VCE Specialist 3/4 Question Thread!

[IndefatigableLover]

I think this is how you do it:







And then it should be just a simple sketching in of the sector from to

Ah geez how could I forget about that property about Arguments >.<
Thanks for the clarifications cooldude123!

[keltingmeith]

Beaten by cooldude, but this was fun. Although, I shaded the wrong section at the end (also, needs a dotted line), so with this method you'd need to shade at the end.

Basically, I changed to Cartesian form, expanded, them modified the axes to suit. Then, slowly made reflections until the axes returned to normal. Hopefully you can decipher the pictures.

[IndefatigableLover]

(Image removed from quote.)

Beaten by cooldude, but this was fun. Although, I shaded the wrong section at the end (also, needs a dotted line), so with this method you'd need to shade at the end.

Basically, I changed to Cartesian form, expanded, them modified the axes to suit. Then, slowly made reflections until the axes returned to normal. Hopefully you can decipher the pictures.

Awesome picture and explanation EulerFan101 (nice whiteboard btw xD)
I just have one question which I'd like clarified:
So for (2), I understand that it's -y+ix however when drawn you drew it from the origin.. why wasn't it translated up on the 'x' axis and then drawn in the same direction (because it looks like -y instead of -y+ix to me)...

[keltingmeith]

Awesome picture and explanation EulerFan101 (nice whiteboard btw xD)
I just have one question which I'd like clarified:
So for (2), I understand that it's -y+ix however when drawn you drew it from the origin.. why wasn't it translated up on the 'x' axis and then drawn in the same direction (because it looks like -y instead of -y+ix to me)...

So, (1) sketches the graph as -y against x, which means we can sketch the point as per usual. Then, in (2), I make the plot y against x, which is why I've had to reflect in the "x"-axis. Since it's just a reflection, there's no need to translate it all.

[Eiffel]

with these questions. Why can you add the horizontal and vertical components? how does that work, e.g. with part a.

a is a vector 3 across 5 up and b is a vector 5 across 3 up, how can you even add these separate paths?

( is a is the part that connects start of 3 to end of 5? or is it actually one that describes to separate vectors? )

[keltingmeith]

with these questions. Why can you add the horizontal and vertical components? how does that work, e.g. with part a.

a is a vector 3 across 5 up and b is a vector 5 across 3 up, how can you even add these separate paths?

( is a is the part that connects start of 3 to end of 5? or is it actually one that describes to separate vectors? )

I'm... Confused by the question, honestly.

a=3i+5j
b=5i+3j
a+b=3i+5j+5i+3j=8i+8j

[Eiffel]

yeah sorry just upto 9A, didnt go into adding vectors so i wasnt sure what you had to do. Looked ahead to 9B and explained it all ~ sorry about that.

[SE_JM]

Hello,
Could someone please help me with these questions?
I'm getting really frustrated with this

Two circles of radii 3 cm and 4 cm have their centres 5 cm apart. Calculate the area of the region common to both circles

and

Two wheels (pulleys) have radii of length 15 cm and 25 cm and have their centres 60 cm apart. What is the length of the belt required to pass tighlt around the pulleys without crossing?

Thank you


This is the questions I posted a few days ago, but no one has answered it for me so far...
Could someone please help me with this questions? It would be much appreciated. I'm posting again so that it doesn't get buried over time

[pi]

Second question is very classic

[SE_JM]

Thank you pi!

[pi]

I'm very rusty in short-cuts, so there is probably an easier way, but here is how to do the first one

(edit: sorry for my messy handwriting too >_<)

(edit2: just realised an easy way to calculate area of triangle is A = (16/5)(12/5) as you know x value of intersection, sigh... knew Heron's formula was overkill)

[vobinhood]

Hey guys! This a not a Specialist problem, but a problem I have with my CAS. 
When I type "cis(5pi/6)", it doesn't spit out an answer in rectangular form, but instead just outputs the input.
When I checked my document settings, under Real or Complex, I even clicked rectangular and made it default - but it still doesn't work.

How do I rectify this problem?

[pi]

cis(x) = e^(ix), try that.

[vobinhood]

It worked! Thank you!

[lzxnl]

If you want justification that cis(x) = e^(ix), here are a few:
Noting that the derivative of cis(x) is -sin x + i cos x, differentiating cis(x)/e^(ix) by quotient rule gives (-sin x + i cos x)*e^(ix) - ie^(ix)*(cos x + i sin x) on the numerator, which if you look carefully is actually 0 as the terms cancel. Hence as the derivative is 0, cis(x)/e^(ix) should be constant. Subbing in x = 0 gives the value of this constant, which is 1. Hence, cis(x)/e^(ix) = 1 and cis(x) = e^(ix)

Alternatively, note that d/dx (cis x) = i*cis(x). cis(x) and e^(ix) both satisfy the differential equation dy/dx = iy and are both 1 when x = 0. As solutions of first order differential equations differ by at most one parameter (e.g. the solution to dy/dx = 1 is y = x + c where c is the parameter in question), and as both of these functions are the same at x = 0, they don't differ at all.