VCE Specialist 3/4 Question Thread!

[BubbleWrapMan]

and , solve simultaneously for and .

[Homer]

not sure how to approach this one

[BubbleWrapMan]

It's the (scalar) resolute of in the direction of .

[jono88]

If tan(theta)=ax/x^2+ab+b^2 and theta=pi/4, find x in terms of a and b.

tan(pi/4)=1

1=ax/x^2+ab+b^2
ax=x^2+ab+b^2 what do i do next :/ pretty noob question to ask but yeh

[BubbleWrapMan]

Rearrange to get a quadratic in , then use quadratic formula.

[jono88]

mmm...that was my problem, rearranging it

[BubbleWrapMan]

[götze]

diff (3x^2-2x)^3

[lzxnl]

Chain rule u=3x^2-2x

[götze]

can somebpdy step by step show me how to do this problem (x^2-2x+3)^-2

[Daenerys Targaryen]

can somebpdy step by step show me how to do this problem (x^2-2x+3)^-2

Do you mean differentiate that?

[götze]

yeah my bad

[b^3]

It's just the chain rule again, let .

Remember with the chain rule we have . Here since we have an expression raised to a power, we can apply this rule by looking at it as and . So the derivative will just be the product of the first differentiated with respect to with the second differentiated with respect to .

[Daenerys Targaryen]

can somebpdy step by step show me how to do this problem (x^2-2x+3)^-2

[götze]

Thanks mate I found out what i was doing wrong