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

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SUMMARY

The discussion focuses on calculating 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. The integral used for the arc length is ∫ √{(dx/dt)² + (dy/dt)²} dt, leading to the expression 2√2 ∫ |sin(t)cos(t)| dt. A critical error identified is the misapplication of the square root function, where √{X²} should be treated as |X|. The integration process needs to correctly account for the absolute values to avoid incorrect evaluations.

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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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