# 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