Are these equal?

y=\dfrac{-1}{m}[br][br]y=\dfrac{1}{-m}[br][br]y= -\dfrac{1}{m}

are these all the same thing?

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Reply 1

Zacken

Original post by thefatone

are these all the same thing?

Yes.

Reply 2

thefatone

Original post by Zacken

Yes.

eyy that was quick ^-^

thanks :smile:

Reply 3

Zacken

Original post by thefatone

eyy that was quick ^-^

thanks :smile:

That's the maths forum for you. :tongue:

No problem.

Reply 4

thefatone

Original post by Zacken

That's the maths forum for you. :tongue:

No problem.

Do you know where i can find reserve papers for Edexcel C1 and C2? they're much more difficult(it also says on the front that the paper is strictly for students outside the UK)

Reply 5

Zacken

Original post by thefatone

Do you know where i can find reserve papers for Edexcel C1 and C2? they're much more difficult(it also says on the front that the paper is strictly for students outside the UK)

Do you men regional papers? If so - those are found here for C1 (all the papers marked (R)) and here for C2 (all the papers marked with an (R) ).

As you can see, there are only 1/2/3 (R) papers available (and btw, they're not all more difficult; that's a myth, it's just normal fluctuations in difficulty.).

Reply 6

Math12345

Original post by thefatone

y=\dfrac{-1}{m}[br][br]y=\dfrac{1}{-m}[br][br]y= -\dfrac{1}{m}

are these all the same thing?

Yes,

y=-\frac{1}{m} = -1 * \frac{1}{m} = \frac{-1}{m} = \frac{-1}{m} * 1 = \frac{-1}{m} * \frac{-1}{-1} = \frac{1}{-m}

(edited 7 years ago)

Reply 7

thefatone

Original post by Zacken

Ah yes that's what i meant :smile:

in which case could you help me do question 9 part b?

https://917c8db4f385e20466384eff57d816b2bed188db.googledrive.com/host/0B1ZiqBksUHNYMVhvRUNrcWJ6LXM/June%202014%20(R)%20QP%20-%20C1%20Edexcel.pdf
C1 2014 june paper

Reply 8

Zacken

Original post by thefatone

Ah yes that's what i meant :smile:

y=3x+k

is tangent to

y = \frac{x^2}{3} + 8

then that means the curves only intersect once, that is - there is only one solution to the equation

3x + k = \frac{x^2}{3} + 8

. Re-arrange this to get a quadratic and then think about the discriminant of this quadratic, what does it mean for the discriminant if there is only one solution?

Reply 9

thefatone

Original post by Zacken

y=3x+k

is tangent to

y = \frac{x^2}{3} + 8

then that means the curves only intersect once, that is - there is only one solution to the equation

3x + k = \frac{x^2}{3} + 8

. Re-arrange this to get a quadratic and then think about the discriminant of this quadratic, what does it mean for the discriminant if there is only one solution?

= to 0?

Reply 10

Zacken

Original post by thefatone

= to 0?

Precisely, and since the discriminant is a function of k only, then that'll let you solve for k.

Reply 11

thefatone

Original post by Zacken

Precisely, and since the discriminant is a function of k only, then that'll let you solve for k.

but i get a negative constant and that's not right....

Spoiler

Reply 12

Math12345

Original post by thefatone

but i get a negative constant and that's not right....

Spoiler

You forgot to multiply 8 by 3 (second line)

Reply 13

thefatone

Original post by Math12345

You forgot to multiply 8 by 3 (second line)

thank you

Reply 14

Notnek

Original post by thefatone

Ah yes that's what i meant :smile:

Zacken's method is fine. But it's good to look at different methods and the more "common" method here would be :

The gradient of

y=3x+c

is 3 so this is a tangent at the point on the curve where

\frac{dy}{dx} = 3

.

Solve

\frac{dy}{dx} = 3

to find the coordinate of this point and you can use that to find

c

Reply 15

thefatone

Original post by notnek

Zacken's method is fine. But it's good to look at different methods and the more "common" method here would be :

The gradient of

y=3x+c

is 3 so this is a tangent at the point on the curve where

\frac{dy}{dx} = 3

.

Solve

\frac{dy}{dx} = 3

to find the coordinate of this point and you can use that to find

c

I'm not quite sure what to do?

do you sub 3 into the curve c to find the y-coordinate of the point they meet? then find the x-co-ord ?

... not realyl sure what o do though

in any case the way zacken did i did initially, i just asked for help to verify where i went wrong :smile:

Reply 16

Notnek

Original post by thefatone

No you would find

\frac{dy}{dx}

first and then solve

\frac{dy}{dx}=3

Reply 17

thefatone

Original post by notnek

No you would find

\frac{dy}{dx}

first and then solve

\frac{dy}{dx}=3

i'm not sure what you would do with that :/
please elaborate on this "common" method

Reply 18

Notnek

Original post by thefatone

i'm not sure what you would do with that :/
please elaborate on this "common" method

The gradient of the tangent is 3 since it's equation is in the form

y=mx+c

where

m=3

. This means that the gradient is 3 on the curve at the point (call it P) where the tangent meets the curve. Draw a sketch if this is confusing.

The next step is to find the coordinates of P. So the question is now:

The gradient of the curve with equation

y=\frac{1}{3}x^2+8

at the point P is 3. Find the coordinates of P.

This a standard C1 question that you should be able to solve.

Once you have found the coordinates of P, you can substitute them into

y=3x+k

to find

k

. You can do this since P lies on the tangent.

(edited 7 years ago)

Reply 19

thefatone

Original post by notnek

The gradient of the tangent is 3 since it's equation is in the form

y=mx+c

where

m=3

y=\frac{1}{3}x^2+8

at the point P is 3. Find the coordinates of P.

This a standard C1 question that you should be able to solve.

Once you have found the coordinates of P, you can substitute them into

y=3x+k

to find

k

. You can do this since P lies on the tangent.

.... i don't really get what to do???

i'm not getting the bit where i do something with the gradient and find the co-ordinates of the point of intersection.