MATHS study thread. (2 Viewers)

[klaw]

Edit:wrong answer

 

[Yip]

Actually klaw, the answer is 5,

a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=1-2(-2)=5

 

[GaDaMIt]

Originally posted by klaw
3. a^2+b^2+c^2=(a+b+c)³-3ab-3ac
=(a+b+c)²-3(ab+ac)
=1-3(-2)
=7

yeah wtf? how does that work?

 

[klaw]

oh yeah... I screwed my expansion up... was doing weird things hahahaha I kept going back and forth between ³ and ² for some reason... hahaha

 

[klaw]

Wrong answer again

 

[Yip]

Incorrect again klaw, but that one was quite tricky. FOr that one since a, b and c are roots, u substitute them in to get a system of 3 simulatenous equations:

6a^3+6a^2-12a+3=0
6b^3+6b^2-12b+3=0
6c^3+6c^2-12c+3=0

Adding the equations,

6(a^3+b^3+c^3)+6(a^2+b^2+c^2)-12(a+b+c)+9=0
6(a^3+b^3+c^3)+6(5)-12(-1)+9=0
6(a^3+b^3+c^3)=-51
therefore a^3+b^3+c^3=-51/6

 

[klaw]

hmm... but that's what I did.... oh... somehow I got 3+3+3=12.... and then I divided the equation by 6 hahahahaha... why are you asking these questions when you know the answer anyway?

 

[klaw]

Incorrect again klaw, but that one was quite tricky. FOr that one since a, b and c are roots, u substitute them in to get a system of 3 simulatenous equations:

6a^3+6a^2-12a+3=0
6b^3+6b^2-12b+3=0
6c^3+6c^2-12c+3=0

Adding the equations,

6(a^3+b^3+c^3)+6(a^2+b^2+c^2)-12(a+b+c)+9=0
6(a^3+b^3+c^3)+6(5)-12(-1)+9=0
6(a^3+b^3+c^3)=-51
therefore a^3+b^3+c^3=-51/6

6(a^3+b^3+c^3)+6(a^2+b^2+c^2)-12(a+b+c)+9=0
2(a^3+b^3+c^3)+2(a^2+b^2+c^2)-4(a+b+c)+3=0
But that gives a different answer to the above equation... why?

 

[Yip]

"why are you asking these questions when you know the answer anyway?"
2 reasons, 1. to check that my solutions are correct
2. just to help the ppl in the forum with revision for their yearlys
"But that gives a different answer to the above equation... why?"
It doesnt from 2(a^3+b^3+c^3)+2(a^2+b^2+c^2)-4(a+b+c)+3=0 u get a^3+b^3+c^3=-17/2=-51/6
Its just that i forgot to simplify ~.~

 

[klaw]

yeah hahahah I was like wtf... why diff answers?

 

[insert-username]

Just quickly: my maths teacher uses "derive" and "differentiate" interchangeably, so that's why I've used "derive". Also... I need to be awake when I write questions, it seems.

What is the equation of the tangent to the parabola f(x) = x² + x - 2 at the point where x = -4?


I_F

 

[Yip]

What is the equation of the tangent to the parabola f(x) = x² + x - 2 at the point where x = -4?

At (-4,10),
equation of tangent is
y-10=m(x+4)
y=mx+(4m+10)
mx+(4m+10)=x^2+x-2
x^2+x(1-m)-(4m+12)=0
For there to be one solution, Discriminant=0
(1-m)^2+4(4m+12)=0
D=m^2+14m+49=0
(m+7)^2=0
m=-7
equation of tangent: y=-7x-18

 

[HamuTarou]

What is the equation of the tangent to the parabola f(x) = x² + x - 2 at the point where x = -4?

f(-4)=16-4-2=10
f'(x)=2x+1
f'(-4)=-8+1=-7
y-10=-7(x+4)
y=-7x-18

yip, i think ur ans is wrong...the eqn cant be positive :S

 

[mitsui]

i think hamutarou is rite..
wait..
no more maths!!!
on to physic and eco for me...*cram**cram*

 

[Yip]

ehehehe^^'''' calc error~.~

 

[klaw]

this thread is cursed... so many silly mistakes...

 

[insert-username]

Tee hee. Errors all over the shop.


Find the focal length, the equation of the directrix, and the coordinates of the focus of the parabola with equation x² = 12y


Find the equation of the normal to the tangent to the parabola x² = 8y at the point where x = 4


I_F

 

[klaw]

Tee hee. Errors all over the shop.


Find the focal length, the equation of the directrix, and the coordinates of the focus of the parabola with equation x² = 12y


Find the equation of the normal to the tangent to the parabola x² = 8y at the point where x = 4


I_F

x² = 12y
a=3
Focal length:3
Directrix:y=-3
Focus: (0,3)

 

[klaw]

x² = 8y
y=x²/8
dy/dx=x/4
when x=4, dy/dx=1
.:Gradient of normal=-1

when x=4, y=2
equ of normal:y-2=-1(x-4)
.:x+y-6=0

 

[mitsui]

this thread is cursed... so many silly mistakes...

ohh oops.. must coz i made this thread.. =P
i am like prone to silly mistakes...
today's exam, i drawn a number plane when they asked for number line.. =_=..
lolz