Question

1. no calculator the line normal to the curve $y = \sqrt{16 - x}$ at the point $(0, 4)$ has slope (a) 8 (b) 4 (c) $\frac{1}{8}$ (d) $-\frac{1}{8}$ (e) $-8$

Explanation:

Step1: Find the derivative of the curve

The function is \( y = \sqrt{16 - x}=(16 - x)^{\frac{1}{2}} \). Using the chain rule, the derivative \( y' \) is \( \frac{1}{2}(16 - x)^{-\frac{1}{2}}\times(-1)=-\frac{1}{2\sqrt{16 - x}} \).

Step2: Evaluate the derivative at \( x = 0 \)

Substitute \( x = 0 \) into the derivative: \( y'(0)=-\frac{1}{2\sqrt{16 - 0}}=-\frac{1}{2\times4}=-\frac{1}{8} \). This is the slope of the tangent line at \( (0,4) \).

Step3: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. So if the tangent slope is \( -\frac{1}{8} \), the normal slope is \( \frac{-1}{-\frac{1}{8}} = 8 \).

Answer:

A. 8