Arc length with parametric curves

Question

Find the exact length of the curve

x= 2+12t^2

y= 8+8t^3

0≤ t≤ 2

Answer 1

Arc length formula L = ∫₀²√[x'²(t) + y'²(t)]dt = ∫₀²√[(24t)² + (24t²)²]dt = ∫₀²24t√ (1 + t²)dt = 12 · 2/3(1 + t²)^3/2₀² =

8(5^3/2 - 1) = 81.44

Answer 2

The equation for arc length of a parametric curve:

a

S = ∫ √ ( (dx/dt)² + (dy/dt)² ) dt

b

In the our case x(t) = 2 + 12t² and y = 8 + 8t³ , a = 2 , b = 0

dx/dt = 24t

dy/dt = 24t²

We will focus on the inside of the integral:

√ ( (dx/dt)² + (dy/dt)² ) = √ ( 24²( t² ) + 24²( t⁴ ) ) = √ ( 24t²( 1 + t² ) ) = 24t * √ (1 + t²)

We will move the 24 outside of the integral which leaves use with:

2

S = 24 * ∫ t * √ (1 + t²) dt

0

We will now use u-substitution:

Let u = √ (1 + t²), This means du/dt = t / √ (1 + t²)

We can solve for t * dt since we have that term in our integral.

t * dt = du * √ (1 + t²)

If we substitute our value for t * dt into our integral we get,

2

S = 24 * ∫ √ (1 + t²) * √ (1 + t²) du

0

Notice that u = √ (1 + t²), this means:

2

S = 24 * ∫ u² du

0

It is important to remember that we are no longer integrating over t and so we must change our interval of integration. This means we have to find the value of u at each of our t values.

When t = 0, u = √ ( 1 + (0)² ) = 1,

When t = 2, u = √ ( 1 + (2)² ) = √5

Now our integral is:

√5

S = 24 * ∫ u² du

1

Integrate:

S = 24 ( ( 1/3 * ( √5 )³ ) - ( 1/3 * ( 1 )³ ) )

Factor out 1/3:

S = 8 ( ( √5 )³ - ( 1 )³ )

Simplify:

S = 8 ( 5 √5 - 1)

This is the exact answer.

Answer 3

A starting point is to write the arc length , A, in the form:

A = ∫ [ (dx)^2 + (dy)^2 ]^1/2

Then, in terms of the parameter t , dx = 24 t dt and dy = 24 t^2 dt

A = 24 ∫ [t^2 + t^4 ]^1/2 dt = 24 ∫ [ 1 + t^2]^1/2 t dt

The limits of integration are 0 and 2. The integrand has a simple anti derivative so

A = 24 (1/3) [ 1 + t^2 ]^3/2 |₀² = 8 ( 5^3/2 -1) ~ 81,44