Question

The slope of the line normal to the graph of 4 sin x – 9 cos y = -9 at the point (pi, 0) is

Answer 1

First take the derivative of the function.

= 4cosx -9(-siny)(dy/dx) = 0
= 4cosx +9siny(dy/dx) =0
dy/dx = -4cosx/9siny

now plug in the values (pi ,0) as your (x ,y)
solve for the value and take the reciprocal. That will be your normal slope.

Answer 2

Good start, Mebs! But look what happens if we substitute y=0 into that expression. How would you get around that issue?

Answer 3

i personally dont know. but i used my calc to solve the rest and got 0. which is the correct answer according to my answer key.

Answer 4

Hi, guy,

Mebs found the derivative dy/dx from the given implicit function. One interpretation of this derivative is that it's the slope (where defined) of the tangent line to any point on the graph of the function. At the point (pi, 0), this slope, dy/dx, becomes
\[\frac{ dy }{ dx }=\frac{ -4\cos 0 }{9\sin \pi }= \frac{ -4 }{ 0 },\]
which is undefined. That means that the tangent line to the graph at (pi,0) is vertical (since the slope of a vertical line is undefined). The normal line to the curve at that point is therefore horizontal, and therefore it has a slope of zero.