what did you guys say for the modeling assumption? I had no idea, i said that if you put fruit in the bowl, it could hold a greater volume as it is over the top hahaha i assume this is wrong

Reply 46

sam del rey

Original post by jackanator

what did you guys say for the modeling assumption? I had no idea, i said that if you put fruit in the bowl, it could hold a greater volume as it is over the top hahaha i assume this is wrong

I said the bottom of the bowl wasn't flat lol.

Reply 47

Rusty1407

Original post by jackanator

what did you guys say for the modeling assumption? I had no idea, i said that if you put fruit in the bowl, it could hold a greater volume as it is over the top hahaha i assume this is wrong

I said you must assume the fishbowl is completely spherical, but in the real world the fishbowl must have a flat bottom and top.

I messed up the differential equation once. I started with ln(sqrtV), instead of 2V^1/2. Do you think I was lose all 10 marks? Or will I get some for working through, but using the wrong equation?

Reply 48

1MacG

Did anyone say anything about the thickness of the bowl being negligible?

Reply 49

Mr Tall

Original post by 1MacG

Did anyone say anything about the thickness of the bowl being negligible?

me!

cant see it being wrong tbh....

Reply 50

Matt857

Could someone put up an unofficial mark scheme or even just post your answers if you think they're right lol. Specifically for questions 3 and 5?

Reply 51

stephen_mcgarry

Original post by Matt857

Could someone put up an unofficial mark scheme or even just post your answers if you think they're right lol. Specifically for questions 3 and 5?

Q3) I got 1072/45 = 23.8 (3sf)
Q5) 7.28PM (19.28)

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Reply 52

jackanator

Original post by Mr Tall

me!

cant see it being wrong tbh....

that is so much more logical. i was thinking that, but then i tought the fruit was a better idea haha such a stupid answer when i look back

Reply 53

dairepatrick1

could you create a mark scheme please?

Reply 54

K995

Not a hard paper. Kicking myself for losing marks in the last question though :frown:

Reply 55

Matt857

did anyone else get -1/2 ln(5-x) - 3/2 ln(x+1) for q1?
i'm aware that could be horribly wrong as i seem to find a way to mess them up every time

(edited 9 years ago)

Reply 56

GingerCodeMan

Sorry for the delay everyone; it takes quite a bit of time to type out the mark scheme :P

Mark scheme is in first post.

Let me know if you have any problems with the mark scheme or if you notice any typos etc.

By the way, I stated the assumption was that the curved surface does not contribute to the volume/negligible thickness.

Reply 57

dairepatrick1

Here are my answers:
1i: 1/(2(5-x)) - 3/(2(1+x))
1ii: -1/2 (ln(5-x)) - 3/2 ln(1+x)

2) AB: 4i +9j
2ii) r = 3 + lamdba 4
-2 9
2iii) proved lambda was 2 for both

3) 23.8

4i) 24.0
4ii) assumed that the bottom of the bowl is flat and the insides are flat

5) 19.46 pm

6i) 25(sinx - 1.29)
6ii) -625 cot(x-1.29)

7 i) f-1 : x -> tan-1 x range = -0.666< x < 0.666
7ii) gf : x -> |tanx| range : 0< x < pi/4
7iii) reflection in y axis of right hand side of tan onto left

8) found t to be 3pi/4 and coordinates to be (3pi/2 + 1) (-2sqrt2

Reply 58

stephen_mcgarry

1)i) 1/2(5-x) - 3/2(1+x)
ii) (-1/2)ln|5-x| (-3/2)ln|1+x|
2) I'll do it shortly don't have answers here, edit post later
3)1072/45 = 23.8 (3sf)
4)i) 24.0 (3sf)
ii) maybe the bowl must be perfectly symmetrical, must be smooth, thickness should be ignored? I'm not 100% sure sorry
5)7.28PM (19.28)
6)i) R=25 alpha=1.29 (3sf)
ii) -(cot(x-1.29))/625 + c
7)i) f^-1:x > arctan(y) [tan^-1(y)] domain (this is a guess) -1<=x<=1 range - pi/4<=f^-1(x)<=pi/4
ii) gf:x > |tanx| range 0<=gf(x)<=1
iii) u shaped curve, Centre point at 0,0 and goes up to pi/4,1 and -pi/4,1 (I'll attach a picture in next post)
8) (2pi,-4) I think you are supposed to disprove this point as, somebody stated, when you sub t into dy/dx, it does not equal sqrt2. The other coordinates are ((5pi-2)/2,-2sqrt2)

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Reply 59

ellipsis909

right heres my attempt at a basic mark scheme:

Q1) i) a= 1/2 b= 3/2
ii) -1/2ln(5-x) -3/2ln(1+x) + c

2 i) 4i + 9j
ii) r= (3i-2j) + y(4i+9j)
iii) y= 2 therefore point on line (y is lambda)
3 i) 23.8

4 i) 7.63pi (units^3)
ii) i said bowl needs flat bottom (not sure tho)
5 i got 7.27pm (i think its 7.28pm so i guess i loose a mark :/)

6 i)R= 25 a= 1.287
ii) i got -cot(x-1.287)/625
7 i)F-1(x) = tan-1(x) 0<f-1(x)<1 -pi/2<x<pi/2
ii) gf(x) = modulus(tanx) gf(x)>0
iii) graph it positive y quadrant domain -pi/4<x<pi/4
8) coordinates(5pi-1, -2root2) (not sure on this though)