Hi guys! How do we find the normal to the curve in this problem?
**Find the equations of the tangent to the curve y = x - x3 at (1,0), and the normal to the same curve at (1/2, 3/8). Find where the tangent and normal intersect.**
First I found the derivative and then calculated the gradient of the tangent. After it, I determined the equation of the tangent:
y = x- x3
dy/dx = 1 - 3x2 >>>>>(1,0)
m = 1 - 3 = -2
y - y0 = m.(x - x0)
y - 0 = -2.(x - 1)
y = -2x + 2
Normally, I would do that:
The normal to the curve = -1/m
= -1/-2
= 1/2
But I know it's false and "the normal to the same curve at (1/2, 3/8)" makes me confused. Do I need to find the slope of the tangent at that point to find the normal? Like :
m = 1 - 3x2 >>>>>>(1/2, 3/8)
m = 1 - 3.(1/2)2
m = 1/4
the normal = -1/m
= -1/1/4
= -4
y - y0 = m.(x - x0) >>>>>>>(1/2, 3/8)
y - 3/8 = -4.(x - 1/2)
y = -4x + 19/8
??
Any help would be appreciated.
**Find the equations of the tangent to the curve y = x - x3 at (1,0), and the normal to the same curve at (1/2, 3/8). Find where the tangent and normal intersect.**
First I found the derivative and then calculated the gradient of the tangent. After it, I determined the equation of the tangent:
y = x- x3
dy/dx = 1 - 3x2 >>>>>(1,0)
m = 1 - 3 = -2
y - y0 = m.(x - x0)
y - 0 = -2.(x - 1)
y = -2x + 2
Normally, I would do that:
The normal to the curve = -1/m
= -1/-2
= 1/2
But I know it's false and "the normal to the same curve at (1/2, 3/8)" makes me confused. Do I need to find the slope of the tangent at that point to find the normal? Like :
m = 1 - 3x2 >>>>>>(1/2, 3/8)
m = 1 - 3.(1/2)2
m = 1/4
the normal = -1/m
= -1/1/4
= -4
y - y0 = m.(x - x0) >>>>>>>(1/2, 3/8)
y - 3/8 = -4.(x - 1/2)
y = -4x + 19/8
??
Any help would be appreciated.
Last edited: