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Re: HSC 2012 Marathon :) Yes heh, Im lacking creativity right now, so I decided to make an easy one Post a question!

Re: HSC 2012 Marathon :) v^2=5(4-x^2) \\ \frac{1}{2}v^2=\frac{5}{2}(4-x^2) \\ \left(\frac{1}{2}v^2\right)\frac{d}{dx}=-5x \\ \therefore \ddot{x}=-5x \therefore \text{Motion is Simple Harmonic, since acceleration is in the form} \ \ \ \ddot{x}=-n^2x Period=\frac{2\pi}{n} \\ \\ n=\sqrt{5} \\...

Thanks to the most random comment ever made in this forum

\log_b{a}=\log_a{b} \\ \frac{\log{a}}{\log{b}}=\frac{\log{b}}{\log{a}} \\ \\ \text{Case 1:}\log^2{a}=\log^2{b} \\ \\ \log{a}=+\log{b} \\ \\ \log{a}-\log{b}=0 \\ \\ \log{\frac{a}{b}}=0 \\ \\ \therefore \frac{a}{b}=1 \ \ \ \ a=b???? \\ \\ \\ \text{Case 2:}\log^2{a}=\log^2{b} \\ \\ \log{a}=-\log{b}...

It cant hurt to try Mathematics....

Cant see the pictures man. 20. \ \frac{d}{dx}\frac{10}{\ln{x}} \\ \\ = \frac{d}{dx}10(\ln{x})^{-1} \\ = 10 \times \frac{1}{x} \times (ln^{-2}{x}) \\ = \frac{10}{x\ln^2{x}} EDIT: 23. \ \frac{d}{dx} \frac{2-3\ln{x}}{2+\ln{x}}.....[1] \\ \\ u=2-3\ln{x} \ \ \ u'=\frac{-3}{x} \\ v=2+3\ln{x} \ \...

So we can assume that it starts at the end point? I dont think you can.

Is that the whole question, because for part b, the max speed occurs at the centre of oscillation (equilibrium), which is at x=0 according to acceleration=-49x So what I have attempted to do is integrate using (1/2 v^2)d/dx = -49x However we have no information to find +C Because we dont know...

http://community.boredofstudies.org/showthread.php?t=234261 Very comprehesive guide :D

\text {Yes, very much so.}

Well you have to bound the commands with the commands [$tex] and at the end [/tex$] There are NO dollar signs here. I just did that otherwise it will think Im writing latex

Haha, yes I like your name too. I gradually learnt latex through: http://www.codecogs.com/latex/eqneditor.php I kept using the buttons to make the image, and gradually I learnt all the commands by heart, and now I just use the tex /tex editor here

f(x)=x^3 \\ f(x+h)=(x+h)^3 \\ \\ \frac{dy}{dx}=\lim_{h\rightarrow 0} \frac{(x+h)^3-x^3}{h} \\ \\ = \lim_{h \rightarrow 0} \frac{3x^2h+3xh^2+h^3}{h} \\ \\ =\lim_{h \rightarrow 0} 3x^2+3xh+h^2 \\ \\ =3x^2 \\ \\ \therefore \frac{dy}{dx}=3x^2 \\ \\ x=1 \Rightarrow \frac{dy}{dx} = 3(1)^2=3 I hope...

There is nothing wrong with the answer because when you normally expand it, the largest coeffiecient is indeed 240, coming from the term \binom{6}{2} \cdot 2^4 Now as you can see, the "k" here is actually 2. So Im guessing it is something to do with the fact that you proved T_{k}>T_{k-1}...

Re: HSC 2012 Marathon :) Yes I got that too, I didnt need to use calculus either. Simply find initial height, then the height at 40 degrees, find their difference, times by 100 to get cm and you get 66.971 cm, at 1cm/s. It means ~67 seconds, minus the 5, then ~62 seconds

Re: HSC 2012 Marathon :) That is what I was aiming for. But anyway, I will probably stay away from word questions for a while, they confuse me as much as they confuse others. So someone else ask a question!

Re: HSC 2012 Marathon :) \\ \\ x''=0 \\ x'=10\sqrt{2} \\ x=10\sqrt{2}t \ \ \ \ [1] \\ \\ y''=-10 \\ y'=-10t+10\sqrt{2} \\ y=-5t^2+10sqrt{2}t \ \ \ \ [2] \\ \\ [1]=[2] \\ y=-5(\frac{x}{10\sqrt{2}})^2+x \\ \\ y=x-\frac{x^2}{40} \\ \\ \frac{dy}{dx}=1-\frac{x}{20} \\ \\ \\...

Re: HSC 2012 Marathon :) Did you use tan(-15) or tan(15) because that was the trick I was aiming at when I said while the particle is decreasing in height Ive rechecked my solution and I am pretty sure its correct. I substituted this into my x in terms of t. x=-20(tan(-15)-1) From...

Re: HSC 2012 Marathon :) I keep getting the same answer of 1.79 seconds EDIT: It is supposed to say 15 degrees to the horizontal, man Im sorry

Re: HSC 2012 Marathon :) A particle is launched in projectile motion from the origin, at an initial velocity of 20 m/s. At an angle 45 degrees. Given that: x''=0 \\ \\ y''=-10 Find the time when the particle is decreasing in height AND is at an angle of 15 degrees to the horizontal. Give...