Qualification > Math

Differentiation doubt

(1/2) > >>

spiderman:
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.

astarmathsandphysics:

--- 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.

--- End quote ---
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

nid404:
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...

astarmathsandphysics:
Corrected nmy post

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

Navigation

[0] Message Index

[#] Next page