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Q: find the coordinates of the two stationary points on the curve with equation

x^2 + 4xy+2y^2 +18

what I've done so far:

differentiation of 4xy (product rule)

v=4x
u= y

dv/dx = 4
du/dx = dy/dx

dy/dx = 4x.dy/dx + 4y

then differentiated the rest to get

dy/dx = 2x +4.dy/dx +4y + 4y.dy/dx

simplifiy = 2x + dy/dx(4y+4) + 4y
=2x + 4y + 4(dy/dx)(y+1)

looking for stationary points so dy/dx = 0

so

0 = (-2x-4y)/4(y+1)

the part I'm stuck on :

do do I say y = -1
and try and sub it into the original equation to get my x values ?
Original post by ADotCross
Q: find the coordinates of the two stationary points on the curve with equation

x^2 + 4xy+2y^2 +18


What is the actual equation

Without an = sign there is no curve so .......
Original post by ADotCross
Q: find the coordinates of the two stationary points on the curve with equation

x^2 + 4xy+2y^2 +18

what I've done so far:

differentiation of 4xy (product rule)

v=4x
u= y

dv/dx = 4
du/dx = dy/dx

dy/dx = 4x.dy/dx + 4y

then differentiated the rest to get

dy/dx = 2x +4.dy/dx +4y + 4y.dy/dx

simplifiy = 2x + dy/dx(4y+4) + 4y
=2x + 4y + 4(dy/dx)(y+1)

looking for stationary points so dy/dx = 0

so

0 = (-2x-4y)/4(y+1)

the part I'm stuck on :

do do I say y = -1
and try and sub it into the original equation to get my x values ?


where has the firest dy/dx come from.?
and x^2 + 4xy+2y^2 +18 is not an equation.
Reply 3
Original post by ADotCross
Q: find the coordinates of the two stationary points on the curve with equation

x^2 + 4xy+2y^2 +18




That isn't an equation!

Can you post the correct question please?
Reply 4
This is what I'm talking about q1
(edited 9 years ago)
Original post by ADotCross
This is what I'm talking about q1


Just tell us what it equals
Reply 6
Did I attach the photo ?
Original post by ADotCross
Did I attach the photo ?


Ok, so it equals 0

Look again at your working

You have differentiated 4xy correctly then missed out the x on the next line
Reply 8
So once I got the full differentiation which should be

dy/dx = (-2x-4y)/ (4x+4y)

i have no clue what to do next
Original post by ADotCross
So once I got the full differentiation which should be

dy/dx = (-2x-4y)/ (4x+4y)

i have no clue what to do next


You know that dy/dx is 0
Reply 10
0 = (-2x-4y) / 4(y+x)

I can't see what I can use to get the coordinates as I still got x and y I the brackets
Original post by ADotCross
0 = (-2x-4y) / 4(y+x)

I can't see what I can use to get the coordinates as I still got x and y I the brackets


That tells you that x = - 2y

Put that into your original equation
Reply 12
Oh
Reply 13
I subbed it in and I got y= plus and minus root 3

working out :

(-2y)^2 + 4(-2y)y + 2y^2 +18

all simplifies to plus or minus 3^1/2
Original post by ADotCross
I subbed it in and I got y= plus and minus root 3

working out :

(-2y)^2 + 4(-2y)y + 2y^2 +18

all simplifies to plus or minus 3^1/2


How did you get that

The square root of 9 is 3
Reply 15
Original post by TenOfThem
How did you get that

The square root of 9 is 3

Nvm I made a silly error
Reply 16
The coordinates should be x = 6 and -6

thankx for your help
Original post by ADotCross
The coordinates should be x = 6 and -6

thankx for your help


Ok

:smile:

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