You are Here: Home >< Maths

# C2 - coordinate geometry etc watch

1. Heya, this is a question from the Edexcel Practice paper A4, I have never seen anything like this before (with logs and e's in) and also I think there's a bracket missing out of the question which makes it even more confusing!

The curve C with equation y=p+qex , where p and q are constants, passes through the point (0,2). At the point P (ln2,p+2q on C, the gradient is 5.
a) Find the value of p and the value of q
The normal to C at P crosses the x-axis at L and the y-axis at M.
b) Show that the area of triangle OLM, where O is the origin is approx 53.8

I have the mark scheme in from of me and I dont understand it, the answer for a is:
q=2.5
p=-0.5

Im not really sure am I meant to differentiate the curve - and if so, how so .....

Thanks !
2. a) dy/dx = qex

At the point P, dy/dx = 5

=> 5 = q(eln2)
5 = 2q
q = 2.5

We also know the curve passes through (0,2), so sub these values of x and y into the equation of the curve:

2 = p + 2.5(e0)
p = -0.5

b) m1m2 = -1, where m1 = 5
m2 = -1/5

P(ln2 , 4.5)

Equation of normal at P: y - 4.5 = (-1/5)(x-ln2)

y + x/5 = 4.5 + (1/5)ln2

At L, y=0 => x/5 = 4.5 + (1/5)ln2
x = 22.5 + ln2
=> L(22.5 , 22.5 + ln2 )

At M, x=0 => y = 4.5 + (1/5)ln2
=> M(0 , 4.5 + (1/5)ln2 )

Now you have a triangle of height 4.5 + [(1/5)ln2] units, and base [22.5 + ln2] units, so the area is:

(1/2)[4.5 + (1/5)ln2][22.5 + ln2] = 53.8 units2
3. a)
y' = qex
at P,
x = ln2
ex = 2
=====
at P,
m = qex
5 = q*2
q = 2.5
=====

b)
at A(0,2),
y = 2 = p + qex
2 = P + qe0
2 = p + q
2 = p + 2.5
p = -0.5
======
4. dy/dx = qex
ok, yes I know im being stupid, its just the first line I dont get...
5. hm, I was never taught how to differentiate y=p+qe^x
C2 exam next week too
6. and can you quickly explain what 'e' is and how it relates to ln and stuff, thank youuuu
7. hm, I was never taught how to differentiate y=p+qe^x
C2 exam next week too
edexcel? i will love you forever if you tell me I dont have to know it for the exam!
lol although it would be nice to find out what it means... and what is goin on !
8. How would you differentitate y = ex?
dy/dx = ex

p and q are just constants, so the derivative of qex is just qex. And when you differentiate a constant on its own, you get 0, hence,
d/dx[p + qex] = qex
9. How would you differentitate y = ex?
dy/dx = ex
why dont you just do it normally, and so get dy/dx =xe^(x-1) ?
10. (chants) times by the power, lower the power by 1 ...

Im on edexcel franks, and Ive never come across differentiant y = a^x or anything
11. Im on edexcel franks, and Ive never come across differentiant y = a^x or anything
I dont reckon we need to know it. Why on earth is it on this edexcel practice paper tho !!!?!

Mockel - so is this just another rule you have to learn - like differentiation and integration, something you do when the equation is 'to the power of x' ?
12. Ive never come across differentiant y = a^x or anything
You should know how to do it for P3 or higher.
y=a^x
lny=ln(a^x)
lny=xlna
(1/y)(dy/dx)=lna
(dy/dx)=ylna
(dy/dx)=(a^x)(lna)
13. Erm...yeah.
Differentiate ex (with respect to x), and you get ex.
You definitely had to know that for P2. Don't really know about the C's, but I'd think so too
14. oooook, I may just start another little thread and ask peeps.
Thanks everybody for your help Stars the lot of you!
franks x
15. I know you don't need to know it for C2 OCR
16. You learn this in C3 for all exam boards.

e is a number, a constant 2.71... and basically it is the base of natural logarithms, which you'll learn about in C3.
17. (Original post by franks)
why dont you just do it normally, and so get dy/dx =xe^(x-1) ?
You'd do this if you were differentiating with respect to e, but you are not, you have a constant to the power of x, not x to the power of a constant... so the normal method to diffrentiate doesnt work.
18. Its a C3 thing, but e is the exponential function. e^x is the curve (between 2^x and 3^x) where at (0,1) the gradient is 1.

Essentially, at x=1, 3^x = 1 and 2^x = 1 etc... if you work out the gradients for these lines at this point, 3^x has a gradient of 1.099 and 2^x has a gradient 0.693... the gradient of e (2.71828...) at x=1 is 1.

The inverse of this is ln|x| the natrual logarithm.

Just to clear up any interests. But for now, just concentrate on the exams you have

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 15, 2005
The home of Results and Clearing

### 1,605

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Sheffield Hallam University
Tue, 21 Aug '18
2. Bournemouth University
Wed, 22 Aug '18
3. University of Buckingham
Thu, 23 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams