Sorry for starting a new thread on C3 but I will edit this post to include the paper and solutions 12:00 Tuesday, if I am awake and if the thread is unlocked.... this way the solutions will be easily to find.
Here is a link to the C3 June 10 thread on the post where I uploaded the solutions, there was some good discussion and useful tips in the pages before it.
In iteration, once we have two consecutive results to the appropriate degree of accuracy, can we stop there, or should we carry on to see if there is greater convergence?
In iteration, once we have two consecutive results to the appropriate degree of accuracy, can we stop there, or should we carry on to see if there is greater convergence?
I just look at what degree of accuracy the question asks for like 4 significant figures, for example. Then i just keep pressing equals on the calculator until the numbers on the calculator relevant to 4sf don't change. Works for me, i guess.
C3 exam is in a months time so I thought it would be good to discuss tips and exam practice? Who has finished their exam revision? Also does anyone have old P2 and P3 exam papers from before 2000?
Yeah I dont tend to worry about 'revision' for maths, as it's basically just practicing on exam papers. So probably just going to work my way through papers from 2006 :P
That was what I was planning to do but I didn't do good in C2 so I didn't want to make the same mistake again for C3 especially since C3 counts towards an A* I don't want to be resitting that. So I'm doing all the past papers.
Not dreading this exam, but need at least 95/100 UMS to make the C4 exam less stressful! How about we collect some of the harder questions for extra practice? One I found in a recent paper is:
If x=tany, show that dxdy=x2+11
Made me think, but once you spot the trick it's simple. 5 marker there .
Not dreading this exam, but need at least 95/100 UMS to make the C4 exam less stressful! How about we collect some of the harder questions for extra practice? One I found in a recent paper is:
If x=tany, show that dxdy=x2+11
Made me think, but once you spot the trick it's simple. 5 marker there .
You're right it is actually simple just a bit of identities, it took me a long time to do this. This would be the question I'm bogged on for most of the exam. dx/dy=sec^2(y) dy/dx=1/sec^2(y) dy/dx=1/tan^2(y)+1 dy/dx=1/x^2+1 I need to get better at identities I only know cos^2+sin^2=1 and I convert from that. What year was that paper?
You're right it is actually simple just a bit of identities, it took me a long time to do this. This would be the question I'm bogged on for most of the exam. dx/dy=sec^2(y) dy/dx=1/sec^2(y) dy/dx=1/tan^2(y)+1 dy/dx=1/x^2+1 I need to get better at identities I only know cos^2+sin^2=1 and I convert from that. What year was that paper?
Ahh you did it differently from me, I wouldn't have even thought of doing that. I did the following:
dx/dy=sec2y dy/dx=cos2y
Now, I drew a right angled triangle, angle y, adjacent 1, opposite x (tany = x/1). From that, the hypotenuse is x2+1 from Pythagoras' Theorem.
Hence: dy/dx=cosy⋅cosy=(x2+11)2=x2+11, as required.
I'm doing C3 and C4 in January and need to get an A* overall for myself and my universities as well! Would be a good idea to discuss a few difficult questions on here
I'm doing C3 and C4 in January and need to get an A* overall for myself and my universities as well! Would be a good idea to discuss a few difficult questions on here
Yeah get the hard questions infact solomon press ones if you can. I'll find some hard ones as well. Gonna aim for at least 94 so I beat my other modules.
Ahh you did it differently from me, I wouldn't have even thought of doing that. I did the following:
dx/dy=sec2y dy/dx=cos2y
Now, I drew a right angled triangle, angle y, adjacent 1, opposite x (tany = x/1). From that, the hypotenuse is x2+1 from Pythagoras' Theorem.
Hence: dy/dx=cosy⋅cosy=(x2+11)2=x2+11, as required.
I never even knew you could do that my spatial awareness is not that good, probably why I flopped C2. Sometimes identities are so long to find so your way is generally foolproof.