VCE Specialist 3/4 Question Thread!

[TrueTears]

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Could someone please help me on this one?!

on what? lol

[b^3]

(Image removed from quote.)

Could someone please help me on this one?!

on what? lol

If you click modify you can copy the image link and see it. Don't know why it didn't come up thought.

EDIT: Or just quote it and copy the link.

[Bhootnike]

edited - dont know why it didnt come up! links up in the prev. post!

here it is anyways: http://imageshack.us/photo/my-images/528/spec2.jpg

[dc302]

Chp7, MQ, pg 314,

Differential Equation question.

I got a bit stuck on it for a while

Q9. Verify that satisfies the differential equation.

When a question asks you to verify something, what they mean is they want you to show that the 2 statements agree with each other. In this case, you just need to substitute y=e^sin2x into the differential equation and show that LHS = RHS.

[TrueTears]

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http://imageshack.us/photo/my-images/528/spec2.jpg


Could someone please help me on this one?!

here's a hint to help you start off

since they want you to prove something look at the required expression for clues:

|c|^2, that reminds you of c.c, so try c.c and then use the fact that a+b = -c to expand and simplify

[Bhootnike]

If you do that, you get: c^2 = (a+b)^2 yea?
so, c^2 = a^2 + 2ab + b^2
tf, c^2 = a^2 +2(abcosB) + b^2

so you end up with c^2 = a^2 +b^2 + 2abcosB

should be negative before the 2abcos B :s

[TrueTears]

no watch out, c = -(a+b)

so c.c = -(a+b).-(a+b)

[Bhootnike]

no watch out, c = -(a+b)

so c.c = -(a+b).-(a+b)

then wouldnt it end up as:   c^(2)=-a^(2)-2ab-b^(2)  ,
so c^(2)=-a^(2)-b^(2)-2(abcosB)
?

[TrueTears]

you cant square vectors, there is no such thing as c^2 if c is a vector.

you have to dot product the expressions

[kamil9876]

No the algebra is correct, the only confusion arises from what is.

Let's call the angle between between the vectors and by

Then (draw the diagram)

Thus hence the negative sign.

[TrueTears]

No the algebra is correct, the only confusion arises from what is.

Let's call the angle between between the vectors and by

Then (draw the diagram)

Thus hence the negative sign.

if he writes c^2 and the expressions on a sac or exam, marks will definitely be taken off

[kamil9876]

Bad notation yes but that's not my point, my point is that the lack of a negative sign didn't come from some incorrect algebra but rather getting the angle mixed up.

[TrueTears]

i know, just clearing things up - which is why i tempted him to redo his workings again using the correct notation

[Bhootnike]

okay so in terms of notation, i shouldve written:

|c|^2 = |a|^2 +2ab + |b|^2
i.e c.c = a.a = 2a.b + b.b ?

you cant square vectors, there is no such thing as c^2 if c is a vector.

you have to dot product the expressions

okay
but arithmetically, that'd mean you end up with : |c|^2 = |a|^2 +|b|^2 + 2a.bcosB

No the algebra is correct, the only confusion arises from what is.

Let's call the angle between between the vectors and by

Then (draw the diagram)

Thus hence the negative sign.

I drew the diagram. but i didnt introduce theta. i just used B haha.. 
can you explain why theta was needed to be introduced into this? i dont get that bit

[kamil9876]

is the angle in the actual triangle. In the formula the angle is the angle between the vectors and .

See diagram for which angle is which.