Finding the tangent and normal of a trigonometric equation?

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Homework Help Overview

The discussion revolves around finding the tangent and normal lines to the trigonometric equation \(3x + \sin(3x)\) at the point \(P(0,0)\), as well as determining when the tangent line is horizontal.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find the derivative of the function to determine the slope at the given point. There is uncertainty regarding the correct form of the derivative, with some questioning the application of the chain rule. Others express confusion about the interpretation of the term \(\sin(3x)\) and its differentiation.

Discussion Status

The conversation includes attempts to clarify the derivative and the application of the chain rule. Some participants provide guidance on how to find the slope at the point of interest, while others explore the implications of the derivative's form.

Contextual Notes

There is a mention of needing to apply the chain rule correctly, and participants are encouraged to reference calculus resources for clarification. The discussion reflects a mix of understanding and confusion regarding the differentiation process.

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[itex]^{2}[/itex]+ 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:

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