Finding the tangent and normal of a trigonometric equation?

esploded
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Homework Statement


a. find the tangent and normal line at P (0,0)
b. find when the tangent is horizontal

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

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).
 
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esploded said:

Homework Statement


a. find the tangent and normal line at P (0,0)
b. find when the tangent is horizontal

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

The Attempt at a Solution


Find the derivative- y'=3 (?) or it it 3x+cos(3x)?

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

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).

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

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

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.
 
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
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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