quick question (2 Viewers)

[bloodysunday]

Can somebody help me with this? I don't quite get it:

Find the equation of the normal to the curve y = xlnx at the point (e,e).

Cheers

 

[smallcattle]

y'= lnx + 1

then sub in x

 

[bloodysunday]

thanks guys

 

[Xayma]

dy/dx=ln x*1+x*1/x
=ln x+1

at y=e
m=ln e +1
=2
∴ m_n=-.5
(y-e)=-.5(x-e)
2y-2e=x+e
x-2y+3e=0

 

[username]

y = xlnx

u = x and u dash = 1 v = lnx and v dash = lnx

therefore

y(dash) = lnx + xlnx

therfore the gradient to the tangent is lne + elne

which 1 + e

then solve.

 

[smallcattle]

y = xlnx

u = x and u dash = 1 v = lnx and v dash = lnx

therefore

y(dash) = lnx + xlnx

therfore the gradient to the tangent is lne + elne

which 1 + e

then solve.

lol... no.. its ln x + 1

v' = 1/x

 

[ptitsa]

dy/dx=ln x*1+x*1/x
=ln x+1

at y=e
m=ln e +1
=2
∴ m_n=-.5
(y-e)=-.5(x-e)
2y-2e=x-e
x-2y+e=0

2y-2e = x + e (as opposed to x - e)
x + e - 2y + 2e = 0
x - 2y + 3e = 0

......? *worried*

 

[bloodysunday]

perhaps:

y-e=-x/2+e/2
2y-2e=-x+e
x+2y-3e=0

 

[jumb]

2y-2e = x + e (as opposed to x - e)
x + e - 2y + 2e = 0
x - 2y + 3e = 0

......? *worried*

You didnt expand the -'s properly. (neither did Xayma)

Should be 2y-2e = -x + e

 

[bloodysunday]

yeah its one of those damn minus signs again, they piss me off when i overlook them

 

[Xayma]

You didnt expand the -'s properly. (neither did Xayma)

Should be 2y-2e = -x + e

Opps I did it as tangent first until I reread the question and being lazy I just edited it instead of deleting and redoing it.

 

[username]

samll cattle, thanks stupid mistake. I was thinking e^x is e^x

 

[lfc_reds2003]

whatr r u worrying about??!!

perhaps:

y-e=-x/2+e/2
2y-2e=-x+e
x+2y-3e=0

CORRECT!!

GL y'all
muahahaha im an accelerant !!!!