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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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