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December 20, 2014

December 20, 2014

Posted by **Mishaka** on Saturday, December 17, 2011 at 3:51pm.

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**, Saturday, December 17, 2011 at 4:12pmLooks 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**, Saturday, December 17, 2011 at 4:14pmOne 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**, Saturday, December 17, 2011 at 4:18pmFor 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**, Saturday, December 17, 2011 at 4:19pmeven 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**, Saturday, December 17, 2011 at 4:23pmYes, 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**, Saturday, December 17, 2011 at 4:24pmGraph 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**, Saturday, December 17, 2011 at 4:24pmUsing 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**, Saturday, December 17, 2011 at 4:26pm-1/oo

is very very small :)

- Calculus (Normal Line, please check my work) -
**Mishaka**, Saturday, December 17, 2011 at 4:29pmWhere 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) -
**Damon**, Saturday, December 17, 2011 at 4:34pmUndefined 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**, Saturday, December 17, 2011 at 4:39pmIf dy/dx is very very large

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

- Calculus (Normal Line, please check my work) -
**Mishaka**, Saturday, December 17, 2011 at 4:42pmI 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**, Saturday, December 17, 2011 at 4:51pmI 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**, Saturday, December 17, 2011 at 4:59pmI 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**, Saturday, December 17, 2011 at 5:19pmSure, that is a perfectly good way to look at it.

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