Part 2: The Unit Tangent Vector

If r( t) denotes the position of an object at time t, then r( t+Dt) -r(t) is the displacement of the object between times t and t+Dt for some small Dt.  The distance Ds the object travels between t and t+Dt satisfies

Ds » ||r( t+Dt) -r( t) ||

when r( t) is smooth and Dt is sufficiently close to 0.  

That is, over the short period of time from t to time t+Dt, the rate of change of the object satisfies

rate »  Ds Dt »  ||r( t+Dt) -r( t) || Dt

ds dt = lim Dt® 0  ||  r(t+Dt) -r( t) Dt || = ||v||

Pragmatically, it is important to note that the speed is a scalar -- in particular, the square root of an inner product.  To keep such square roots from becoming overly tedious, we will see many problems in which either algebra or trigonometry can be used to reduce an expression under a radical to a perfect square.

EXAMPLE 3    Find the speed of the object with position r( t) = á 3sin( 2t) ,5cos(2t) ,4sin( 2t) ñ in feet at time t in seconds.       

Solution: To do so, we first compute the velocity:

v = á 6cos( 2t) ,-10sin( 2t),8cos( 2t) ñ

The speed is then the magnitude of the velocity:

ds dt  =  [ 6cos( 2t) ] 2+[-10sin( 2t) ] 2+[ 8cos( 2t) ]2  =  36cos2( 2t) +100sin2( 2t) +64cos2( 2t)  =  100cos2( 2t) +100sin2( 2t)  =  10 feet per second

       

Speed is also denoted by v (given that we typically use the italic to denote the magnitude of a vector denoted by the same letter in bold typeface). In addition, at points where the speed is non-zero, we define the unit tangent vector T(t) to be the unit vector in the direction of v:

T = 1 v  v

(A curve is regular if v is non-zero at every point.)

As a result, v = vT, which shows that velocity can be written as the product of its speed and direction:       

EXAMPLE 4    Find the unit tangent vector to r(t) = á e^t, 2t, 2e^-t ñ .      

Solution: To do so, we first compute the velocity:

v( t) =  d dt áet,2t,2e-t ñ = á et,2,-2e-t ñ

Then we find the magnitude of the velocity

|| v|| =    ( et)2+( 2)2+( -2e-t)2   =    (et)2 +4+4(e-t)2

Notice now that since e^te^-t = 1, the quantity under the square root can be factored into a perfect square:

|| v|| =  ( et+2e-t)2  = et+2e-t

We then divide the velocity by the magnitude to obtain the unit tangent vector:

T( t) = 1 et+2e-t v =   et et+2e-t ,  2 et+2e-t ,  -2e-t et+2e-t