Find Length of Curve x=cos(2t), y=3sin(2t): 0 or Not?

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Homework Help Overview

The discussion revolves around finding the length of a parametric curve defined by the equations x=cos(2t) and y=3sin(2t) over the interval t from 0 to π. Participants are exploring the implications of the curve being closed and whether this affects the calculated length.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the length but arrives at a result of zero, questioning if this is typical for closed curves. Other participants inquire about the setup of the integral and the procedures used in the calculations.

Discussion Status

Participants are actively discussing the setup of the integral for arc length, with some suggesting that the integral may be more complex than initially presented. There is a mix of interpretations regarding the correct approach to calculating the length, and guidance has been offered regarding the mathematical formulation.

Contextual Notes

There are indications of confusion regarding the implications of the curve being closed and how this relates to the length calculation. Participants are also addressing potential misunderstandings about the arc length formula and its application in this context.

sibiryk
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I need to find length of the curve x=cos(2t), y=3sin(2t), t[0,p]

I'm getting a length of this curve to be equal zero.
I think it is because this curve is closed and start point = end point.
Is it normal?
How can I get a length of this curve?
 
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No you should still be getting a length, what procedure are you using? Show the integral you set up.
 
length of a arc is given by

s = # * r

# is the internal angle in rads

r is radias
 
I got

s(t)=Integral (2sin(2t)-6cos(2t))dt

integral from 0 to pi
 
I remember the arc length integral being more complex than that.. can you explain how you set it up/
 
The length of a curve f given in parametric form can be calculated with

[tex]\int_a^b {\left\| {\frac{{d\vec f}}{{dt}}} \right\|dt}[/tex]

Which is, written out in 2 variables:

[tex]\int_a^b {\sqrt {x'\left( t \right)^2 + y'\left( t \right)^2 } dt}[/tex]
 
A tip: [tex]\sqrt {x^2 } = \left| x \right| \ne x,\left| x \right| = x \Leftrightarrow x \ge 0[/tex]
 

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