What is the equation of the normal line to the curve y=cube root(x^2-1) at x=3?
We know that the product of slopes of two perpendicular lines is -1. Therefore, we'll determine the slope of the tangent line to the given curve, at x = 3.
We'll find the slope m using derivative:
dy/dx = (1/3)*[(x^2 - 1)^(-2/3)]*(2x)
We'll calculate the slope at x=3
dy/dx = m = (1/3)*[(9 - 1)^(-2/3)]*(6)
m = 2/8^(2/3)
m = 2/4
m = 1/2
The slope of the normal line is m1:
m*m1 = -1=> m1 = -1/(1/2) = -2
Now, we'll calculate the y coordinate at x=3:
y = cube root(9-1) = cube root 8 = 2
The equation of a line that passes through a point and it has a slope is:
y - y1 = m1*(x - x1)
Comparing, we'll get:
y - 2 = -2(x-3)
Therefore, the equation of normal line is: y = -2x + 8.