VCE Specialist 3/4 Question Thread!

[Daenerys Targaryen]

Use that as a guide, do the same with
Expand each dot product and collect like terms

[Daenerys Targaryen]

[Stick]

Stuck on part a for both of these questions:

[Will T]

Stuck on part a for both of these questions:

(Image removed from quote.)

(Image removed from quote.)

14.

(Arc length)
(Chord length)

Then equate them and you're done.

15.

The area of the triangle is hence the kite's area is twice that.
The area of the circle is
Equating them and converting theta to radians yields the result.

[zvezda]

a) solve [sec(x)]^2 + tan(x) = 1
b) hence, solve the above equation when it is greater than 1.

Ive got part a but im stuck on part b.
cheers in advance

edit: forgot to chuck in the domain. Domain = (0,2pi]

[brightsky]

we know that 1 + tan^2(x) = sec^2(x).
so the inequation becomes (1+tan^2(x)) + tan(x) > 1
tan^2(x) + tan(x) > 0
tan(x) [tan(x) + 1] > 0
so possibilities are:
tan(x) > 0 and tan(x) + 1 > 0
OR
tan(x) < 0 and tan(x) + 1 < 0
solve each pair of simultaneous inequations over the given domain.

[zvezda]

we know that 1 + tan^2(x) = sec^2(x).
so the inequation becomes (1+tan^2(x)) + tan(x) > 1
tan^2(x) + tan(x) > 0
tan(x) [tan(x) + 1] > 0
so possibilities are:
tan(x) > 0 and tan(x) + 1 > 0
OR
tan(x) < 0 and tan(x) + 1 < 0
solve each pair of simultaneous inequations over the given domain.

thats exactly what i was doing. But i sort of tested it out with a simple quadratic like a(a+1)>0.
normally, you would get a>0 and a<-1, but by doing it the way you have up there ^, you get a>0 and a>-1???

[BubbleWrapMan]

thats exactly what i was doing. But i sort of tested it out with a simple quadratic like a(a+1)>0.
normally, you would get a>0 and a<-1, but by doing it the way you have up there ^, you get a>0 and a>-1???

You don't normally get a>0 and a<-1 -- you get a>0 OR a<-1. You need to make the distinction between 'and' and 'or.'

The reason you arrive at this


tan(x) > 0 and tan(x) + 1 > 0
OR
tan(x) < 0 and tan(x) + 1 < 0


is because for a product of two numbers to be positive, they have to be the same sign.

So you need tan(x) > 0 and tan(x) + 1 > 0 to be true simultaneously, or tan(x) < 0 and tan(x) + 1 < 0 to be true simultaneously.

For the case where tan(x) > 0 and tan(x) + 1 > 0, this is equivalent to tan(x) > 0 and tan(x) > -1, which simplifies to tan(x) > 0 (in other words we need tan(x) to be both greater than 0 and -1, which can only happen if it's greater than 0)

For the case where tan(x) < 0 and tan(x) + 1 < 0, this is equivalent to tan(x) < 0 and tan(x) < -1, and this simplifies to tan(x) < -1 (same logic as before)

Combining the two cases, we get tan(x) < -1 or tan(x) > 0

[zvezda]

You don't normally get a>0 and a<-1 -- you get a>0 OR a<-1. You need to make the distinction between 'and' and 'or.'

The reason you arrive at this


tan(x) > 0 and tan(x) + 1 > 0
OR
tan(x) < 0 and tan(x) + 1 < 0


is because for a product of two numbers to be positive, they have to be the same sign.

So you need tan(x) > 0 and tan(x) + 1 > 0 to be true simultaneously, or tan(x) < 0 and tan(x) + 1 < 0 to be true simultaneously.

For the case where tan(x) > 0 and tan(x) + 1 > 0, this is equivalent to tan(x) > 0 and tan(x) > -1, which simplifies to tan(x) > 0 (in other words we need tan(x) to be both greater than 0 and -1, which can only happen if it's greater than 0)

For the case where tan(x) < 0 and tan(x) + 1 < 0, this is equivalent to tan(x) < 0 and tan(x) < -1, and this simplifies to tan(x) < -1 (same logic as before)

Combining the two cases, we get tan(x) < -1 or tan(x) > 0

Oh ok. yeah that makes sense, of course.
thanks for the help guys. much appreciated

[Stick]

Probably just making a stupid error here since it seems like a pretty simple question. My calculator is spitting out complex solutions however. :|

[pi]

What was your working?

[Stick]

Well I was a bit concerned by the word 'zero' but I assumed that both 1+i and 1-i were solutions. Then I applied a simultaneous equations method.

[BubbleWrapMan]

It doesn't have real coefficients so you can't infer that z = 1 - i is a zero

[Stick]

That makes sense. I knew it was a simple error somewhere.

[zvezda]

are three linearly dependent vectors ALWAYS coplanar?
Not necessarily would they?
unless my definition of coplanar is somehow different to the universal one