Posted by Mishaka on .
The slope of the line normal to the graph of 4 sin x + 9 cos y = 9 at the point (pi, 0) is:
Derivative: 4cosx  9siny(dy/dx) = 0
(dy/dx) = (4cosx) / (9siny)
(dy/dx) = (4) / 0
Normal line = 1 / (4/0)
Does this mean that the slope of the normal line is undefined, or did I do my calculations incorrectly???

Calculus (Normal Line, please check my work) 
Damon,
Looks to me like the slope is vertical at (pi,0)
Therefore the normal is horizontal
dy/dx of normal = 0 
Calculus (Normal Line, please check my work) 
Steve,
One branch of the graph is an oval, and at (pi,0) does indeed have a vertical tangent. So, the normal line has zero slope.
Line is just y=0
go by wolfram dot com and ask it to graph your equation. You'll see you were right. 
Calculus (Normal Line, please check my work) 
Mishaka,
For the tangent line at (pi, 0), I find that the slope is 4/0, which is undefined. So, wouldn't this make the slope of the normal line undefined as well??? Did I miss something?

Calculus (Normal Line, please check my work) 
Steve,
even better, have wolfram plot
z = 4*sin(x) + 9*cos(y)
and check "show contour lines"
One of those contour lines will be at z=9, and will look show why your graph looks the way it does. 
Calculus (Normal Line, please check my work) 
Damon,
Yes, you are missing something. A vertical line has undefined (infinite) slope.
However the normal to that vertical line is simply a line with zero slope, a horizontal line. 
Calculus (Normal Line, please check my work) 
Damon,
Graph it to see as Steve suggested.
By the way Steve, did you see the message from Bob Pursley? 
Calculus (Normal Line, please check my work) 
Mishaka,
Using strictly the derivative (because wolfram isn't working for me), how can you prove that the slope of the normal line is 0?

Calculus (Normal Line, please check my work) 
Damon,
1/oo
is very very small :) 
Calculus (Normal Line, please check my work) 
Mishaka,
Where did you get infinity? Using differentiation, I found that:
dy/dx = (4cosx) / (9(siny))
When I put (pi, 0) into this equation, the denominator is 0, making the slope of the tangent line undefined. And since the slope of the normal line is 1/(slope of the tangent line), the normal line's slope would also be undefined. Is this is wrong, could you please show me stepbystep (using derivatives, please), how you came to your answer? 
Calculus (Normal Line, please check my work) 
Damon,
Undefined means very large magnitude, like infinite.
1/1 = 1
1/.1 = 10
1/.01 = 100
.
.
.
1/10^6 = 1,000,000
1/10^10 = 10,000,000,000
etc
as the denominator goes to zero, the term goes to infinity 
Calculus (Normal Line, please check my work) 
Damon,
If dy/dx is very very large
then 1/(dy/dx) is very very small 
Calculus (Normal Line, please check my work) 
Mishaka,
I apologize if I seem like I don't understand what you are trying to explain, but you have really confused me on what I thought was a more simple problem. I would really appreciate it if you could check my original answer using differentiation, instead, as this is the method I used to begin with.

Calculus (Normal Line, please check my work) 
Damon,
I did.
You correctly found that the slope of the function was undefined or infinitely large at the desired point.
You failed to make the connection that "undefined" means "huge". 1/zero is undefined and is the limit of 1/a very small number which is huge.
If the slope of the function is huge, the slope of the normal is tiny.
if m = 1/zero
then m' = 1/m = zero/1 = 0 
Calculus (Normal Line, please check my work) 
Mishaka,
I think I may be understanding what you are saying. So, to put this in very simple turns, rather than assuming that 1 / (4/0) was just undefined, I should have taken it out farther; for example, dividing by a fraction is the same as multiplying by its reciprocal, which would be this:
1 x (0/4) = 1 x 0 = 0
This better simplifies the answer for me and explains why the slope of the normal line is zero. Thanks! 
Calculus (Normal Line, please check my work) 
Damon,
Sure, that is a perfectly good way to look at it.