# Homework Help: Parametric Arclength

1. Aug 16, 2013

### sassafrasaxe

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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

2. Aug 16, 2013

### HallsofIvy

Yes, that's correct.

3. 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.

Same point

How did you get this? with $du= sec(\sigma)tan(\sigma)d\sigma)dx$and [itex]v= tan(\sigma)$, as you say, that last integrand is $sec(\sigma)tan^2(\sigma)d\sigma$, not $tan^3(\sigma)sec(\sigma)d\sigma$. With $tan^2(\sigma) sec(\sigma)d\sigma$, I would proceed by putting everything in terms of sine and cosine: $$\int tan^2(\sigma) sec(\sigma)d\sigma= \int \frac{sin^2(\sigma)}{cos^2(\sigma)}\frac{1}{cos(\sigma)}d\sigma= \int \frac{sin^2(\sigma)}{cos^3(\sigma)}d\sigma$$ With that odd power of cosine, I would multiply both numerator and denominator by$cos(\sigma)[/itex] to get
$$\int \frac{sin^2(\sigma)}{cos^4(\sigma)}cos(\sigma)d\sigma= \int\frac{sin^2(\sigma)}{(1- sin^2(\sigma))^2} cos(\sigma)d\sigma$$
and use the substitution $u= sin(\sigma)$, $du= cos(\sigma)d\sigma$ so that the integral becomes
$$\int \frac{u^2}{(1- u^2)^2}du$$

3. Aug 16, 2013

### verty

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. Aug 16, 2013

### sassafrasaxe

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.