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.
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 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
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,
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