Finding the tangent and normal of a trigonometric equation?

1. Sep 29, 2011

esploded

1. The problem statement, all variables and given/known data
a. find the tangent and normal line at P (0,0)
b. find when the tangent is horizontal

Equation: 3x+sin3x at (0,0)

3. The attempt at a solution
Find the derivative- y'=3 (?) or it it 3x+cos(3x)?
But from there, I don't know how to find the rate of change to create a new equation. I usually find the slope, get the x and y point values (for this problem, the points are given) and use y-y1=m(x-x1).

2. Sep 29, 2011

LCKurtz

Neither. Of course it has two terms but you need the chain rule for the second term.

You just need the slope at (0,0) so put in x = 0 once you have the correct derivative and you will have your m.

3. Sep 29, 2011

esploded

I'm not sure I follow. Is sin3x a product of sin and 3x?
I thought sin3x was a single term.

4. Sep 29, 2011

LCKurtz

It is sin(3x). The derivative of sin(x) is cos(x). But you don't have sin(x). Look in your calculus book about the chain rule.

5. Sep 29, 2011

Allenman

chain rule:
if you have
f(g(x))

take the derivative of the inside "g(x)", and multiply it by the derivative of the outside "f(x)":
g'(x)f'(g(x))

So say we have a similar problem:

2x$^{2}$+ x + cos(5x)

The derivative would be

4x + 1 - 5sin(5x)

Plugging in zero would give you
1 - 5sin(0)

sin(0)=0

so the answer for this example would be 1
now try the same method on your problem

Last edited: Sep 29, 2011