Question

The equation of the normal line to the curve y = cube root of (x^2 - 1) at the point where x = 3 is

Answer 1

First, you need to find the slope of the curve. You'll find it 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 = -1/(1/2) = -2
Now, we'll calculate the y coordinate at x=3
y = cube root(9-1) = 2
The equation of normal line is:
y - 2 = -2(x-3)
y = -2x + 8