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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: DannyN on April 14, 2011, 06:14:08 pm

Title: DannyN's SM question thread =D
Post by: DannyN on April 14, 2011, 06:14:08 pm: hey guys thought it would be a better idea to make a thread since im always stuck :P

Vectors:
Given that OP=p, OQ=q and the points O,P,Q are not collinear, which one of the following points, whose position vectors are given, is not collinear with P and Q

a)1/2p+1/2q
b)3p-2q
c)p-q
d)1/3p+2/3q
or
e)2p-q

thanks~
Title: Re: DannyN's SM question thread =D
Post by: evaever on April 14, 2011, 08:33:06 pm: c)
Title: Re: DannyN's SM question thread =D
Post by: Greatness on April 14, 2011, 09:09:10 pm: Quote from: evaever on April 14, 2011, 08:33:06 pm
c)
Maybe explain it :P
p-q=-(q-p) which is the vector from Q to P. Which is therefore not colinear with either P or Q.
Title: Re: DannyN's SM question thread =D
Post by: evaever on April 14, 2011, 10:37:27 pm: Quote from: swarley on April 14, 2011, 09:09:10 pm
Quote from: evaever on April 14, 2011, 08:33:06 pm
c)
Maybe explain it :P
p-q=-(q-p) which is the vector from Q to P. Which is therefore not colinear with either P or Q.
?
Title: Re: DannyN's SM question thread =D
Post by: DannyN on April 14, 2011, 10:39:21 pm: Thanks guys ^^ ,okay next question :S

Find the modulus of 1+cos(2theta)+isin(2theta) where 0<theta<pi/2
Title: Re: DannyN's SM question thread =D
Post by: evaever on April 14, 2011, 10:48:07 pm: Quote from: DannyN on April 14, 2011, 10:39:21 pm
Thanks guys ^^ ,okay next question :S

Find the modulus of 1+cos(2theta)+isin(2theta) where 0<theta<pi/2
2cosx
Title: Re: DannyN's SM question thread =D
Post by: DannyN on April 14, 2011, 10:52:27 pm: can you please explain to me briefly how you did that? :P it's amazing how you get the answers so quickly ^^
Title: Re: DannyN's SM question thread =D
Post by: evaever on April 14, 2011, 10:57:57 pm: Quote from: DannyN on April 14, 2011, 10:52:27 pm
can you please explain to me briefly how you did that? :P it's amazing how you get the answers so quickly ^^
draw 1+z where z=cis2x
Title: Re: DannyN's SM question thread =D
Post by: evaever on April 14, 2011, 10:59:27 pm: the arg is x
Title: Re: DannyN's SM question thread =D
Post by: luffy on April 14, 2011, 11:39:26 pm: Just thought I would state things slightly more clearly - Below is how I would do this question. (Sorry if I made any errors).

$1 + \cos{2\theta} + i\sin{2\theta}$

$= 1 + \cos^2{\theta} - \sin^2{\theta} + i(2\sin{\theta}\cos{\theta})$

$= 2\cos^2{\theta} + 2\sin{\theta}\cos{\theta} \times i$

You can then factorise the expression by taking out a $2\cos{\theta}$ .

$= 2\cos{\theta}(\cos{\theta} + i\sin{\theta})$

$= 2\cos{\theta}(\textup{cis}{\theta})$

Recall that in the expression, $z = r\textup{cis}\theta$ , $|z| = r$ That same principle applies to this question, where r = $2\cos{\theta}$ .

$\therefore \textup{let } z = 2\cos{\theta}(\textup{cis}{\theta})$

$|z| = |2\cos{\theta}(\textup{cis}{\theta})|$

$= 2\cos{\theta}$

Hope I helped.
Title: Re: DannyN's SM question thread =D
Post by: brightsky on April 14, 2011, 11:46:15 pm: Another way:
sqrt((1 + cos(2t))^2 + (sin(2t)^2))
= sqrt(1 + 2cos(2t) + cos^2(2t) + cos^2(t))
= sqrt(2 + 2cos(2t))
= sqrt(4 cos^2(t))
= |2cos(t)|
Title: Re: DannyN's SM question thread =D
Post by: DannyN on April 15, 2011, 09:45:50 am: Awesome thanks guys! :D
Title: Re: DannyN's SM question thread =D
Post by: DannyN on April 25, 2011, 01:34:28 pm: another one :P
if acos^2(theta)+bsin^2(theta)=c then tan^2(theta) in terms of a,b and c is:
Title: Re: DannyN's SM question thread =D
Post by: luffy on April 25, 2011, 02:18:24 pm: My method seems a bit long-winded, but this would be my solution:

$a\cos^2{(\theta)} + b\sin^2{(\theta)} =c$

First, solve for $\cos^2{(\theta)}$ , therefore:

$a\cos^2{(\theta)} + b(1 - cos^2{(\theta)}) = c$

$\therefore(a-b)\cos^2{(\theta)} = c - b$

$\therefore\cos^2{(\theta)} = \frac{c - b}{a - b}$

Do the same for $\sin^2{(\theta)}$ , therefore:

$a(1 - \sin^2{(\theta)}) + b\sin^2{(\theta)} = c$

$(b - a)\sin^2{(\theta)} = c - a$

$\sin^2{(\theta)} = \frac{c - a}{b - a}$

Now, solve $\tan^2{(\theta)}$ :

$\tan^2{(\theta)} = \frac{\sin^2{(\theta)}}{\cos^2{(\theta)}}$

$\tan^2{(\theta)} = \frac{c - a}{b - a} \div \frac{c - b}{a - b}$

$\tan^2{(\theta)} = -\frac{c - a}{a-b} \times \frac{a - b}{c - b}$

$\tan^2{(\theta)} = -\frac{c - a}{c - b}$

Lets hope I made no mathematical errors.

Hope I helped.
Title: Re: DannyN's SM question thread =D
Post by: DannyN on April 25, 2011, 02:38:45 pm: thanks luffy! =D very helpful ^_^

stuck on another...
sin(2cos^-1(2x)) is defined for x an element of (root(2)/4,1/2)
it can be shown that sin(2cos^-1(2x)=ax(square root(1-b^2x^2) where a and b are positive constants
find a and b
sorry about the text i dont know how to write the square root sign
Title: Re: DannyN's SM question thread =D
Post by: luffy on April 25, 2011, 03:07:06 pm: Quote from: DannyN on April 25, 2011, 02:38:45 pm
thanks luffy! =D very helpful ^_^

stuck on another...
sin(2cos^-1(2x)) is defined for x an element of (root(2)/4,1/2)
it can be shown that sin(2cos^-1(2x)=ax(square root(1-b^2x^2) where a and b are positive constants
find a and b
sorry about the text i dont know how to write the square root sign

This ones tricky - This would be my solution:

Recall that $\sin{(2\theta)} = 2\sin{(\theta)}\cos{(\theta)}$

Therefore:
$\sin{(2\cos^{-1}{(2x)})} = 2\sin{(\cos^{-1}{(2x)})}\cos{(\cos^{-1}{(2x)})}$

Now, the cos and cos^-1 cancel each other out (for obvious reasons):
$\sin{(2\cos^{-1}{(2x)})} = 4x\sin{(\cos^{-1}{(2x)})}$

Now, I'm not sure if you know how to simplify matters when you have trigonometric functions of an inverse trig function.
In order to understand this, draw a right-angled triangle, and label one side '2x 'and the hypotenuse '1'. Therefore, cos^{-1} of 2x and 1 will produce the adjacent angle. Therefore, sine of that angle will be the third side divided by 1 (i.e. just the third side). So, you must find the value of the third side. (I doubt I explained this well enough - but its best you draw a diagram and see what I mean):

Therefore, the third side, using simple Pythagoras, will be:

$\sqrt{1^2 - (2x)^2}$

$= \sqrt{1 - 4x^2}$

$\therefore \sin{(\cos^{-1}{2x})} = \sqrt{1 - 4x^2}$

$\therefore \sin{(2\cos^{-1}{(2x)})} = 4x( \sqrt{1 - 4x^2})$

Therefore, a = 4 and b = 2.

I apologize in advance if I made any errors or if I didn't explain anything in enough detail. Let me know if I did..

Hope I helped.
Title: Re: DannyN's SM question thread =D
Post by: DannyN on July 11, 2011, 01:12:51 pm: Stuck on another question :P,
i can do parts a,b and c, but d confuses me
the answer for d is --------> N/2 is when the population is increasing most rapidly

i've attached the question below, thanks in advance for the help ^^
Title: Re: DannyN's SM question thread =D
Post by: moekamo on July 11, 2011, 01:55:07 pm: so you should get $y(t) = \frac{N e^{2 t}}{e^{2 t}+3}$ then find the double derivative and equate to 0 to find when the rate of change of population is maximum, should get:

$y'(t) =\frac{6 N e^{2 t}}{(e^{2 t}+3)^2} \implies y''(t) = -\frac{12 N e^{2 t} (e^{2 t}-3)}{(e^{2 t}+3)^3}$

so solving y''(t)=0 gives $t=\frac{1}{2} \ln 3$ , then subbing this into y(t) gives y=N/2

this way is really long and messy so if anyone has a shorter way then please share!
Title: Re: DannyN's SM question thread =D
Post by: DannyN on July 11, 2011, 02:47:30 pm: wow thanks! i forgot about find the double derivative :P

no the method isnt that long if i use the calc ^^, but do you think this would be in the tech-free exam?
Title: Re: DannyN's SM question thread =D
Post by: moekamo on July 11, 2011, 02:57:28 pm: its possible, but unlikely, i mean its just a really long quotient rule when your differentiating. And solving is quite simple because your solving y''(t)=0.
Title: Re: DannyN's SM question thread =D
Post by: DannyN on September 18, 2011, 10:35:36 am: hey guys i need help with another question regarding mechanics

The 5.4kg mass is now placed on a rough plane and a driving force 100N acting upon the particle makes an angle of 20 degrees with the horizontal.

a)if the coefficient of friction between the surface and the object is 0.2, find the acceleration of the object correct to one decimal place.

this is what i did -->

100cos20-0.2R=5.4a
R=5.4g-100sin20
100cos20-0.2(5.4g-100sin20)=5.4a
therefore a=16.7m/s^2

can anyone please confirm whether that is correct?

part b
Find the value of angle for which the acceleration of the particle is maximum, correct to one decimal place.
Considering all the other values are still the same.

my working out is:

100cos(x)-0.2[5.4g-100sin(x)]=5.4a
maximum acceleration occurs when 0.2[5.4g-100sin(x)]=0
so i solved and found the angle was approximately 29 degrees, however when i used this angle to find the acceleration it was less than the acceleration in the previous part, so does this mean my approach for the question was wrong? can someone help me please!

also below is an attachment of the diagram
Title: Re: DannyN's SM question thread =D
Post by: xZero on September 18, 2011, 11:16:17 am: Quote from: DannyN on September 18, 2011, 10:35:36 am
100cos(x)-0.2[5.4g-100sin(x)]=5.4a
maximum acceleration occurs when 0.2[5.4g-100sin(x)]=0
how you know max acceleration occurs when 0.2[5.4g-100sin(x)]=0? try diff the whole expression with respect to x and see if you find a different answer
Title: Re: DannyN's SM question thread =D
Post by: DannyN on September 18, 2011, 01:08:32 pm: sorry, i assumed maximum acceleration occurs when the retarding force is 0 however, i was wrong. and thank you zero! that works!
Title: Re: DannyN's SM question thread =D
Post by: DannyN on October 04, 2011, 12:00:17 am: Stuck on another question, could you guys help me out please? :P
A particle is projected up a rough plane inclined at 40 degrees to the horizontal with an initial speed of 16m/s. It comes instantaneously to rest after 2.3 seconds. Give coefficient of friction between the particle and the plane in three decimal places
Title: Re: DannyN's SM question thread =D
Post by: luken93 on October 04, 2011, 12:10:50 am: Firstly, find acceleration.

$a = \frac{v - u}{t}$

$= \frac{0 - 16}{2.3}$

$= -6.96 m/s^2$

Resolving forces up the plane;

$ma = -mgsin(\theta) - \mu R$

$-6.96m = -mgsin(40) - \mu mgcos(\theta)$

$-6.96 = -gsin(40) - \mu gcos(40)$

$\mu = \frac{-6.96 +gsin(40)}{ -gcos(40)}$

$\mu = 0.052$