- #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