Help me find the arc length of a parametric equation....

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Homework Help Overview

The discussion revolves around finding the arc length of a parametric equation defined by x = (sin(t))^2 and y = (cos(t))^2, with t ranging from 0 to 3π. Participants are exploring the integration process required to compute the arc length using the formula involving the derivatives of x and y with respect to t.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to set up the integral for arc length and express it in terms of trigonometric functions. Questions arise regarding the evaluation of the integral and the implications of using absolute values in the context of the square root of squared terms.

Discussion Status

There is an ongoing exploration of the integration process, with participants questioning the correctness of their steps and the use of absolute values. Some guidance has been offered regarding the treatment of the square root of squared terms, but no consensus has been reached on the next steps for evaluating the integral.

Contextual Notes

Participants note the bounds of integration and express concern about potential errors in their calculations. There is an acknowledgment of the constraints imposed by forum rules regarding the level of assistance that can be provided.

Rijad Hadzic
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1. The problem statement, all variables and given/k
nown data

x = (sin(t))^2 y = (cos(t))^2 t goes from 0 to 3 pi

Homework Equations


∫\sqrt{ {(dx/dt)}^2 + {(dy/dt)}^2 } dt

The Attempt at a Solution


∫\sqrt{ {(2sin(t)cost)}^2 + {(-2cos(t)sin(t))}^2 } dt

∫\sqrt{ 4(sin(t)cos(t))^2 + {(-2cos(t)sin(t))}^2 } dt

∫\sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt

∫\sqrt{ 8(sin(t)cos(t))^2 } dt

2\sqrt 2∫\sqrt{ (sin(t)cos(t))^2 } dt

2\sqrt 2∫{ (sin(t)cos(t)) } dt

keep in mind my bounds are from 0 to 3 pi right now

so I integrate this using u = sin (t) du = cos(t) dt
and get
(root 2)(sin(t))^2 from 0 to pi/3

now when I evaluate I get 0-0

where did my integration go wrong?
 
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Rijad Hadzic said:
1. The problem statement, all variables and given/k
nown data

x = (sin(t))^2 y = (cos(t))^2 t goes from 0 to 3 pi

Homework Equations


∫\sqrt{ {(dx/dt)}^2 + {(dy/dt)}^2 } dt

The Attempt at a Solution


∫\sqrt{ {(2sin(t)cost)}^2 + {(-2cos(t)sin(t))}^2 } dt

∫\sqrt{ 4(sin(t)cos(t))^2 + {(-2cos(t)sin(t))}^2 } dt

∫\sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt

∫\sqrt{ 8(sin(t)cos(t))^2 } dt

2\sqrt 2∫\sqrt{ (sin(t)cos(t))^2 } dt

2\sqrt 2∫{ (sin(t)cos(t)) } dt

keep in mind my bounds are from 0 to 3 pi right now

so I integrate this using u = sin (t) du = cos(t) dt
and get
(root 2)(sin(t))^2 from 0 to pi/3

now when I evaluate I get 0-0

where did my integration go wrong?

Never, never just write ##\sqrt{X^2} = X## because half the time it is wrong. What IS always correct is ##\sqrt{X^2} = |X|.##
 
Ray Vickson said:
Never, never just write ##\sqrt{X^2} = X## because half the time it is wrong. What IS always correct is ##\sqrt{X^2} = |X|.##

so it should be absolute value sin(t)cos(t) ??
 
Rijad Hadzic said:
so it should be absolute value sin(t)cos(t) ??

I thought that was what I said.
 
Ray Vickson said:
I thought that was what I said.
So where do I go from here? can I just call that the absolute value of 2sin since sin and cos have the same range?
 
Rijad Hadzic said:
So where do I go from here? can I just call that the absolute value of 2sin since sin and cos have the same range?

I have already offered all the help I feel I can according to PF rules.
 
Ray Vickson said:
I have already offered all the help I feel I can according to PF rules.

Thanks for the help.
 

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