Calculus Derivatives

Question

What is the slope of the line normal to the curve

y = sqrt(16 - x)

at the point (0, 4)?

1. -1/8
2. -8
3. 1/8
4. 4
5. 8

Answer 1

y = sqrt(16 - x)

First, find the slope of the function. The slope is tne 1st derivative.

To better see how to get the derivative, put the equation in this form:

y = (16 - x) ^1/2

dy/dx = 1/2 (16 - x)^-1/2 *( -1 ) (chain rule)

At the point (0, 4), x = 0, so

dy/dx = 1/2 (16) ^-1/2 (-1) = -1/8

The "line normal to the curve"---this means perpendicular (i.e. at 90 degrees). The slope will be the negative inverse of -1/8: -(1/(-1/8)) = +8

To check, make a plot----if necessary, use an online tool such as DESMOS