How was 9709_w15_qp_22

[Manishke]

To prove it u got to use (x1-x2)/2 , (y2-y1)/2

 

[Manishke]

U sure man. Cause I definitely got a y intersect using that method ?

 

[Manishke]

Anyways this was a messed up paper bro. I used to get 90's straight. After this I'll even settle for a 80

 

[The Sarcastic Retard]

To prove it u got to use (x1-x2)/2 , (y2-y1)/2

differentiation I did.
I got -2(9-2x^2)^-1/2
This is the differentiation I got.
Then I kept x as 2 I got -2.
Then I did, 1/2 for normal.. I didnt got y intercept

 

[The Sarcastic Retard]

Anyways this was a messed up paper bro. I used to get 90's straight. After this I'll even settle for a 80

Ahan! A grade. Coongo.
I will end up with a B!

 

[Manishke]

Bro I think ur differentiation is wrong. U got to multiply it by -4x which is the differentiation of the inside terms

 

[The Sarcastic Retard]

Bro I think ur differentiation is wrong. U got to multiply it by -4x which is the differentiation of the inside terms

Oh f*ck!!

 

[Manishke]

Hey did u get the volume q rite ???

 

[DarkEclipse]

Could u do that thing of showing B is a midpoit of OP ?
How did u do?
We had y = (9 - 2x^2)^1/2 and point was (2,1) can u show me how did u do? Coz m getting 2y =x

You're making me relive a nightmare

You had to show that AP's midpoint was B. AP was a normal to the curve.
Differentiate y = (9 - 2x^2)^1/2, then substitute x for 2. That's your gradient of the curve. Then use m1m2 = -1 to find the normal's gradient. Once you do that,
use y-y1=m(x-x1) to find the equation of the normal. Once you get that, substitute y for 0 (since A lies on the x axis). You'll get the coordinates of A. Then, substitute x for 0 (since y lies on the y axis). You'll get the coordinates of B. Next, find the coordinates of the midpoint of AP, which is the same as B's coordinates.

 

[The Sarcastic Retard]

Hey did u get the volume q rite ???

I left that question.

 

[The Sarcastic Retard]

You're making me relive a nightmare

You had to show that AP's midpoint was B. AP was a normal to the curve
Differentiate y = (9 - 2x^2)^1/2, then substitute x for 2. That's your gradient of the curve. Then use m1m2 = -1 to find the normal's gradient. Once you do that,
use y-y1=m(x-x1) to find the equation of the normal. Once you get that, substitute y for 0 (since A lies on the x axis). You'll get the coordinates of A. Then, substitute x for 0 (since y lies on the y axis). You'll get the coordinates of B. Next, find the coordinates of the midpoint of AP, which is the same as B's coordinates.

Chill bro, we will get the thing we deserve.
Thanks.

 

[Manishke]

Hey dark did u get the volume q ??

 

[DarkEclipse]

Yes. Make x the subject of the equation, then use the volume of revolution equation, only this time, you have to integrate x with respect to y.

 

[The Sarcastic Retard]

Yes. Make x the subject of the equation, then use the volume of revolution equation, only this time, you have to integrate x with respect to y.

70pi/3 ?

 

[Manishke]

Yeah I got that. Sorry was talking about the volume of the wedge thing. Where they poured water !!

 

[DarkEclipse]

I got 16pi.

Please, no more discussion

 

[DarkEclipse]

Yeah I got that. Sorry was talking about the volume of the wedge thing. Where they poured water !!

3 i)? No I didn't.

 

[Manishke]

Ahhh same here. That was a total bummer

 

[Manishke]

Who got the identity part ??

 

[DarkEclipse]

That was a real pain.