Question

How do I find the equation of the normal line, (y=mx+b) to the curve at the given point? For example f(x)=sec(xy) - 1=tan(x/y) at (0,-1). I've already taken the derivative and end up with some really big complicated equation. Then, when I plug in the coordinate points to figure out the slope of the line, I get an answer that isn't possible like 0=-1. Any ideas?

Answer 1

Get the derivative first :)

Answer 2

Already have, but thanks for the tip!

Answer 3

What did you get for the derivative?

Answer 4

Well, if you get it, it's
\[\Large \sec(xy)\tan(xy)[xy'+y]=\sec^2\left[\frac{x}y\right]\left(\frac{y-xy'}{y^2}\right)\]

Answer 5

I actually wouldn't want to isolate yet, first, replace all the x's with 0, and all the y's with -1, and only then to isolate y'
^.^

Answer 6

But then again, never mind :)

Answer 7

I have seen your point...

Answer 8

yeah, i'm not sure where I'm going wrong. It's either 0=1, or 0=-1

Answer 9

Then maybe there simply is no escaping isolating at the first expression?

Answer 10

I don't understand...what's the original equation you're given?

Answer 11

Maybe the tangent is vertical.

Answer 12

If you do isolate, I reckon it's going to be a 1/0 slope, so that's vertical. Then its normal line must be horizontal, no?

Answer 13

Draw the graph of the function and see if it looks like that tan is vertical at the given point. (use wolf)

Answer 14

Better, use wolf to actually get the derivative :D

Answer 15

how in the world did you get that isolated?

Answer 16

I didn't. Wolfram did.

Answer 17

But if plugging x = 0 and y = -1
yields a false equation, it must mean that y' does not exist at that point. So it must be a vertical slope :)

Answer 18

I plugged them in and got y' = -1/0. Does that mean the normal line for that equation would be 0/1, making it a horizontal line? If that's true, then I can just plug in the coordinate point into y=0(x)+b, right?

Answer 19

Yeah, that's right :)
Now you have a point (0 , -1)
and a slope m = 0
Now get the line equation ^.^

Answer 20

I got y=-1.

Answer 21

And there's your normal line :)

Answer 22

Awesome! Thanks for all the help.

Answer 23

Anytime... well not really ^.^

Answer 24

I was gonna say.. I got some more problems if you wanna help

Answer 25

Post them. I'll see what I can do ^.^

Answer 26

I posted one other one, the rates of change one. It's closed now, but think you can help with that one?

Answer 27

Saw it, it makes me dizzy, I think I might do more harm than good If I tackle that :)

Answer 28

Darn, thanks though!

Answer 29

Sorry :)