Let **r**(t)=<f(t),g(t),h(t)> be the position vector of a particle.
The velocity of the particle is given by

and the acceleration is given by

In the following discussion we gain further insight into these formulas.

**Velocity**

We can write the velocity in terms of the
unit
tangent vector
**T**(t). Recall that

The magnitude of the velocity,
|**r**'(t)|, is the speed and is
often denoted v(t). It follows that

Recall that the velocity vector **r**'(t) is tangent to the
space curve **r**(t). This is shown in the plot below for
the position function **r**=<2cos(t),sin(t)>, where the velocity
**r**'(t)=<-2sin(t),cos(t)> and the unit tangent vector are
plotted.

**Acceleration**

Acceleration is given by the formula

Here we have used the product rule for
differentiation of the product
of a scalar function and a vector function. An interesting fact
is that **T'(t)** is orthogonal to **T**(t)!. We prove this
by noting that

since **T**(t) is a tangent vector. Taking the derivative of both
sides above we have

We have used the rule for differentiating the dot product of two vector functions. Noting that the order does not matter in the dot product, we have

Since **T**(t) is tangent to the space curve **r**(t). The
vector **T**'(t) is normal to the space curve. In three dimensions,
there are an infinite number of linearly independent vectors normal
to **r**(t). The direction represented by **T**'(t) is called
the *principal normal* direction. The unit vector

is called the *principal unit normal vector*.

We can write the acceleration:

Recalling that |**T**'(t)|/v(t) is the
curvature, we have

This last formula shows that there are components of acceleration
tangential and normal to **r**(t). These components are

and

**Example**

Let **r**(t)=<2cos(t),sin(t)>. Find the tangential and normal
components of the acceleration.

We have **r**'(t)=<-2sin(t),cos(t)> and

Differentiating this last expression, we have:

From a previous calculation,

Hence,

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