A special feature of a smooth space curve is that they have measurable length. This allows us to locate points along these curves by giving their directed distances s along the curve from some base point. Remember that time is the parameter used for describing a moving body's velocity and acceleration, but s is the parameter used for studying a curve's shape.

FACT:

The length of a smooth curve r (t) = f (t)i + g (t)j + h (t)k, a £ t £ b, that is traced exactly once as t increases from t = a to t = b is

EXAMPLE 1:

Find the length of the curve r (t) = 6t ³ i - 2t ³ j - 3t ³ k, 1 £ t £ 2.

SOLUTION:

First I will find the derivatives of each of the components.

Now I am going to determine the radicand.

Now to find the length of this curve.

EXAMPLE 2:

Find the length of the curve r (t) = (t sin t + cos t)i + (t cos t - sin t)j,

SOLUTION:

EXAMPLE 3:

Find the arc length parameter along the curve r (t) = (cos t + t sin t)i + (sin t - t cos t)j from the point where t = 0 by evaluating the integral

Then find the length of the curve from p /2 £ t £ p .

SOLUTION:

v (t ) = (-sin t + sin t + t cos t )i + (cos t - cos t + t sin t )j

= (t cos t )i + (t sin t )j

To find the length of the curve from p /2 to p , S = S (p ) - S (p /2).

Since the derivatives beneath the radical are continuous (the curve is smooth), then the Fundamental Theorem of Calculus tells us that s is differentiable function of t with derivative

Notice that ds/dt > 0 since, by definition, | v | is never zero for a smooth curve.

Since ds/dt > 0 for the curves we are considering, s is then one-to-one and has an inverse that gives t as a differentiable function of s. The derivative of the inverse is

This makes r a differentiable function of s whose derivative can be calculated with the chain rule to be the following.

FACT:

The unit tangent vector of a differentiable curve r (t) is

EXAMPLE 4:

Find the unit tangent vector for r (t) = (6 sin 2t)i + (6 cos 2t)j + 5t k.

SOLUTION:

v = (12 cos 2t)i + (-12 sin 2t)j + 5k

EXAMPLE 5:

Find the unit tangent vector for r (t) = (2 + t)i - (t + 1)j + t k.

SOLUTION:

v = i - j + k

EXAMPLE 6:

Find the unit tangent vector for r (t) = (t sin t + cos t)i + (t cos t - sin t)j.

SOLUTION:

v = (sin t + t cos t - sin t)i + (cos t - t sin t - cos t)j

= (t cos t)i - (t sin t)j