Arc Length of Parametric Curve: t = 0 to t = 1

In summary, the problem is to find the arc length of a curve given parametrically from t = 0 to t = 1, with the curve given by x = 4t^2 and y = 2t. The parametric arc length equation is used, and a triangle is drawn for trigonometric substitution. However, a mistake is made in the integration by parts step, and the correct integral is found to be \int tan^2(\sigma) sec(\sigma)d\sigma. Using partial fractions, the final answer is determined to be (15/4)sqrt(68) - (14/4).
  • #1
sassafrasaxe
12
0

Homework Statement



Find the arc length of a curve given parametrically from t = 0 to t = 1.
Curve given by x = 4t^2, y = 2t


Homework Equations



[I think] parametric arclength =
integral from t = b to t = a of sqrt( (dx/dt)^2 + (dy/dt)^2)dt


The Attempt at a Solution



dx/dt = 8t, and dy/dt = 2
Then I did S from 0 to 1 of sqrt((8t)^2) + (2)^2)dt

I then drew a triangle for trigonometric substitution, with the tangent case.
Let t = (1/4)tan(σ)
Let dt = (1/4)sec^2(σ) dσ
Let sqrt(64t^2 + 4) = 4sec(σ)
Then S[ sqrt( (8t)^2 + (2)^2) ]dt becomes
S of 4sec(σ) * (1/4)sec^2(σ) dσ
which equals S[ sec^3(σ) ] dσ

Now I solved that with integration by parts, letting u = sec(σ), and dv = sec^2(σ)
so that du = sec(σ)tan(σ)dσ and v = tan(σ)
now S[ sec^3(σ) ]dσ = sec(σ)tan(σ) - S[ tan^3(σ)sec(σ) ]dσ

that last little integral = S[ tan^3(σ)sec(σ) ]dσ = S[ ((sec^2(σ) - 1)*sec(σ)tan(σ)]dσ]
=(1/3)sec^3(σ) - sec(σ)
so the entire S[ sec^3(σ) ]dσ = sec(σ)tan(σ) - (1/3)sec^3(σ) + sec(σ)
so (4/3)S = sec(σ) + sec(σ)*tan(σ)
so S = (3/4)[sec(σ)tan(σ) + sec(σ)]

Now I'm replacing this value with the value given by my triangle for sec(σ) and tan(σ)
so that S = (3/4)[sqrt(64t^2 + 4)*4t + sqrt(64t^2 + 4)] evaluated from t = 0 to t = 1
which would be
(3)sqrt(68) + (3/4)sqrt(68)
minus
(6/4) + 2
=
(15/4)sqrt(68) - (14/4) as my final answer.

Could someone please tell me if I'm doing anything wrong? Maybe if I'm going about this entirely the wrong way? Unfortunately I don't even know the right answer, but even if I can't get the right answer, as long as I understand how to get there, I feel okay.

I would really appreciate it, and I'm sorry about the typing, I'm not very good at typing these problems out. So much thanks,
Sassy
 
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  • #2
sassafrasaxe said:

Homework Statement



Find the arc length of a curve given parametrically from t = 0 to t = 1.
Curve given by x = 4t^2, y = 2t


Homework Equations



[I think] parametric arclength =
integral from t = b to t = a of sqrt( (dx/dt)^2 + (dy/dt)^2)dt
Yes, that's correct.


The Attempt at a Solution



dx/dt = 8t, and dy/dt = 2
Then I did S from 0 to 1 of sqrt((8t)^2) + (2)^2)dt

I then drew a triangle for trigonometric substitution, with the tangent case.
Let t = (1/4)tan(σ)

Let dt = (1/4)sec^2(σ) dσ[/quote]
No big deal but I wouldn't use the word "let" here. You already said "let t= (1/4)tan(σ) and then dt= (1/4)sec^2(σ) follows from that.

Let sqrt(64t^2 + 4) = 4sec(σ)
Same point

Then S[ sqrt( (8t)^2 + (2)^2) ]dt becomes
S of 4sec(σ) * (1/4)sec^2(σ) dσ
which equals S[ sec^3(σ) ] dσ

Now I solved that with integration by parts, letting u = sec(σ), and dv = sec^2(σ)
so that du = sec(σ)tan(σ)dσ and v = tan(σ)
now S[ sec^3(σ) ]dσ = sec(σ)tan(σ) - S[ tan^3(σ)sec(σ) ]dσ
How did you get this? with [itex]du= sec(\sigma)tan(\sigma)d\sigma)dx[itex] and [itex]v= tan(\sigma)[/itex], as you say, that last integrand is [itex]sec(\sigma)tan^2(\sigma)d\sigma[/itex], not [itex]tan^3(\sigma)sec(\sigma)d\sigma[/itex].

that last little integral = S[ tan^3(σ)sec(σ) ]dσ = S[ ((sec^2(σ) - 1)*sec(σ)tan(σ)]dσ]
=(1/3)sec^3(σ) - sec(σ)
so the entire S[ sec^3(σ) ]dσ = sec(σ)tan(σ) - (1/3)sec^3(σ) + sec(σ)
so (4/3)S = sec(σ) + sec(σ)*tan(σ)
so S = (3/4)[sec(σ)tan(σ) + sec(σ)]
With [itex]tan^2(\sigma) sec(\sigma)d\sigma[/itex], I would proceed by putting everything in terms of sine and cosine:
[tex]\int tan^2(\sigma) sec(\sigma)d\sigma= \int \frac{sin^2(\sigma)}{cos^2(\sigma)}\frac{1}{cos(\sigma)}d\sigma= \int \frac{sin^2(\sigma)}{cos^3(\sigma)}d\sigma[/tex]

With that odd power of cosine, I would multiply both numerator and denominator by [/itex]cos(\sigma)[/itex] to get
[tex]\int \frac{sin^2(\sigma)}{cos^4(\sigma)}cos(\sigma)d\sigma= \int\frac{sin^2(\sigma)}{(1- sin^2(\sigma))^2} cos(\sigma)d\sigma[/tex]
and use the substitution [itex]u= sin(\sigma)[/itex], [itex]du= cos(\sigma)d\sigma[/itex] so that the integral becomes
[tex]\int \frac{u^2}{(1- u^2)^2}du[/tex]

Now I'm replacing this value with the value given by my triangle for sec(σ) and tan(σ)
so that S = (3/4)[sqrt(64t^2 + 4)*4t + sqrt(64t^2 + 4)] evaluated from t = 0 to t = 1
which would be
(3)sqrt(68) + (3/4)sqrt(68)
minus
(6/4) + 2
=
(15/4)sqrt(68) - (14/4) as my final answer.

Could someone please tell me if I'm doing anything wrong? Maybe if I'm going about this entirely the wrong way? Unfortunately I don't even know the right answer, but even if I can't get the right answer, as long as I understand how to get there, I feel okay.

I would really appreciate it, and I'm sorry about the typing, I'm not very good at typing these problems out. So much thanks,
Sassy
 
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Likes 1 person
  • #3
Let sqrt(64t^2 + 4) = 4sec(σ)

Should this be 2sec(σ)?

Then also, let t = 1/4 sinh(σ) for a shorter solution, although this is an example of why I disliked integration, it is sometimes very arcane.
 
  • #4
Okay great thank you very much. So I took it from where you left off, HallsofIvy, and I see what you mean that yes, I did mess up with the integral and it should have been tan^2(σ)sec(σ) d(σ).
Picking up where you left off, I suppose I would then need partial fractions?
Wow, this has been quite the problem! Looks deceivingly simple.
 

Related to Arc Length of Parametric Curve: t = 0 to t = 1

1. What is the definition of arc length?

Arc length is the distance along a curve or arc. It is calculated by measuring the length of the curve between two points.

2. How is arc length calculated for a parametric curve?

To calculate arc length for a parametric curve, t values are substituted into the curve equation to get x and y coordinates. The distance between each set of coordinates is then calculated using the distance formula. These distances are then summed up to get the total arc length.

3. What is the significance of t = 0 to t = 1 in the arc length of a parametric curve?

The values of t = 0 to t = 1 represent the starting and ending points of the parametric curve. These values are used to determine the portion of the curve that is being measured for arc length.

4. How is arc length affected by the shape of a parametric curve?

The shape of a parametric curve can greatly affect the arc length. Curves with sharp turns or high curvature will have a longer arc length compared to curves with gentler slopes.

5. Can the arc length of a parametric curve be negative?

No, the arc length of a parametric curve cannot be negative. Arc length is always a positive value as it represents the distance along the curve.

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