Part 2:  The Decomposition of Acceleration   

Since a curve's velocity can be written v = vT where v is the speed and T is the unit tangent vector, the acceleration for the curve is  

a =  d dt ( vT) =   dv dt  T + v dT dt

(4)

Moreover, if we now combine (3) for dT/dt with (4) then we find that the acceleration of a curve parameterized by r(t) is

a = dv dt T  +  kv2 N

(5)

The quantity a_T = dv/dt is the rate of change of the speed and is often called the linear acceleration for the parameterization. The linear acceleration is also known as the tangential component of acceleration because it measures the acceleration in the direction of the velocity.

The quantity a_N = kv² is called the normal component of acceleration because it measures the acceleration applied at a right angle to the velocity. Specifically, the normal component of acceleration is a measure of how fast the direction of the velocity vector is changing.

Moreover, since a·T = dv/dt, the decomposition (5) implies that

||a||2 = aT2+aN2 = ( a·T) 2+k2v4

Thus, k²v⁴ = ||a||²  -  ( a·T) ², so that

k = ||a||2 - ( a·T) 2 v2

(6)

which does not require the calculation of a cross product.  

EXAMPLE 2    Find the curvature of the vector-valued function

r( t) = á sinh(t), t, cosh(t) ñ

Solution: The velocity is v( t) = ácosh( t) ,1,sinh( t) ñ , so that the speed is

v =  cosh2( t) +1+sinh2( t)  =  2cosh2( t)  = Ö2 cosh( t)

As a result, the linear acceleration is

dv dt  =  d dt Ö2cosh( t) = Ö2sinh( t)

The derivative of v( t) then yields the acceleration,

a( t) = á sinh( t) ,0,cosh( t) ñ

and the dot product a·T  is given by

a · T  =  1 Ö2cosh( t) (2sinh(t) cosh( t) ) = Ö2 sinh(t)

Thus, (8) implies that the curvature is

k =   sinh2( t) +cosh2( t)-2sinh2( t) 22cosh2( t)  =  cosh2( t) -sinh2( t) 22cosh2( t)  =  1 4cosh2( t)

since cosh²( t) -sinh²( t) = 1.

Indeed, if the speed v is constant, then dv/dt = a·T = 0 and (6) reduces to

k =   ||a||2 v2   =    a v2

(7)

where a is the magnitude of the acceleration. That is, the curvature of an object moving at a constant speed along a curve is proportional to the magnitude of the acceleration.      

EXAMPLE 3    Find the linear acceleration and curvature of the helix

r( t) = á 3cos( t) ,3sin(t) ,4t ñ

Solution: The velocity and acceleration are, respectively, given by

v( t) = á -3sin( t) ,3cos(t) ,4 ñ ,        a = á -3cos(t) ,-3sin( t) ,0 ñ

It follows that the speed is given by

v = 9sin2( t) +9cos2( t) +16  = 5

Thus, we can use (7).  Since a = 3, the curvature is

k =  a v2 =  3 25

 

Finally, since v is parallel to T, the decomposition (5) implies that

v × a =  dv dt ( v×T) + kv2( v×N) = kv2( v×N)

Since v and N are orthogonal, it follows that || v×N|| = v·1·sin( p/2) = v, so that

|| v×a|| = kv2|| v×N|| = kv3

Finally, solving for k yields another means of computing curvature:

k = || v×a|| v3