Edexcel C3,C4 June 2013 Thread

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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 :smile:

Reply 661

justinawe

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.

Reply 662

justinawe

Original post by Better

Here you go:

Using the Pythagorean theorem, the distance between any point

(x,y)

on Line L (equation

y=mx+c

) and a point P with coordinates

(a,b)

would be:

D^2 = (x-a)^2 + (y-b)^2

Where D is the distance

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

2D \cdot \dfrac{dD}{dx} = 2(x-a) + 2(y-b) \cdot \dfrac{dy}{dx}

D \cdot \dfrac{dD}{dx} = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

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

0 = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

(y-b) \cdot \dfrac{dy}{dx} = -(x-a)

\dfrac{dy}{dx} = -\dfrac{x-a}{y-b} = -\dfrac{1}{\left(\dfrac{y-b}{x-a} \right)}

Now, we know that

\dfrac{dy}{dx}

is equal to the gradient

m

of the line L, and

\dfrac{y-b}{x-a}

is equal to the gradient

m_{1}

of the line joining any point on the Line L and point P.

So,

m = -\dfrac{1}{m_{1}}

m \times m_{1} = -1

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.

Reply 663

Better

Original post by justinawe

Here you go:

Using the Pythagorean theorem, the distance between any point

(x,y)

on Line L (equation

y=mx+c

) and a point P with coordinates

(a,b)

would be:

D^2 = (x-a)^2 + (y-b)^2

Where D is the distance

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

2D \cdot \dfrac{dD}{dx} = 2(x-a) + 2(y-b) \cdot \dfrac{dy}{dx}

D \cdot \dfrac{dD}{dx} = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

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

0 = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

(y-b) \cdot \dfrac{dy}{dx} = -(x-a)

\dfrac{dy}{dx} = -\dfrac{x-a}{y-b} = -\dfrac{1}{\left(\dfrac{y-b}{x-a} \right)}

Now, we know that

\dfrac{dy}{dx}

is equal to the gradient

m

of the line L, and

\dfrac{y-b}{x-a}

is equal to the gradient

m_{1}

of the line joining any point on the Line L and point P.

So,

m = -\dfrac{1}{m_{1}}

m \times m_{1} = -1

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!

Reply 664

justinawe

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! :lol:

No problem, glad I could help.

Reply 665

Better

Original post by justinawe

A god among mortals who has to re-sit C4 due to screwing it up in January? Some god! :lol:

No problem, glad I could help.

Haha you'll do great mate especially with your Maths skills! I am impressed

Reply 666

justinawe

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 :s-smilie:

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.

Reply 667

Better

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 :s-smilie:

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!

Reply 668

justinawe

Original post by Better

Ah, good luck with that :wink:

I've self-studied D1 for the upcoming exams, and it's so incredibly boring!

Reply 669

Gekko123

solve cos2x=1-cosx any ideas??? tried 2cos^2-1 but am stumped

Reply 670

JOR2010

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.

Posted from TSR Mobile

Reply 671

Converse girl

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

Express your answer in the form y = f(x)
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?

Reply 672

justinawe

Original post by Converse girl

I think that's right, I'm a bit too lazy to check though :tongue:

sorry

Use your C2 log rules. What does a*ln(b) equal? What does ln(a)-ln(b) equal?

Reply 673

Kvothe the Arcane

Original post by justinawe

A god among mortals who has to re-sit C4 due to screwing it up in January? Some god! :lol:

No problem, glad I could help.

Nice proof. What did you get in C4 in Jan?

Reply 674

Converse girl

Original post by justinawe

I think that's right, I'm a bit too lazy to check though :tongue:

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

Reply 675

justinawe

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 :redface:

Reply 676

justinawe

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? :wink:

Reply 677

Kvothe the Arcane

Original post by justinawe

70, by far the worst maths mark I've ever received :redface:

How did it average with C3?

Reply 678

justinawe

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 :lol: