VCE Specialist 3/4 Question Thread!

[BubbleWrapMan]

The first part would be more or less the same, but your transformed integral will end up as instead of  .

[Kanye East]

How do I do express this as a single sine?
see attachment below

[lzxnl]

sin x + sin y = 2 sin ((x+y)/2) * cos ((x-y)/2)
sin x + sin (x+b) = 2 sin (x+b/2) * cos b/2

Alternatively, sin x + sin (x+b) = sin x + sin x cos b + sin b cos x
= (1+cos b) sin x + sin b cos x = r sin (x+a) = r sin x cos a + r sin a cos x
here, r cos a = 1+cos b
r sin a = sin b
so r^2 = (1+cos b)^2 + (sin b)^2 = 1+1+2 cos b = 2+2 cos b = 2*(1+cos b) = 4*cos^2 b/2
r = 2 cos b/2 due to domain of b
tan a = sin b/(1+cos b) = (2 sin b/2 cos b/2)/(2cos^2 b/2) = tan b/2
a=b/2
Hence sin x + sin (x+b) = 2 cos b/2 * sin(x+b/2)

Knowing the formula helps, but it's not essential (:

[Homer]

A solid sphere of radius 6 has a cylindrical hole of radius 1cm bored through its centre. What is the volume of the remainder of the sphere?

[b^3]

A solid sphere of radius 6 has a cylindrical hole of radius 1cm bored through its centre. What is the volume of the remainder of the sphere?

Volume of Curves Question

[zvezda]

Hey,
Im having a little trouble with ex 9E q3a in essentials, and i need a little bit of guidance.
Thanks in advance

[b^3]

Place an axis with being the vertical and being the horizontal, with the origin at the bottom of the tank. The equation of the circle at the end of the tank then becomes .
Now if you look at the dotted rectangle, the area of this is , which is what we need in terms of . The length of this rectangle is , so we need the width. At any height , the width will be two times the value. So we will need in terms of .

Spoiler

Now that means will become

Spoiler

So our rate will then become

Spoiler

[zvezda]

Place an axis with being the vertical and being the horizontal, with the origin at the bottom of the tank. The equation of the circle at the end of the tank then becomes .
Now if you look at the dotted rectangle, the area of this is , which is what we need in terms of . The length of this rectangle is , so we need the width. At any height , the width will be two times the value. So we will need in terms of .

Spoiler

Now that means will become

Spoiler

So our rate will then become

Spoiler

Thanks heaps for that, i dont think that that wouldve occurred to me. Appreciate it

[Professor Polonsky]

This might be of interest.

my spesh teacher told us a lot of things about what Phillip swedosh, the spesh chief examiner said regarding special techniques. Yes you can use techniques beyond the spesh course (integration by parts and vector cross product for example) provided you show all working and get EVERYTHING 100% correct. So if your answer and working are totally right, you're doing fine.
However, if your answer is wrong, you won't get many, if any, working out marks for using 'exotic' maths.

[Kanye East]

To express z=-1+i in polar form, i am a bit confused with finding the angle.
so i know that r=sqrt(2)
that's fine
z=r cis(theta)
z=sqrt(2) cis(theta)
tan (theta) = im(z)/re(z)
tan (theta) = 1/-1
theta=arctan(-1)
theta=-pi/4
THIS IS WHERE I AM CONFUSED.
the solution says that theta is 3pi/4 but i thought the principle argument is any angle between -pi<x<pi so technically, you can leave it as -pi/4
Why do you have to ad pi to -pi/4?

[BubbleWrapMan]

Plot the point on an Argand diagram.

[lzxnl]

Alternatively, finding the polar form of -1+i just means to solve the equations r cos t = -1 and r sin t = 1. Obviously r = sqrt 2.
If t = -pi/4, the cosine term is positive and the sine term is negative. That is evidently not what we're seeing here. If we add pi to t, getting 3pi/4, the equations are satisfied.

[Kanye East]

OH so the argand diagram and Cartesian graph with the polar form should have the exact same graph?
If this is a fundamental concept of complex numbers, we didn't learn this in eng math and if not,  i didn't realise it
Thanks guys 

[Kanye East]

Convert this from polar to cartesian form:
            sqrt(2)
z=______________
          cis(3pi/4)

[Conic]

Note that and then use



Then it should be simple to convert to cartesian form: