Finding length (calculus)

Question

a) find the length of the curve (x, y) = (3t^2,2t^3),0 ≤ t ≤ 2.

b) Set up (do not evaluate) an integral that gives the length of the curve

y= 2 ln(sin(x/2)), π/2 ≤ x ≤ π

Answer 1

The length of a curve (also referred to as the length of an arc) is given by the formula:

L = ∫_a^b √(1 + f’(x)²)

(a)

We are given

x = 3t²

y = 2t³

0 ≤ t ≤ 2

We know that f’(x) = dy/dx

dy/dx = dy/dt/dx/dt

          = 6t² / 6t

          = t

L = ∫₀² √(1+t²)dt

  ≈ 2.96

A calculator was used to get this. The integration is quite messy.

(b)

y = 2ln(sin(x/2))

Using the chain rule,

y’ = cot(x/2)

Using the same formulas as in part (a)

L = ∫_π/2^π√(1+cot²(x))

Answer 2

From a pure math standpoint, the parameter t can represent anything. However, if you think of this as a motion problem, you can simply think of it if you're integrating speed.

Recall that velocity is a vector, and is equal to <dx/dt, dy/dt>. The magnitude of this vector is called "speed" and is

√((dx/dt)²+(dy/dt)²).

Integrating speed with respect to time (between the start and finish times) provides you with the total distance traveled during that time:

∫ (✓((dx/dt)²+(dy/dt)²)))dt. Make sure to use the given t-values as the boundaries of the integral.