# Calculus (Normal Line, please check my work)

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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) -

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) -

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) -

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) -

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) -

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) -

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) -

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) -

-1/oo

is very very small :)

• Calculus (Normal Line, please check my work) -

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 step-by-step (using derivatives, please), how you came to your answer?

• Calculus (Normal Line, please check my work) -

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) -

If dy/dx is very very large
then -1/(dy/dx) is very very small

• Calculus (Normal Line, please check my work) -

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) -

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) -

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) -

Sure, that is a perfectly good way to look at it.

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