velocity and acceleration - three dimensional

Question

Velocity and Acceleration - Three Dimensional Space

In this part we need to take a look at the velocity and acceleration of a moving object.   

From Calculus I we are familiar with that given the position function of an object that the velocity of the object is the 1^st derivative of the position function and the acceleration of the object is the 2^nd derivative of the position function. 

Thus, given this it shouldn't be too surprising that whether the position function of an object is specified by the vector function  r→(t) then the velocity and acceleration of the object is illustrated by,

v^→ (t) = ^→r'(t)

a^→ (t) = ^→r'' (t)

Note: The velocity and acceleration are as well going to be vectors also.

In the study of the motion of objects the acceleration is frequently broken up into a tangential component, a_T, and the normal component denoted as a_N.  The tangential component is the part or element of the acceleration which is tangential to the curve and the normal component is the part of the acceleration which is normal or orthogonal to the curve.  If we do this we can write down the acceleration as,

a^→ = a_T T^→+ a_NN^→

where T^→ and N^→ stands for the unit tangent and unit normal for the position function.

If we illustrate v = ||v^→ (t)|| then the tangential and normal components of the acceleration are specified by,  

a_T = v' =^→r' (t). ^→r''(t) /(||r' (t)||)

a_N = kv² = ||?r' (t) *^→r" (t)|| / ||r' (t)||

in which k is the curvature for the position function.

There are two (2) formulas to employ here for each component of the acceleration and when the second formula may seem excessively complicated it is frequently the easier of the two.  In the tangential component, v, might be messy and calculating the derivative may be unpleasant.  In the normal component we will previously be computing both of these quantities in order to get the curvature and thus the second formula in this case is certainly the easier of the two.