ as the hypotenuse, $3$ as the opposite side and $\sqrt{x^2-9}$ and $y$ as the appropriately placed angle) as you'll need to work out $cos(y)$ for the derivative.

I haven't finished the examples completely but just shown the groundwork of how to approach the problem.

TrueTears · « **Reply #54 on:** December 29, 2008, 07:30:59 pm »

yeah, im just beginning the differentiations for inverse trig functions, i think i get how to do the basic differentiations now, but theres one question in the an example i dont quite get.

Differentiate with respect to x
$sin^{-1}(x^2-1)$

Let $y = sin^{-1}(x^2-1)$ and $u = x^2-1$

so $\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}}$ x $2x$

$\implies \frac{2x}{\sqrt{1-(x^2-1)^2}}$

after simplification it becomes $\frac{2x}{\sqrt{x^2}\sqrt{2-x^2}}$

then this is where i don't understand. My book says $\frac{dy}{dx} = \frac{2}{\sqrt{2-x^2}}$ if $0 < x< \sqrt{2}$

and $\frac{dy}{dx} = \frac{-2}{\sqrt{2-x^2}}$ if $-\sqrt{2} < x < 0$ .

How do u get 2 answers for $\frac{dy}{dx}$ ? And how do u get the restrictions for $x$ , ie $0 < x< \sqrt{2}$ and $-\sqrt{2} < x < 0$ .

Matt The Rat · « **Reply #55 on:** December 29, 2008, 07:42:40 pm »

$\frac{2x}{\sqrt{x^2}\sqrt{2-x^2}}$

$=> \frac{2x}{x\sqrt{2-x^2}}$

$=> \frac{2}{\sqrt{2-x^2}}$

$= \frac{dy}{dx}$

The domain of the derivative is to do with where the $sin^{-1}$ function is defined.

As its domain initially was within $[-1,1]$ then

$-1 < x^2 - 1 < 1$

$\therefore -\sqrt{2} < x < \sqrt{2}$

The terminals are not included as you would be dividing by 0 in the derivative.

TrueTears · « **Reply #56 on:** December 29, 2008, 07:46:36 pm »

ahh yes i think i understand, but wat about the 0, how do u know to split it at $0 < x < \sqrt{2}$ and $-\sqrt{2} < x < 0$

and how come $\frac{dy}{dx} = \frac{-2}{\sqrt{2-x^2}}$ if $-\sqrt{2} < x < 0$ . How do u get the negative?

humph · « **Reply #57 on:** December 29, 2008, 09:28:38 pm »

Should be
$\frac{dy}{dx} = \frac{2x}{\sqrt{x^2}\sqrt{2-x^2}} = \frac{2x}{|x|\sqrt{2-x^2}}$
Now remember that
$\frac{x}{|x|} = \begin{cases} 1 & \text{if $x > 0$}\\ -1 & \text{if $x < 0$}\\ \text{undefined} & \text{if $x=0$} \end{cases}$

Ahmad · « **Reply #58 on:** December 29, 2008, 10:12:15 pm »

Quote from: humph on December 28, 2008, 09:59:26 pm

$\frac{\sin(\alpha)}{1-\cos(\alpha)}$

This is probably perfectly acceptable, but it might be pursued further:

$\frac{\sin(\alpha)}{1-\cos(\alpha)} = \frac{2\sin(\alpha/2) \cos(\alpha/2)}{1-(1 - 2\sin^2(\alpha/2))} = \cot(\alpha/2)$

TrueTears · « **Reply #59 on:** December 30, 2008, 04:36:04 pm »

ah thanks.

also,

1. show that the parabola with equation $ax^2+bx+c$ has no points of inflexion.

2. Determine the points of inflexion for the following function:

$(x^2-1)(x^2+1) = y$

$\implies \frac{dy}{dx} = 4x^3$

$\implies \frac{d^2y}{dx^2} = 12x^2$

so point of inflexion occurs when $\frac{d^2y}{dx^2} = 0$ $\implies x = 0$ . But this is not a point of inflexion, why is that?

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