Question

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

Answer 1

Take the derivative, find the slope, take the negative reciprocal to find the orthogonal slope, then plug in the point to find the y coordinate, then use point slope form y-y_0=m(x-x_0) to find your line.

Answer 2

\[y=\sqrt[3]{x^2-1}\]
\[y'=(x^2-1)^{-\frac{2}{3}}*2x\]
The slope of the normal line is the negative reciprocal of this, so at
(3,2)=(x,y)
y'=24
Normal line slop = -1/24
Put this into point-slop form for a line:
y-2=(-1/24)(x-3)

Answer 3

I'm a bit confused after we took the derivative... where is the (3,2) coming from?

Answer 4

Just plug 3 into the original function. It is the point we are finding the normal at.

Answer 5

So what isn't making sense?

Answer 6

the reciprocal part

Answer 7

For any line:
y=mx
The line:
y=-(1/m)x is perpendicular / normal /orthogonal where they intersect.
Try playing with it a bit, sketching it.

Answer 8

sorry, now I dont see how y' = 24

Answer 9

Just calculate y'(x) where x=3

Answer 10

I keep getting 3/2

Answer 11

You are correct. I forgot to put a negative sign in my exponent above, and haven't checked my work. it is 3/2.