# The derivative: Notation and Definition; finding normal to a curve

#### Nougat

##### New member
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 - x[SUP]3[/SUP] 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- x[SUP]3[/SUP]

dy/dx = 1 - 3x[SUP]2[/SUP] >>>>>(1,0)
m = 1 - 3 = -2

y - y[SUB]0[/SUB] = m.(x - x[SUB]0[/SUB])
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 - 3x[SUP]2[/SUP] >>>>>>(1/2, 3/8)
m = 1 - 3.(1/2)[SUP]2[/SUP]
m = 1/4

the normal = -1/m
= -1/1/4
= -4

y - y[SUB]0[/SUB] = m.(x - x[SUB]0[/SUB]) >>>>>>>(1/2, 3/8)
y - 3/8 = -4.(x - 1/2)
y = -4x + 19/8

??

Any help would be appreciated.

Last edited:

#### stapel

##### Super Moderator
Staff member
**Find the equations of the tangent to the curve y = x - x[SUP]3[/SUP] at (1,0)...
For this part, do like you did: Find the derivative, evaluate at x = 1 to find the slope, and then find the equation of the line with that slope and passing through the point (x, y) = (1, 0).

...and the normal to the same curve at (1/2, 3/8).
For this part, use the derivative equation you found for the tangent-line computations. But, in this case, evaluate at x = 1/2, because you're working with a totally different point on the curve. Find the slope, and then find the perpendicular slope. Then find the equation of the line with that perpendicular slope and passing through the point (x, y) = (1/2, 3/8).

Find where the tangent and normal intersect.**
Find the intersection point of the two line equations you've found.

Note: You can kind of check your work graphically: Draw the original curve, locate the two points, draw a tangent line (approximation) at the first point, a perpendicular line (approximation) at the second point, and see where these two lines cross. Your sketch should put you in the general neighborhood of the correct algebraic solution.

#### Nougat

##### New member
This really helped me. Thanks!