Calculus 2 Parametric and Arc Length

Question

A simple curve is defined parametrically by

x = 2t and y = t^3 + 1/3t for 1<= t <= 2

Show that the curve is smooth on the given interval and then find its arc length.

Answer 1

Hi David:

Here we go!

Parametric equations of the curve: x = 2t

y = t³ + t/3 for 1 <= t <= 2

Substitute t = x/2 into the expression for y: y = (x/2)³ + (1/3)(x/2)

= x³/8 + x/6

This is a standard smooth cubic curve with no discontinuities and everywhere differentiable in [2, 4]

Arc length = ∫ [2, 4] √( 1 + (dy/dx)² ) dx with the integration limits now set to x values in [2, 4]

= ∫ [2, 4] √( 1 + (3x²/8 + 1/6)² ) dx

= ∫ [1, 2] √(1 + 9x⁴/64 + x²/8 + 1/36) dx

= ∫ [1, 2] √(9x⁴/64 + x²/8 + 37/36) dx

Unfortunately, the expression inside the square root isn't a perfect square so this definite integral can't be solved without the use of much more complex math functions.

Therefore, we must stop here.

Answer 2

Hi David:

Here we go!

Parametric equations of the curve: x = 2t

y = t³ + t/3 for 1 <= t <= 2

Substitute t = x/2 into the expression for y: y = (x/2)³ + (1/3)(x/2)

= x³/8 + x/6

This is a standard smooth cubic curve with no discontinuities and everywhere differentiable in [2, 4]

Arc length = ∫ [2, 4] √( 1 + (dy/dx)² ) dx with the integration limits now set to x values in [2, 4]

= ∫ [2, 4] √( 1 + (3x²/8 + 1/6)² ) dx

= ∫ [1, 2] √(1 + 9x⁴/64 + x²/8 + 1/36) dx

= ∫ [1, 2] √(9x⁴/64 + x²/8 + 37/36) dx

Unfortunately, the expression inside the square root isn't a perfect square so this definite integral can't be solved without the use of much more complex math functions.

Therefore, we must stop here.