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Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: spiderman on October 28, 2009, 03:49:41 am

Title: Differentiation doubt
Post by: spiderman on October 28, 2009, 03:49:41 am

Hey guys,
Pls could you'll help me with this sum:

-: Fin the equation to the normal to the curve y=3x^2 -2x-1 which is parallel to the line y=x-3.

Title: Re: Differentiation doubt
Post by: astarmathsandphysics on October 28, 2009, 07:50:35 am

Quote from: spiderman on October 28, 2009, 03:49:41 am

Hey guys,
Pls could you'll help me with this sum:

-: Fin the equation to the normal to the curve y=3x^2 -2x-1 which is parallel to the line y=x-3.

dy/dx=6x-2
since it is perpendicular  to y=x-3 the gradient=dy/dx=-1/1 s0 6x-2=-1 so x=1/6
y=3(1/6)^2-2(1/6)-1=-5/4 so the point is (1/6,-5/4)
y+5/4=-1(x-1/6) so 12y+15=-12x+2 so 12x+12y=-13
Didn't read the question then put x=-1/6 somewhere

Title: Re: Differentiation doubt
Post by: nid404 on October 28, 2009, 11:59:07 am

the answer to this is different
It's 12y=12x-17

In y=x-3
m=1
therefore gradient will be same for point on the curve 3x2- 2x-1

6x-2=1
x=1/2

Using this too.....i get a wrong answer

Even m confused wid this one...

Title: Re: Differentiation doubt
Post by: astarmathsandphysics on October 28, 2009, 02:38:24 pm

Corrected nmy post

Title: Re: Differentiation doubt
Post by: astarmathsandphysics on October 28, 2009, 02:39:55 pm

Do you mean the normal is parallel to y=x-2 or the tangent is parallel to y=x-3

Title: Re: Differentiation doubt
Post by: nid404 on October 28, 2009, 02:48:40 pm

Quote from: astarmathsandphysics on October 28, 2009, 02:39:55 pm

Do you mean the normal is parallel to y=x-2 or the tangent is parallel to y=x-3

tangent parallel to y=x-3....i think. we have to find normal to tangent on the curve(when the curve's gradient is same as y=x-3)

Title: Re: Differentiation doubt
Post by: astarmathsandphysics on October 28, 2009, 02:57:10 pm

It works out worse for that.