Calculus question?

Question

Im having trouble with the following problem. Can someone please provide an explanation on how to get the following answer?

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


Thank You.

Answer 1

First, find the derivative: y = (x^2 - 1)^(1/3) --> y' = (1/3)*(x^2 - 1)^(-2/3)*(2x).

Plug x=3 into y' to find the slope of the tangent line at that point.

Go back to the original problem and find the value of y at the point x=3.

Finally, plug those values into the Point-Slope formula for a line: Y - y = m*(X - x).

Answer 2

The normal is perpendicular to tangent. If we know the slope of the tangent, we can find out the slope of the normal since product of their slopes is -1

the slope of the tangent to the curve is the derivative of the curve

y = (x^2 -1)^1/3

differentiate implicitly

y' = 1/3(x^2-1)^(-2/3) (2x)

= (2/3)(x)/(x^2-1)^2/3

y'(3) = (2/3)(3)/(9 - 1)^2/3

=2/8^2/3

= 2/4

= 1/2

so the slope of tangent at x = 3 is 1/2

the point of tangency = (x, y(x))

so substituting x = 3 in the curve

y = (9-1)^(1/3)

y = (8)^1/3 = 2

The tangent and normal both passe through (3,2)

the slope of normal = -1/1/2 = -2

the equation of normal is

y - y1 = m(x-x1)

substituting y1= 2, m= -2 and x1 = 3

y - 2 = -2(x - 3)

y - 2 = -2x + 6

y + 2x = 8 is the equation of Normal