You can find them in the SQA website, go to mathematics, principal assesors report, then the year. I think 50% was a B and 63% was an A, can't remember what C was.

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# THE EXAM!

Started by AM4R, May 15 2007 03:27 PM

30 replies to this topic

### #21

Posted 15 May 2007 - 06:09 PM

### #22

Posted 15 May 2007 - 06:18 PM

Coolio, cheers dude

### #23

Posted 15 May 2007 - 08:28 PM

ad absurdum, agreed with all your answers (except 14 which i got wrong) apart from 13b) which i got k = 2, are you pretty confident u were right?

### #24

Posted 15 May 2007 - 08:50 PM

ad absurdum, agreed with all your answers (except 14 which i got wrong) apart from 13b) which i got k = 2, are you pretty confident u were right?

Relatively, but not entirely. What I done:Which looks like it might be right, but there's a good chance I might have made a mistake somewhere

HMFC - Founded 1874, beefing the Cabbage since 1875

### #25

Posted 15 May 2007 - 08:53 PM

thought it was pretty easy to be honest, would doubt if the boundaries went below what they're meant to be

### #26

Posted 15 May 2007 - 08:56 PM

i think the difference was i worked out d2y/dx^2 by doing d2y/dt^2 divided by d2x/dt^2

which gave d2y/dx^2 as sin(2t)/cos(2t).

maybe thats wrong but i just though with parametric you couldnt just differentiate dy/dx again?

which gave d2y/dx^2 as sin(2t)/cos(2t).

maybe thats wrong but i just though with parametric you couldnt just differentiate dy/dx again?

### #27

Posted 15 May 2007 - 09:13 PM

ad absurdum, the derivative of cot(x) is (-cosec2 (x)) so why do you have cosec3(x). Also, it was -cot(x) which was being differentiated so the result would be +cosec2(x).

### #28

Posted 15 May 2007 - 09:29 PM

Does anyone know if the sqa give you marks if you make a mistake at one point and continue it through, using the correct techniques but getting the wrong answers. For example in Q.14 i said k=1, which resulted in both negative growth and height. Thanks.

### #29

Posted 15 May 2007 - 09:43 PM

i think the difference was i worked out d2y/dx^2 by doing d2y/dt^2 divided by d2x/dt^2

which gave d2y/dx^2 as sin(2t)/cos(2t).

maybe thats wrong but i just though with parametric you couldnt just differentiate dy/dx again?

That seems like a very good approach, butI'm not totally sure on if that would work. If you take a simplified view of the chain rule (i.e., "cancelling" on multiplication), that would give you d^2y/d^2x, rather than d^2y/dx^2 - maybe the simplified analogue breaks down here though. However, at t=0, this would give d^2y/dx^2 as 0, when it should be undefined (the parametric equations describe a circle, and the tangent at t=0 is the line x=1, so calculating the rate of change of the gradient at this point should involve division by zero).which gave d2y/dx^2 as sin(2t)/cos(2t).

maybe thats wrong but i just though with parametric you couldnt just differentiate dy/dx again?

What I done was said . So you differentiate dy/dx with respect to t, then divide by dx/dt.

HMFC - Founded 1874, beefing the Cabbage since 1875

### #30

Posted 15 May 2007 - 10:59 PM

That looks pretty similar to an example in my notes.

I differentiated dy/dt and squared dt/dx then multiplied the answers since I'd forgotten how to do the other thing. I remember looking at the notes, thinking that'll never come up and ignoring it.

I differentiated dy/dt and squared dt/dx then multiplied the answers since I'd forgotten how to do the other thing. I remember looking at the notes, thinking that'll never come up and ignoring it.

### #31

Posted 16 May 2007 - 10:04 AM

I thought bits of the exam were ok, the rest was horrid though!! I got the roots question to work eventually, took forever cause i kept getting an answer which didnt work. I really hope i passed, i need to get a B!

Grrrrrr!!

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