Author Topic: dcc help me thread (Read 31491 times) Tweet Share

dcc · « **Reply #120 on:** June 18, 2008, 09:19:20 pm »

Let $u = \pi (h + 1) \implies \dfrac{du}{dh} = \pi$

Let $v = \left(\log_{e}(h + 1)\right)^2 + h$

Making the substitution $z = \log_{e}(h + 1) \implies v = z^2$

$\therefore \; \dfrac{dz}{dh} = \dfrac{1}{h + 1}$

$\therefore \; \dfrac{dv}{dz} = 2z$

By the chain rule, we get: $\dfrac{dv}{dh} = \dfrac{dz}{dh} \cdot \dfrac{dv}{dz} = 2z \cdot \dfrac{1}{h + 1} = \dfrac{2\log_{e}(h + 1)}{h + 1}$

By the product rule, therefore we have:

$\dfrac{dV}{dh} = u \cdot \dfrac{dv}{dh} + v \cdot \dfrac{du}{dh}$

Which gives: $\dfrac{dV}{dh} = \pi (h + 1) \cdot \left(\dfrac{2\log_{e}(h + 1)}{h + 1} + 1\right) + \pi \left(\left(\log_{e}(h + 1)\right)^2 + h\right)$

$\rightarrow \dfrac{dV}{dh} = \pi \left(1 + 2h + log_{e}(h + 1)\left(2 + log_{e}(h + 1)\right)\right)$

droodles · « **Reply #121 on:** June 21, 2008, 10:17:33 pm »

hi mao

Mao · « **Reply #122 on:** June 21, 2008, 10:19:05 pm »

hey droodles

if we draw a line from C vertically downwards, and the point of intersection is X, then CX will be the same length as the dashed vertical line of BC to O

CX also exist in a right angled triangle, with a hypotenuse of 4 and angle of $\theta$

$\implies |CX|=4\sin \theta$

if we now call the where the dashed verticle line meets BC be Y, then $OY=CX=4\sin\theta$

OY also exist in a right angled triangle, with a hypotenuse of 4 and a side of $4\sin\theta$ , hence the other side, CY, can be found using pythagoras:

$\implies |CY|=\sqrt{16-16\sin^2 \theta}=4\sqrt{1-\sin^2\theta}=4\cos\theta$

hence $|BC|=2|CY|=8\cos\theta$

alternatively, you can not be an idiot like me and recognize that $\angle BCO=\theta$ , and get $4\cos \theta$ straight away for CY....
damn myself.

b)

area of trapezium is $A=\frac{1}{2}(A+B)\cdot H$

in this case, base is 8, top is $8\cos\theta$ , and height is [as we have already worked out] $4\sin\theta$

$A=\frac{1}{2}\cdot (8+8\cos\theta)\cdot 4\sin\theta=16\left(\sin\theta + \frac{1}{2}\sin 2\theta\right)$ (double angle formula)

now, it seems sensible to put some kind of domain restrictions on $\theta$ , and it seems obvious that it should be a positive angle, and not be more than a 90 degree angle $\implies 0\le \theta \le \frac{\pi}{2}$

differentiating A:

$\frac{dA}{d\theta}=16(\cos\theta + \cos 2\theta)$
solving for 0

$2\cos^2 \theta + \cos \theta -1=0 \implies \cos \theta = \frac{-1\pm \sqrt{1+8}}{4}=\left\{ -1, \frac{1}{2}\right\}$

now, looking back to our domain, it is obvious to see that $1 \ge \cos \theta \ge 0$ , hence, we discard the -1 option

and $\theta = \frac{\pi}{3}$

$\implies A=16\cdot \left(\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot \frac{\sqrt{2}}{2}\right)=12\sqrt{2}$

* Mao thanks lwine for pointing out the mistake, if only i can bloody notice the plus sign!!!

lwine · « **Reply #123 on:** June 21, 2008, 10:29:45 pm »

Also, in addition to Mao, note that $sin(90-\theta)=\dfrac{BC}{2\times 4}\Rightarrow cos(\theta)=\dfrac{BC}{8}$
Therefore $BC=8cos(\theta)$

The second part of this question looks fairly nasty:
The area of a trapezium $\Rightarrow A=\left(\dfrac{BC+AD}{2}\right)h$

With a method similar to what I used earlier to determine BC, h can be found to equal $4sin(\theta)$

Therefore: $\Rightarrow A=\left(\dfrac{8cos(\theta)+8}{2}\right)4sin(\theta)$
$=16sin(\theta)cos(\theta)+16sin(\theta)$
$\dfrac{dA}{d\theta}=-16sin^2(\theta)+16cos^2(\theta)+16cos(\theta)$
$=-16(1-cos^2(\theta))+16cos^2(\theta)+16cos(\theta)$
$=32cos^2(\theta)+16cos(\theta)-16$
Therefore $cos(\theta)=-1 \mbox{ or }\dfrac{1}{2}$
Can't be -1 (because that's equal to 90 degrees), so take $acos(\dfrac{1}{2})=\dfrac{\pi}{3}$ .
Sub that back into the previous Area equation, to get $12\sqrt{3}$ as the answer.

Mao, please check if I'm correct

Argh its still not the same as yours, I think, what did I do wrong, Mao?!
Also, are double angle formulas in the methods course??
Sorry, I've edited the answer a few times now

droodles · « **Reply #124 on:** June 21, 2008, 10:30:15 pm »

Mao · « **Reply #125 on:** June 21, 2008, 10:32:04 pm »

Quote from: lwine on June 21, 2008, 10:29:45 pm

Also, in addition to Mao, note that $sin(90-\theta)=\dfrac{BC}{2\times 4}\Rightarrow cos(\theta)=\dfrac{BC}{8}$
Therefore $BC=8cos(\theta)$

or mao sucks

i admit that

Mao · « **Reply #126 on:** June 21, 2008, 11:32:01 pm »

Quote from: lwine on June 21, 2008, 10:29:45 pm

Also, in addition to Mao, note that $sin(90-\theta)=\dfrac{BC}{2\times 4}\Rightarrow cos(\theta)=\dfrac{BC}{8}$
Therefore $BC=8cos(\theta)$

The second part of this question looks fairly nasty:
The area of a trapezium $\Rightarrow A=\left(\dfrac{BC+AD}{2}\right)h$

With a method similar to what I used earlier to determine BC, h can be found to equal $4sin(\theta)$

Therefore: $\Rightarrow A=\left(\dfrac{8cos(\theta)+8}{2}\right)4sin(\theta)$
$=16sin(\theta)cos(\theta)+16sin(\theta)$
$\dfrac{dA}{d\theta}=-16sin^2(\theta)+16cos^2(\theta)+16cos(\theta)$
$=-16(1-cos^2(\theta))+16cos^2(\theta)+16cos(\theta)$
$=32cos^2(\theta)+16cos(\theta)-16$
Therefore $cos(\theta)=-1 \mbox{ or }\dfrac{1}{2}$
Can't be -1 (because that's equal to 90 degrees), so take $acos(\dfrac{1}{2})=\dfrac{\pi}{3}$ .
Sub that back into the previous Area equation, to get $12\sqrt{3}$ as the answer.

Mao, please check if I'm correct

Argh its still not the same as yours, I think, what did I do wrong, Mao?!
Also, are double angle formulas in the methods course??
Sorry, I've edited the answer a few times now

yes, my bad. I dont know how i did what i did [its all fixed now]

I told you, I suck

droodles · « **Reply #127 on:** June 23, 2008, 09:56:22 pm »

dcc · « **Reply #128 on:** June 23, 2008, 10:01:25 pm »

droodles · « **Reply #129 on:** June 24, 2008, 04:50:19 pm »

simplify:
$\dfrac {5^{2n}\times 4^{3n}}{50^{3n}\times16^{-n}} \div (\frac {125}{16})^{n}$

Glockmeister · « **Reply #130 on:** June 24, 2008, 05:15:55 pm »

droodles · « **Reply #131 on:** June 26, 2008, 06:47:31 pm »

$e^{x-2}=5e^{6-x}$

prove via algebra that $x= 4+0.5log_e(5)$

Mao · « **Reply #132 on:** June 26, 2008, 06:55:19 pm »

Quote from: droodles on June 26, 2008, 06:47:31 pm

$e^{x-2}=5e^{6-x}$

prove via algebra that $x= 4+0.5log_e(5)$

$\begin{align*} e^{x-2} &= 5e^{6-x} \\ ln(e^{x-2}) &=ln(5)+ln(e^{6-x}) \\ x-2 &= ln(5)+(6-x)\\ 2x &= ln(5)+8 \\ x &= \frac{1}{2}ln(5) + 4 \end{align*}$

droodles · « **Reply #133 on:** June 26, 2008, 06:57:46 pm »

GOD DAMN IT THAT QUESTION WAS ON MY SAC FUCK SHIT ARGHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

it's ok

droodles · « **Reply #134 on:** July 08, 2008, 07:19:23 pm »

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