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Over9000 · « **Reply #555 on:** September 06, 2009, 09:17:51 pm »

Quote from: TrueTears on September 06, 2009, 09:17:00 pm

ok.

Then according to /0, the VCAA examiners were wrong.

Rofl

/0 · « **Reply #556 on:** September 06, 2009, 09:38:15 pm »

Quote from: Over9000 on September 06, 2009, 09:17:51 pm

Quote from: TrueTears on September 06, 2009, 09:17:00 pm
ok.

Then according to /0, the VCAA examiners were wrong.
Rofl

forget it

Mao · « **Reply #557 on:** September 06, 2009, 10:57:28 pm »

I don't think they assess the exam at that level of technicality.

TrueTears · « **Reply #558 on:** September 14, 2009, 12:20:45 am »

How to find $\lim_{t \to 0} \frac{3-\sqrt{9-4t^2}}{2t}$

and why is $\lim_{t \to 0} \frac{3+\sqrt{9-4t^2}}{2t}$ undefined?

/0 · « **Reply #559 on:** September 14, 2009, 12:56:21 am »

For the top you can use L'hopital's rule, since the indeterminate form $\frac{0}{0}$ is reached. For the bottom one, you have $\frac{6}{0}$ , which is not an indeterminate form and diverges to infinity.

TrueTears · « **Reply #560 on:** September 14, 2009, 01:01:54 am »

Oh I get the 2nd one now, yeah what about not using L'hopitals?

Mao · « **Reply #561 on:** September 14, 2009, 01:04:21 am »

Quote from: TrueTears on September 14, 2009, 12:20:45 am

How to find $\lim_{t \to 0} \frac{3-\sqrt{9-4t^2}}{2t}$

and why is $\lim_{t \to 0} \frac{3+\sqrt{9-4t^2}}{2t}$ undefined?

be aware that for the question you are applying this to, the limit is only the left-hand limit (as specified by the domain $t\in (-\infty,0]$ , y must also be negative, and 't', in this case, carries the values of y), i.e. $t \to 0^-$ , hence the second limit is equivalent to $\lim_{t \to 0^-} \frac{6}{t} = -\infty$

TrueTears · « **Reply #562 on:** September 14, 2009, 01:05:07 am »

Quote from: Mao on September 14, 2009, 01:04:21 am

Quote from: TrueTears on September 14, 2009, 12:20:45 am
How to find $\lim_{t \to 0} \frac{3-\sqrt{9-4t^2}}{2t}$

and why is $\lim_{t \to 0} \frac{3+\sqrt{9-4t^2}}{2t}$ undefined?

be aware that for the question you are applying this to, the limit is only the left-hand limit (as specified by the domain $t\in (-\infty,0]$ ), i.e. $t \to 0^-$ , hence the second limit is equivalent to $\lim_{t \to 0^-} \frac{6}{t} = -\infty$

Yeap, thanks Mao!

Mao · « **Reply #563 on:** September 14, 2009, 01:11:02 am »

The first one can be kind of 'squeezed'.

$y = \frac{2t}{t^2 + 1}$ , $t\in (-\infty,0] \implies y \le 0$

$\lim_{y \to 0^-} \frac{3 - \sqrt{9-4y^2}}{2y} = L$

firstly, $\sqrt{9-4y^2} \le \sqrt{9}$ , this implies $L \ge \frac{0}{2y} = 0$

secondly, we note that $L \in t$ since it is an expression for the inverse and $0^- \in y$ . Hence another condition on L is $L \in (-\infty, 0] \equiv L \le 0$

combining the two conditions $L\ge 0$ and $L \le 0$ , we arrive at $L = \lim_{y \to 0^-} \frac{3 - \sqrt{9-4y^2}}{2y} = 0$

TrueTears · « **Reply #564 on:** September 28, 2009, 01:28:40 am »

Solve this differential equation:

$x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + (4 + \pi^2)y = \log_e(x)$

Given $y(1) = 0$ and $y(e) = 0$

sachinmachin · « **Reply #565 on:** September 28, 2009, 07:57:42 am »

Quote from: TrueTears on September 28, 2009, 01:28:40 am

Solve this differential equation:

$x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + (4 + \pi^2)y = \log_e(x)$

Given $y(1) = 0$ and $y(e) = 0$

Shouldn't an equation for "y" be specified, so that we are able to do the derivative/double derivative??
I dont think theres any other way you can do it.

Mao · « **Reply #566 on:** September 28, 2009, 10:44:24 am »

that is a second order non-homogenous. [absolutely not in VCE]

$x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + (4 + \pi^2) y = \log_e (x)\; \; \; (1)$

It can be seen here that a particular solution is in the form $y_p (x) = \frac{\log_e (x)}{4 + \pi^2} + B$ , where the first and second derivative multiplied by 'x' and 'x^2' coefficients will give constants. Solving this gives $B = -\frac{4}{(4+\pi^2)^2}$

Now, to solve the homogenous part, $\frac{d^2y}{dx^2} + \frac{5}{x} \frac{dy}{dx} + \frac{4+\pi^2}{x^2} y = 0$
since the coefficients are not constants (they are functions of x), we cannot simply use the characteristic equation. To derive a solution from this will be very very hard, and at this point most problem solvers will go forth and use power series, or specifically, the Forbenius method. This stuff you learn in 2nd/3rd year at uni.

So yeah... bitch of a question.

Mathematica tells me the general solution is $\frac{-4 + (4 + \pi^2) \log_e x}{(4 + \pi^2)^2} + \frac{<br /> C_2 \cos(\pi \log_e x) + C_1 \sin(\pi \log_ex)}{x^2}$

And for this boundary value problem, there's no solution.

TrueTears · « **Reply #567 on:** September 28, 2009, 07:37:53 pm »

Thanks Mao, I was just reading some stuff on non-homogenous DE's and tried to fiddle around with change of variables etc. Didn't work out =S

Again thanks!

TrueTears · « **Reply #568 on:** October 09, 2009, 12:17:30 am »

What's the systematic way of working out $\tan(\theta) = \sqrt{3}-2$

Damo17 · « **Reply #569 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:

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

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