VCE Specialist 3/4 Question Thread!

[Dominatorrr]

Use a double angle formula to show that the exact value of cos(pi/8)=sqrt(2+sqrt(2))/2
Explain why any values are rejected.

[aznxD]

Use a double angle formula to show that the exact value of cos(pi/8)=sqrt(2+sqrt(2))/2
Explain why any values are rejected.

Use cos(2x) = 2cos^2(x) - 1
Sub x=pi/8 and solve for cos(x)=pi/8
Reject the negative solution as pi/8 is in the first quadrant.

[Bhootnike]

Find the area between the curve the x-asus and given lines for each of the following functions:



i know theres an easier way to do this, but i started off on the other foot, so i dont wanna start over the other way :p ( TJERE IS NO EASIER WAY TO DO THIS W.O CALC! GRRRR)
attached is a diagram of the situation described above, where the pink area shows what i want, and the green area shows the area which adds onto the pink area, to make a rectangle.

now, to find out the pink area, i thought i'd make a rectangle, so i found that:  DISREGARD AND READ NEXT POST PLZZZZ
length of rectangle =

width of rectangle =                   - i found inverse function here

therefore total area of rectangle =   x    

this = 0.0465501..

Now to find green area, i wasnt really sure, but i thought that the area would be :


this = 1

therefore area of pink bit = area of rectangle - area of green bit
= 0.04655 - 1
= -0.04655
thus, area = 0.04655 units squared.

which is the answer

but, just wanted to confirm here that what i've done is right? or is it a fluke through misunderstanding of the concept?!

i know i couldve just done  and gotten my answer, but yeah, :S  N.B. that'd involve using calc and 1 line working!

ta !

[Bhootnike]

lulz, how dafuck did i get a length and width in terms of an integral? ladies and gentleman, i just shot myself in the foot..

anyways, New length =
new width =

thus area of rectangle = (using calc)  0.07014893454
Area of the green bit was 0.5

thus area of pink bit = -0.429... bla bla

WHICH IS WRONG!


help please. im dying.. in vain, i cannot let integral calculus overcome my pride. BRUUUAHH

[brightsky]

area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

[Bhootnike]

i dont mean to be picky, but could you please  give the working in latex or upload a pic?!
i cant really workout some of the lines e.g (6 - pi*sqrt(3))/12)

thx

[Bhootnike]

area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

i think ive got it, dw then

So, 2 questions.
1) Why are the terminals to ? i thought it was the other way around
2) How did you get the area of the rectangle to be, ?
i thought the length was

EDIT.

i think when you typed that out brightsky you werent thinking of latex coding ? cause i thought you meant the terminals were from pi/6 to 0.  :p

and i get why the length of the rectangle isnt what i said it is. !
so nvm!

[Mr. Study]

The position of an object, r metres at time t, is given by: =2cos(3t)+2sin(3t). Show that the object moves in a circular path of radius 2.

Here are my steps:






Is there any way to do this question quicker? (I had to cut out a few lines)

[brightsky]

well you have:
x = 2cos(3t)
y = 2sin(3t)
so x^2 + y^2 = 4(cos^2(3t) + sin^2(3t) = 4

[Planck's constant]

area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

accordingly, depending on your choice, your integral becomes either

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6

[Bhootnike]

area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

accordingly, depending on your choice, your integral becomes either

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6

i don't think that works for that specific question tho?

[Planck's constant]

area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

accordingly, depending on your choice, your integral becomes either

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6

i don't think that works for that specific question tho?

The second one does (gives you the pink)
The first one gives you the green

[Planck's constant]

area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

accordingly, depending on your choice, your integral becomes either

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6

i don't think that works for that specific question tho?

The second one does (gives you the pink)
The first one gives you the green

The key point is that you don't need the 'rectange' because the rectangle pops out naturally when you use area between two curves. It's a neater method and you don't have to give messy explanations like 'The area of the shaded region is the value of the integral blah blah blah minus the area of the rectangle ABCD (refer diagram) blah blah blah'

[ashoni]

Having trouble with a few questions about partial fractions

6. Express the following in partial fractions.
 g. (3x+1) / (x^3+x)
 h. (3x^2+8) / x(x^2+4)
 t. (x^3) / (x+1)(x-1)

Thanks!

[b^3]

Remember for partial fractions when you get a quadratic factor that cannot be expressed as linear factors you have you use

So for the first one we start off with
Now that will result in the form I originally spoke about above, and the other x term will just give us the normal result.

So we are looking at
Which then gives

Now if we equate the coefficients we get
C=3, A=1
A+B=0
B=-A
B=-1
So that gives

The second one can be approached in the same method.

Now for the last one, because the greatest power of the numerator is equal to or greater than the greatest power of the denominator, you will have to long divide first, then apply the methods for partial fractions. So for the numerator the greatest power is and for the denominator the greatest power will be , so we have to long divide first.


So that brings us to
Then from there just apply the normal rules again.

Then add the other x back in later.