278 CHAPTER 15. MOTION ON A SPACE CURVE

15.1.1 Some Simple TechniquesRecall the formula for acceleration is

a= aTT +aNN (15.3)

where aT = d|v|dt and aN = κ |v|2. Of course one way to find aT and aN is to just find

|v| , d|v|dt and κ and plug in. However, there is another way which might be easier. Take the

dot product of both sides with T. This gives,

a ·T = aTT ·T +aNN ·T = aT .

Thusa= (a ·T )T +aNN

and soa− (a ·T )T = aNN (15.4)

and taking norms of both sides,

|a− (a ·T )T |= aN .

Also from (15.4),a− (a ·T )T

|a− (a ·T )T |=

aNN

aN |N |=N.

Also recall

κ =|a×v||v|3

, a2T +a2

N = |a|2

This is usually easier than computing T ′/∣∣T ′∣∣. To illustrate the use of these simple obser-

vations, consider the example worked above which was fairly messy. I will make it easierby selecting a value of t and by using the above simplifying techniques.

Example 15.1.5 Let R(t) =(cos(t) , t, t2

)for t ∈ [0,3]. Find the speed, velocity, curva-

ture, and write the acceleration in terms of normal and tangential components when t = 0.Also find N at the point where t = 0.

First I need to find the velocity and acceleration. Thus

v = (−sin t,1,2t) , a= (−cos t,0,2)

and consequently, T = (−sin t,1,2t)√sin2(t)+1+4t2

. When t = 0, this reduces to

v (0) = (0,1,0) , a= (−1,0,2) , |v (0)|= 1, T = (0,1,0) .

Then the tangential component of acceleration when t = 0 is

aT = (−1,0,2) · (0,1,0) = 0

Now |a|2 = 5 and so aN =√

5 because a2T +a2

N = |a|2. Thus√

5 = κ |v (0)|2 = κ ·1 = κ .Next lets find N . From a= aTT +aNN it follows

(−1,0,2) = 0 ·T +√

5N