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# Edexcel C3,C4 June 2013 Thread watch

1. (Original post by Better)
Oh I get you so if I was to "rotate" if I can call it that the green line, I can see its longer.

So in essence its just like a Mathematical truth or something? like Pythagoras?
Yeah i think it's very much an intuitive thing, although no doubt there is a mathematical proof for it
2. (Original post by Better)
Quick Question just for understanding sakes.

I know the fact that the closet distance from lets say Line L to point P, would be the perpendicular distance, but WHY is this?

Can anyone care to explain or provide a link. I know it's applicable in C4 Vectors, but it just came up in a Physics 5 Question I did and I realized I don't understand the logic behind why it is that way.

THANK YOU
I don't have much time right now, but I'll post a simple proof for this later, if you want.
3. (Original post by Better)
Quick Question just for understanding sakes.

I know the fact that the closet distance from lets say Line L to point P, would be the perpendicular distance, but WHY is this?

Can anyone care to explain or provide a link. I know it's applicable in C4 Vectors, but it just came up in a Physics 5 Question I did and I realized I don't understand the logic behind why it is that way.

THANK YOU
Here you go:

Using the Pythagorean theorem, the distance between any point on Line L (equation ) and a point P with coordinates would be:

Where D is the distance

Differentiating both sides of the equation with respect to x, we get:

We know that at the minimum value of D, dD/dx=0. So we set dD/dx=0;

Now, we know that is equal to the gradient of the line L, and is equal to the gradient of the line joining any point on the Line L and point P.

So,

As the product of the gradient of two perpendicular lines is -1, we can conclude that the closest distance from a Line L to a point P would be the perpendicular distance.
4. (Original post by justinawe)
Here you go:

Using the Pythagorean theorem, the distance between any point on Line L (equation ) and a point P with coordinates would be:

Where D is the distance

Differentiating both sides of the equation with respect to x, we get:

We know that at the minimum value of D, dD/dx=0. So we set dD/dx=0;

Now, we know that is equal to the gradient of the line L, and is equal to the gradient of the line joining any point on the Line L and point P.

So,

As the product of the gradient of two perpendicular lines is -1, we can conclude that the closest distance from a Line L to a point P would be the perpendicular distance.
You are a god among mortals.

Thank you very much, I will try my best to get my head around it hahaha!
5. (Original post by Better)
You are a god among mortals.

Thank you very much, I will try my best to get my head around it hahaha!
A god among mortals who has to re-sit C4 due to screwing it up in January? Some god!

No problem, glad I could help.
6. (Original post by justinawe)
A god among mortals who has to re-sit C4 due to screwing it up in January? Some god!

No problem, glad I could help.
Haha you'll do great mate especially with your Maths skills! I am impressed
7. (Original post by Better)
Haha you'll do great mate especially with your Maths skills! I am impressed
I certainly hope so! I think jan was a one-off, I just got too fixated on a particular question, ended up wasting a whole lot of precious time, and proceeded to panic after that

I definitely won't make the same mistake again though, and since I've got FP2/FP3 coming up in June as well, hopefully that will make C4 seem easier.
8. (Original post by justinawe)
I certainly hope so! I think jan was a one-off, I just got too fixated on a particular question, ended up wasting a whole lot of precious time, and proceeded to panic after that

I definitely won't make the same mistake again though, and since I've got FP2/FP3 coming up in June as well, hopefully that will make C4 seem easier.
Yeah mate lets do this! I'll do the same, with maths I don't get so nervous, I just rush through the papers too fast, I make so many basic errors as I go through. In Physics I get sooo nervous, and I get fixated on questions; its horrible. But I'm working on that haha

About to do a D1 Past Paper, wish me luck brah!
9. (Original post by Better)
Yeah mate lets do this! I'll do the same, with maths I don't get so nervous, I just rush through the papers too fast, I make so many basic errors as I go through. In Physics I get sooo nervous, and I get fixated on questions; its horrible. But I'm working on that haha

About to do a D1 Past Paper, wish me luck brah!
Ah, good luck with that I've self-studied D1 for the upcoming exams, and it's so incredibly boring!
10. solve cos2x=1-cosx any ideas??? tried 2cos^2-1 but am stumped
11. (Original post by Gekko123)
solve cos2x=1-cosx any ideas??? tried 2cos^2-1 but am stumped
Express cos2x in terms of cos only, you should get a quadratic.

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12. can any one help please

i have part 1 of the question it was expressing 13-2x/(2x-3)(x+1) in partial fractions and i got
4/(2X-3)-3/(x+1)

and then part b) says b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation

dy/dx = [y(13-2x)]/[(2x-3)(x+1)], x > 1.5

what do i do

i got lny=2ln(2x-3)-3ln(x+1) but i dont know if its right and what do i do with the x an y values?
13. (Original post by Converse girl)

i have part 1 of the question it was expressing 13-2x/(2x-3)(x+1) in partial fractions and i got
4/(2X-3)-3/(x+1)

and then part b) says b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation

dy/dx = [y(13-2x)]/[(2x-3)(x+1)], x > 1.5

what do i do

i got lny=2ln(2x-3)-3ln(x+1) but i dont know if its right and what do i do with the x an y values?
I think that's right, I'm a bit too lazy to check though sorry

Use your C2 log rules. What does a*ln(b) equal? What does ln(a)-ln(b) equal?
14. (Original post by justinawe)
A god among mortals who has to re-sit C4 due to screwing it up in January? Some god!

No problem, glad I could help.
Nice proof. What did you get in C4 in Jan?
15. (Original post by justinawe)
I think that's right, I'm a bit too lazy to check though sorry

Use your C2 log rules. What does a*ln(b) equal? What does ln(a)-ln(b) equal?
so is the y=4 at x=2 is just there then? yeah i have that part with the logs

thanks mr Lazy
16. (Original post by reubenkinara)
Nice proof. What did you get in C4 in Jan?
70, by far the worst maths mark I've ever received
17. (Original post by Converse girl)
so is the y=4 at x=2 is just there then? yeah i have that part with the logs

thanks mr Lazy
The y=4 at x=2 part is to find the constant of integration. You haven't forgotten your "+c" now, have you?
18. (Original post by justinawe)
70, by far the worst maths mark I've ever received
How did it average with C3?
19. (Original post by reubenkinara)
How did it average with C3?
I got 95 for C3. Would have been 100, but I somehow ran out of time at the end and so missed out the final bit of the last question, which was worth 4 raw marks

Obviously I'll be retaking C4 for the A*. It's not that C4 is that much harder than C3 though, I just had an extremely bad day personally.
20. (Original post by justinawe)
The y=4 at x=2 part is to find the constant of integration. You haven't forgotten your "+c" now, have you?
nooooooo i didnt okay fine i did

thanks makes all sense now

i think i need to tattoo +c on myself so i wont forget the it

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