AQA Core 4 - Monday 10th June 2013 (AM) - Official Thread

Original post by bugsuper

5+ lambda
3
-3lambda

That's the confusing bit for me, I know this may sound weird but, isn't lambda just a constant, how is it related to C?

Reply 203

bugsuper

8

Lambda is a variable that describes the whole line.

What we're doing is saying that, because the point C is on the line, for some value of lambda, r = the point C

do you see what I mean? In this way you can write C with lambda before you know what lambda is, and then use that in your calculations

Reply 204

amish123

Original post by Last Chance

That's the confusing bit for me, I know this may sound weird but, isn't lambda just a constant, how is it related to C?

What lambda is telling you, is basically how far up or down you are travelling on a particular line. The bigger the value for lambda, the further away you are from the position vector. In other words, you can give a point which lies on a line a generic position vector, in terms of lambda. Once you've found how far up or down the point is (the value of lambda), you can put that back into the generic position vector and you'll be able to find the co-ordinates for C.

Reply 205

Original post by bugsuper

Lambda is a variable that describes the whole line.

What we're doing is saying that, because the point C is on the line, for some value of lambda, r = the point C

do you see what I mean? In this way you can write C with lambda before you know what lambda is, and then use that in your calculations

I don't understand, sorry! This stuff isn't in my book and I can't find anything online.

Reply 206

Original post by amish123

What lambda is telling you, is basically how far up or down you are travelling on a particular line. The bigger the value for lambda, the further away you are from the position vector. In other words, you can give a point which lies on a line a generic position vector, in terms of lambda. Once you've found how far up or down the point is (the value of lambda), you can put that back into the generic position vector and you'll be able to find the co-ordinates for C.

Sorry... I'm having trouble grasping this, I'll explain how I usually do questions

(c) The points B and C lie on l such that the distance AC is equal to the distance AB . Find
the coordinates of C. (5 marks)

What information can I obtain here is that I think?

What I've gathered

AC = root 30, calculated previously
AC = -OA + OC
B lies on the line as it's on the vector equation
C lies on the line

http://filestore.aqa.org.uk/subjects/AQA-MPC4-W-QP-JUN08.PDF

I don't understand how you start

Reply 207

bugsuper

8

The person below my post explained it pretty well.

All a parametric line is is a vector on the line, and a parameter that tells you how far along the line you are.

You know that point C is on the line - so you know it's lambda * the direction vector away from that first point they give you.

You don't know how far along it is, but you can write the vector as a "general point", like I did above. Then you can use that in your calculations, along with information the question gives you, to work out what its value is.

I can't put it any simpler than that.

Reply 208

Original post by bugsuper

You know that point C is on the line - so you know it's lambda * the direction vector away from that first point they give you.

You don't know how far along it is, but you can write the vector as a "general point", like I did above. .

Oh okay, I understand that.

!! Was a lot simpler than I thought it would be. :redface:

Thanks both.

Reply 209

ABaz

Original post by GeneralOJB

Some hard de/vectors questions for A* peeps

Hey, thanks for these. Have you got the mark scheme?

Reply 210

bornab3

Can someone help me with part bii. of this question please? I think I know how to do it but I'm not sure and the mark scheme doesn't have the answer to that part of the question on it for some reason. The paper is the Jan12 one.

(k is 72, and the equation of the tangent is "y=-16x+16" if that helps.)

Screen Shot 2013-06-08 at 20.46.48.png

Reply 211

Ravster

Original post by bornab3

Can someone help me with part bii. of this question please? I think I know how to do it but I'm not sure and the mark scheme doesn't have the answer to that part of the question on it for some reason. The paper is the Jan12 one.

(k is 72, and the equation of the tangent is "y=-16x+16" if that helps.)

Screen Shot 2013-06-08 at 20.46.48.png

Plug x=3/2 into your y=-16x+16 to get a value for y then plug the corresponding x and y values into part a and you'll see that it equals 72 thus confirming it intersects at the x=3/2.

Reply 212

Bord3r

10

Original post by bornab3

Can someone help me with part bii. of this question please? I think I know how to do it but I'm not sure and the mark scheme doesn't have the answer to that part of the question on it for some reason. The paper is the Jan12 one.

(k is 72, and the equation of the tangent is "y=-16x+16" if that helps.)

Screen Shot 2013-06-08 at 20.46.48.png

x=3/2\ \: and\ \ y=16-16x[br] \Rightarrow y=16-16(\frac{3}{2})[br]y = -8[br][br]sub in x=3/2 y=-8 to xy^{2}+3y=k[br] \Rightarrow (\frac{3}{2})(-8)^{2}+3(-8)=k [br]\Rightarrow 72=k[br]therefore tangent intersects line[br]

(edited 10 years ago)

Reply 213

alwayshope

Hi all,

Please can someone help me to understand the double angle formulae?

In class we've been taught that cos2A=cos^2A - sin^2A
and cos^2A + sin^2A=1

However, on page 5 of the AQA formula booklet, under "Hyperbolic Functions", it says: cosh^2x - sinh^2x=1
and cosh2x=cosh^2x + sinh^2x

I don't understand how both these sets of formulae are compatible? They can't both be true! What effect does "h" have on the formulae?

Please help!

Reply 214

Prepotency

7

Original post by alwayshope

Hi all,

Please can someone help me to understand the double angle formulae?

In class we've been taught that cos2A=cos^2A - sin^2A
and cos^2A + sin^2A=1

However, on page 5 of the AQA formula booklet, under "Hyperbolic Functions", it says: cosh^2x - sinh^2x=1
and cosh2x=cosh^2x + sinh^2x

I don't understand how both these sets of formulae are compatible? They can't both be true! What effect does "h" have on the formulae?

Please help!

cos/sin etc are trigonometric functions. cosh/sinh are hyperbolic functions, they're very different!

Cosh/Sinh is only used in AQA FP2. It will never come up in AQA Core 4.

(edited 10 years ago)

Reply 215

alwayshope

Are any Core 4 formulae in the booklet then, or do need to memorise them? Thanks so much for the quick reply :smile:

Reply 216

LittleMissNoface

Does anyone have some crazy hard stuff which may be higher level yet relevant to exam?
I know Core 4 seems to mix things up a little and it just throws whateva it can at you?

Any awesome trig stuff?

Reply 217

Prepotency

7

Original post by LittleMissNoface

Does anyone have some crazy hard stuff which may be higher level yet relevant to exam?
I know Core 4 seems to mix things up a little and it just throws whateva it can at you?

Any awesome trig stuff?

Have you tried the Elmwood papers? They're meant to be slightly tougher. Otherwise, you could always try a STEP I question :tongue:

Reply 218

Ravster

Original post by alwayshope

Are any Core 4 formulae in the booklet then, or do need to memorise them? Thanks so much for the quick reply :smile:

Only the following C4 identities are in the booklet:

sin(A+-B) =
cos(A+-B)=
tan(A+-B)=

Everything else you'll need to memorise, including C3 stuff.

Reply 219

LittleMissNoface