Author Topic: BEC'S methods questions (Read 127942 times) Tweet Share

Collin Li · « **Reply #30 on:** January 02, 2008, 09:32:48 pm »

Just so I don't appear like a robot:

No problem (to all the thanks you've given)! I enjoy doing these, it's a fun way to keep my mind exercised during the lazy holidays

bec · « **Reply #31 on:** January 02, 2008, 09:34:17 pm »

haha ok that's good to hear! well here's the last bit of excercise your mind will be getting from me tonight...

i was working out a problem and got to a point where i had:
$-1.1t+lnt=ln\frac{0.57}{3}$
my best mate ti-89 got the answer out in 2 seconds but how would i find the solution without a calc? or is it impossibly complicated?

Collin Li · « **Reply #32 on:** January 02, 2008, 09:37:26 pm »

You can't do this in VCE without the help of a calculator. You can do it without the CAS (TI-89), however, just incase you're doing non-CAS Maths Methods.

(I don't even know how to do this without a calculator, but I'm sure Ahmad would)

bec · « **Reply #33 on:** January 02, 2008, 09:42:18 pm »

yeah i'm doing cas
phewww...although i'm still learning how to work the calculator
and to be honest, if I don't have to know how to solve that equation without my gc i'm happy not knowing. doing unit 3 in 3 days has been hard enough for me, i need no more challenges!

Collin Li · « **Reply #34 on:** January 02, 2008, 09:46:45 pm »

That's pretty dedicated! I didn't even know about the chain rule until a few weeks into term 1.

Ahmad · « **Reply #35 on:** January 02, 2008, 10:03:17 pm »

Quote from: bec on January 02, 2008, 09:34:17 pm

haha ok that's good to hear! well here's the last bit of excercise your mind will be getting from me tonight...

i was working out a problem and got to a point where i had:
$-1.1t+lnt=ln\frac{0.57}{3}$
my best mate ti-89 got the answer out in 2 seconds but how would i find the solution without a calc? or is it impossibly complicated?

$-1.1t + \log_e t = \log_e \frac{0.57}{3}$

$\implies e^{-1.1t + \log_e t} = \frac{0.57}{3}$

$\implies t\cdot e^{-1.1t} = \frac{0.57}{3}$

$\implies (-1.1t)\cdot e^{(-1.1t)} = \frac{-1.1\cdot 0.57}{3}$

$\implies -1.1t = W( \frac{-1.1\cdot 0.57}{3} )$

$\implies t = -\frac{1}{1.1} W( \frac{-1.1\cdot 0.57}{3} ) \approx 0.25019$

Where W is the Lambert W function

enwiabe · « **Reply #36 on:** January 03, 2008, 01:21:11 am »

Ahmad... what the FUCK... *faints*

This guy is maths god.

HEAR ME NOW:
Ahmad: Official Mathematics GOD of FSN.

Collin Li · « **Reply #37 on:** January 03, 2008, 01:29:15 am »

Heh, the Lambert W function is just the inverse of $f(x) = xe^x$ . It's not a very meaningful exact form, but it has a lot of applications, so computers would be trained to compute it well. I had that in mind, but I wasn't sure whether you could get a closed-form answer to it.

Disclaimer: none of this is VCE material.

AppleXY · « **Reply #38 on:** January 03, 2008, 02:17:59 pm »

The Lambert W function is an awesome piece (Y) Ahmad showed a great example of it

bec · « **Reply #39 on:** January 03, 2008, 07:55:56 pm »

interesting...i admit i didn't understand a word of it...but it's interesting anyway haha

taking it down a notch, can anyone explain this to me?

Find, correct to 2 decimal places, the area enclosed by the curve with equation $y=tanx$ , the x-axis and the lines $x =-\frac{pi}{4}$ and $x=\frac{pi}{4}$

I know how to do it once I've got the integral, but I can't figure it out - I'm assuming you have to make it $\int\frac{sinx}{cosx}dx$ but even with that i'm struggling...

merci!

Collin Li · « **Reply #40 on:** January 03, 2008, 08:37:33 pm »

It is possible in VCE Specialist Maths to evaluate this integral.

However, this is Mathematical Methods, and the question says "to 2 decimal places." This means, get out your calculator and plug in $\tan x$ and then integrate from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$ . No need to integrate this. (EDIT: Not fully incorrect, see post below)

For curiosity's sake: here is how you would integrate $\tan x$ in Specialist Maths (a quite common integral)

$\int\tan x\,dx = \int\frac{\sin x}{\cos x}\,dx$

$\mbox{Let }u = \cos x \implies \frac{du}{dx} = -\sin x$

$\implies \int\frac{\sin x}{\cos x}\,dx = -\int\frac{1}{u}\,du = -\log_e|u| + C = -\log_e|\cos x| + C$

Collin Li · « **Reply #41 on:** January 03, 2008, 08:45:07 pm »

Actually, I was wrong! I think it's talking about the area. When looking for the area, we have to remember that we take the negative areas, make them positive, and then add up all the positive areas and previously-negative areas together.

$\tan x$ has an x-intercept at $x=0$ in between $-\frac{\pi}{4}$ and $\frac{\pi}{4}$ , so you have to integrate from $-\frac{\pi}{4}\mbox{ to }0$ , take a note of that number (3-4 decimal places will do), and then integrate from $0\mbox{ to }\frac{\pi}{4}$ . Take the positive value of both of the "areas" and add them up.

On a side-note, if you are smart, you will notice that $\tan x$ is an odd function, that is: $\tan x = -\tan (-x)$ . This means that the magnitude of $\displaystyle\int^{\frac{\pi}{4}}_{0} \tan x\,dx$ is equal to $\displaystyle\int^{0}_{-\frac{\pi}{4}} \tan x\,dx$ .

That is: $\left|\displaystyle\int^{\frac{\pi}{4}}_{0} \tan x\,dx\right| = \left|\displaystyle\int^{0}_{-\frac{\pi}{4}} \tan x\,dx\right|$ . You could just integrate from $0\mbox{ to }\frac{\pi}{4}$ , make it positive (should already be), then double it.

Sorry! A bit more confusing this time. $:-\$

bec · « **Reply #42 on:** January 03, 2008, 10:39:40 pm »

Quote from: coblin on January 03, 2008, 08:45:07 pm

if you are smart, you will notice that $\tan x$ is an odd function, that is: $\tan x = -\tan (-x)$ . This means that the magnitude of $\displaystyle\int^{\frac{\pi}{4}}_{0} \tan x\,dx$ is equal to $\displaystyle\int^{0}_{-\frac{\pi}{4}} \tan x\,dx$ .

haha well i didn't notice. does that mean i'm not smart?
i understand what you're saying though...makes sense! i kind of wish i was doing spesh when i see things like this though - it's interesting stuff! although i think i'm preaching to the converted when i say that...

bec · « **Reply #43 on:** January 03, 2008, 11:00:53 pm »

if anyone has mathsworld, this is question 1 on page 335...
I got the first few parts but i'm stuck on part (d) and (e)

The entrance to a disused train tunnel of constant cross-sectional area is shown in the diagram. (The diagram is of a parabola, $y=-\frac{1}{6}x^{2}+2x$ Domain [0,12] )
For the next few parts I had to find out:
...Area under the curve (48m²)
...Volume of concrete required to fill the tunnel if it's 8m long (383m³)

And then...
"The council gets a quote for filling the tunnel with concrete and finds that it is too expensive. (don't you love the little anecdotes?) The engineer investigates possible alternatives and determines that access to the tunnel can be prevented if the tunnel is completely filled with concrete to a height of 3 metres and then wire mesh used to cover the remaining area at the entrance and exit.

(d) What volume of concrete will be required if the council agrees to the engineer's proposal?

(e) What area of mesh is required?

Is that clear? Because haha this picture isn't very! (can you tell i'm an art student?)
Basically it's saying, find the area of the concrete bit in my masterpiece below and multiply it by 8. How can you get the area though? Then for e, find the area of the top of the arch formed by the parabola and multiply that figure by 2 - does that sound right?

edit: the picture didn't show for some reason but it was just a diagram of my interpretation of the question: the graph of $y=-\frac{1}{6}x^{2}+2x$ Domain [0,12] and $y=3$ on the same set of axes with pretty little labels informing of "mesh" and "concrete". you're missing out, trust me.

Collin Li · « **Reply #44 on:** January 03, 2008, 11:11:49 pm »

You have to find the area in between the curves $y=-\frac{1}{6}x^2 + 2x$ and $y=3$ . Then subtract that area from 48 (because the area in between those curves gets you the top part - where the mesh is supposed to be), and multiply that area by 8 to get the volume. You have to find when the curves intersect (2 solutions), and integrate from there to there.

The area of mesh required is just the area you found in part d times 2 (2 for entrance and exit).

I will draw a graph if need-be, but I'm getting sloppy tonight.

EDIT: This post has been edited for errors, please re-read it

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