Find the length of the curve.

Question

x=9t-3t^3, y=9t^2, 0<_t<_3

Answer 1

Arc Length = ∫_{(0 to 3)} √ [(dx/dt)² + (dy/dt)²]dt

= ∫_{(0 to 3)} √[(9-9t²)² + (18t)²]dt

= 9∫_{(0 to 3)} √ [(1 - t²)² + 4t²] dt

= 9∫_{(0 to 3)} √ (t⁴ + 2t² + 1) dt = 9∫_{(0 to 3)} √(t² + 1)²dt

= 9∫_{(0 to 3)} (t² + 1)dt = 9[(1/3)t³ + t]_{(0 to 3)} = 108

Answer 2

Use the relationship that:

and use a variation of the arc length equation:

dy/dt = 18t

dx/dt = 9 - 9t²

So dy/dx = 18t/(9 - 9t²) = 2t/(1 - t²)

Since you also need dx for the integral, dx/dt = 9 - 9t² so dx = (9 - 9t²) dt

Then plug into the integral.

Answer 3