Prelim 2016 Maths Help Thread (4 Viewers)

Green Yoda · May 23, 2016

leehuan said:
We ALL ALREADY know.

And learn to adapt to different ways of writing the same thing

davidgoes4wce · May 23, 2016

eyeseeyou said:
Ans are (for v=v(x))
1. x=3
2. max occurs when a=0, e^x/2(cosx-sinx)=0...tanx=1,...x=pi/4
3. a. x''=-1024/(x^3) = pi/4
b. the particle slows down to x=8

a=a(x)
1.a. v=2e^(-x/2)
b. v approaches 0 (as in v arrow 0)
3. a. 4squrroot of pi/3
b. Squreroot of 2pi/3

For some reason I didn't get an answer of x=3 for the first one.

eyeseeyou · May 23, 2016

davidgoes4wce said:
For some reason I didn't get an answer of x=3 for the first one.

I think it's the wrong answers

Did anyone else not get x=3 for the first one? Shuuya? Leehuan? Integrand (where r u)?

eyeseeyou · May 23, 2016

leehuan said:
We ALL ALREADY know.

And learn to adapt to different ways of writing the same thing

Calm down man I was just reminding ppl

eyeseeyou · May 23, 2016

leehuan said:
Well have you learnt pretty much all the other 2U topics? Cause motion is usually the LAST taught 2U topic

Also I'm teasing because Shuuya's handwriting was what sparked this whole debate and lead Nailgun to believing she was a guy

How about you stop worrying about other people's genders. THere's nothing interesting about it okay

eyeseeyou · May 23, 2016

BTW Shuuya, I think this is great practice before your trials

davidgoes4wce · May 23, 2016

$Here is my working out$

$v=\frac{cos^2 x}{1+sin x cos x }$

$\frac{dx}{dt}=\frac{cos^2 \ x}{1+sin x \ cos x }$

$Rearranging the terms$

$\frac{1+sin x \ cos \ x}{cos^2 x} dx=dt$

$\frac{1}{cos^2 x}+\frac{sin x \ cos x}{cos^2 \ x} \ dx = dt$

$\int \frac{1}{cos^2 x}+\frac{sin x \ cos x}{cos^2 \ x} \ dx= \int \ dt$

$\int \ sec^2{x} + \frac{sin x}{cos x} = \int \ dt$

$tan \ x - ln |cos x| + C = t$

$$ Given that the particle goes through the origin at P (0,0) we can solve for C$

$\therefore C=0$

$$ By letting $ x=\frac{\pi}{4} ,$

$tan \frac{\pi}{4}-ln |cos \frac{\pi}{4}| = t$

$1-ln |\frac{1}{\sqrt{2}}| =t$

eyeseeyou · May 23, 2016

davidgoes4wce said:
$Here is my working out$

$v=\frac{cos^2 x}{1+sin x cos x }$

$\frac{dx}{dt}=\frac{cos^2 \ x}{1+sin x \ cos x }$

$Rearranging the terms$

$\frac{1+sin x \ cos \ x}{cos^2 x} dx=dt$

$\frac{1}{cos^2 x}+\frac{sin x \ cos x}{cos^2 \ x} \ dx = dt$

$\int \frac{1}{cos^2 x}+\frac{sin x \ cos x}{cos^2 \ x} \ dx= \int \ dt$

$\int \ sec^2{x} + \frac{sin x}{cos x} = \int \ dt$

$tan \ x - ln |cos x| + C = t$

$$ Given that the particle goes through the origin at P (0,0) we can solve for C$

$\therefore C=0$

$$ By letting $ x=\frac{\pi}{4} ,$

$tan \frac{\pi}{4}-ln |cos \frac{\pi}{4}| = t$

$1-ln |\frac{1}{\sqrt{2}}| =t$

Shuuya did you get this right?

Shuuya · May 23, 2016

eyeseeyou said:
Shuuya did you get this right?

I haven't done it

davidgoes4wce · May 23, 2016

eyeseeyou said:
Some more motion questions:

Motion defined by v=v(x)

2.The velocity of a particle moving on the x-axis is given by v= (squareroot of sinx) e^(x/2), 0≤x≤pi/3, where x is the displacement of the particle from the origin
a.Show that the acceleration of the particle in terms of its displacement can be expressed by a=1/2 e^(-x) (cosx-sinx)
b.Hence find the displacement of the particle at which the maximum speed occurs

$$ With this question are you sure it is not $ v=\sqrt{sin \ x} \ e^{\frac{-x}{2}}$

$In your question you have used $ e^{\frac{x}{2}}$ (a positive x instead of a negative)$

leehuan · May 23, 2016

eyeseeyou said:
How about you stop worrying about other people's genders. THere's nothing interesting about it okay

Then you shouldn't worry about if she's done a topic either.

leehuan · May 23, 2016

Ok I should apologise. There is the factor of hunger that's fueling my temper right now.

eyeseeyou · May 23, 2016

leehuan said:
Then you shouldn't worry about if she's done a topic either.

just to make sure she's doing it right (I still keep thinking Shuuya is a guy even though she's a girl)

eyeseeyou · May 23, 2016

leehuan said:
Ok I should apologise. There is the factor of hunger that's fueling my temper right now.

It's okay

eyeseeyou · May 26, 2016

Help me with this easy question: sin2A/(1+cos2A)

Shuuya · May 26, 2016

eyeseeyou said:
Help me with this easy question: sin2A/(1+cos2A)

What's the question lol (are you asking for it to be simplified?)

eyeseeyou · May 26, 2016

Shuuya said:
What's the question lol (are you asking for it to be simplified?)

yup I'm just a bit blind right now and can't seeem to figure it out

Drongoski · May 26, 2016

Shuuya · May 26, 2016

eyeseeyou said:
Help me with this easy question: sin2A/(1+cos2A)

Oops I got beaten

eyeseeyou · May 26, 2016

Shuuya said:
Oops I got beaten

Thanks Shuuya

Prelim 2016 Maths Help Thread (4 Viewers)

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