# Find the equation of the normal line to the curve y^6 + secx - 2siny = 2y + x + π at the origin.?

y^6 means "y to the 6th power"

Relevance

First, find the slope of the tangent line. Use implicit differentiation.

6y^5y' + tanx secx -2cosyy' = 2y' + 1 + 0

Solving for y',

y' = (1 - tanx secx)/(6y^5 - 2cosy - 2)

Plug in the point (0,0) to get

y' = 1/(-4)

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal.

m=4

Since the normal line passes through the origin, the normal line is y=4x.

On further inspection, the origin does not even lie on the curve. Are you sure the problem is as stated? It is a much more difficult question if they ask for normal lines containing the origin.

• Anonymous

y^6 - 2y - 2 sin y = x - sec x + pi

Implicit differentiation.

6 y^5 y' - 2 y' - 2 y' cos y = 1 - sec x tan x

y' (6 y^5 - 2 cos y - 2) = 1 - sec x tan x

dy/dx = (1 - sec x tan x) / (6 y^5 - 2 cos y - 2)

When x=0,

y^6 - 2y - 2 sin y + 1 - pi = 0

Use Newton's method to find y.

y = -0.5421355

y is the y-intercept of the normal line.

Use x=0, y=-0.5421355 to find dy/dx @ x=0.

dy/dx = (1 - sec x tan x) / (6 y^5 - 2 cos y - 2)

dy/dx = -0.2503626

m = slope of the normal line = -(dy/dx)^(-1)

m = 3.994207

Equation of the normal line:

y = 3.994207 x - 0.5421355