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

In summary, when solving an integral involving the absolute value of a function, it is important to use the correct notation and account for the range of the function in order to avoid errors in the final answer.
  • #1
Rijad Hadzic
321
20
1. The problem statement, all variables and given/k
nown data

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

Homework Equations


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

The Attempt at a Solution


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

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

∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt[/itex]

∫[itex] \sqrt{ 8(sin(t)cos(t))^2 } dt[/itex]

[itex]2\sqrt 2 [/itex]∫[itex] \sqrt{ (sin(t)cos(t))^2 } dt[/itex]

[itex]2\sqrt 2 [/itex]∫[itex] { (sin(t)cos(t)) } dt[/itex]

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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  • #2
Rijad Hadzic said:
1. The problem statement, all variables and given/k
nown data

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

Homework Equations


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

The Attempt at a Solution


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

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

∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt[/itex]

∫[itex] \sqrt{ 8(sin(t)cos(t))^2 } dt[/itex]

[itex]2\sqrt 2 [/itex]∫[itex] \sqrt{ (sin(t)cos(t))^2 } dt[/itex]

[itex]2\sqrt 2 [/itex]∫[itex] { (sin(t)cos(t)) } dt[/itex]

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|.##
 
  • #3
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) ??
 
  • #4
Rijad Hadzic said:
so it should be absolute value sin(t)cos(t) ??

I thought that was what I said.
 
  • #5
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?
 
  • #6
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.
 
  • #7
Ray Vickson said:
I have already offered all the help I feel I can according to PF rules.

Thanks for the help.
 

1. What is the definition of arc length?

The arc length of a parametric equation is the distance along the curve from one point to another, measured along the curve itself.

2. How do you find the arc length of a parametric equation?

To find the arc length, you must first determine the parametric equations for x and y in terms of a third variable, typically t. Then, use the following formula:
Arc Length = ∫ab √(dx/dt)2 + (dy/dt)2 dt
where a and b are the starting and ending values of t.

3. Can you provide an example of finding arc length for a parametric equation?

For the parametric equation x = 2t, y = t2, the starting and ending values of t are 0 and 1, respectively. Plugging these values into the formula, we get:
Arc Length = ∫01 √(2)2 + (2t)2 dt
= ∫01 √4 + 4t2 dt
= ∫01 √4 + 4t2 dt
= ∫01 √4 + 4t2 dt
= ∫01 √4 + 4t2 dt
= ∫01 √4 + 4t2 dt
= ∫01 2 + 4t2 dt
= 2t + 4t3/3 + C
Evaluating at the limits of integration, we get the final answer of 2 + 4/3 = 6/3 = 2 units.

4. Are there any special cases to consider when finding arc length of a parametric equation?

Yes, there are a few special cases to keep in mind. If the parametric equations are given in terms of polar coordinates, the formula for arc length changes to: Arc Length = ∫ab √(r)2 + (r')2 dt. Additionally, if the parametric equations contain a trigonometric function, you may need to use trigonometric identities to simplify the integral.

5. How is the arc length of a parametric equation used in real-life applications?

The arc length of a parametric equation is a useful tool in many fields of science and engineering, particularly in studying the motion of objects. For example, it can be used to calculate the distance traveled by a particle along a curved path, or the length of a wire in an electrical circuit. It is also used in calculus to find the surface area of 3D objects and to solve optimization problems.

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