276 CHAPTER 15. MOTION ON A SPACE CURVE

and N in the case where T ′ ̸= 0 is called the osculating1 plane. It identifies a particularplane which is in a sense tangent to this space curve.

The important thing about this is that it is possible to write the acceleration as the sumof two vectors, one perpendicular to the direction of motion and the other in the directionof motion.

Theorem 15.1.3 For R(t) the position vector of a space curve, the acceleration is givenby the formula

a=d |v|dt

T +κ |v|2N ≡ aTT +aNN . (15.1)

Furthermore, a2T +a2

N = |a|2.

Proof:

a=dvdt

=ddt

(R′)=

ddt

(|v|T ) =d |v|dt

T + |v|T ′ = d |v|dt

T + |v|2 κN.

This proves the first part.For the second part,

|a|2 = (aTT +aNN) · (aTT +aNN)

= a2TT ·T +2aNaTT ·N +a2

NN ·N = a2T +a2

N

because T ·N = 0. ■From 15.1 and the geometric properties of the cross product,

a×v = κ |v|2N ×v

Hence, using the geometric description of the cross product again using that the anglebetween N and T is 90◦,

|a×v|= κ |v|2 |v| , κ =|a×v||v|3

=|v×a||v|3

(15.2)

Finally, it is good to point out that the curvature is a property of the curve itself, anddoes not depend on the parametrization of the curve. If the curve is given by two differentvector valued functions R(t) and R(τ), then from the formula above for the curvature,

κ (t) =

∣∣T ′ (t)∣∣|v (t)|

=

∣∣ dTdτ

dτ

dt

∣∣∣∣ dRdτ

dτ

dt

∣∣ =∣∣ dT

dτ

∣∣∣∣ dRdτ

∣∣ ≡ κ (τ) .

From this, it is possible to give an important formula from physics. Suppose an objectorbits a point at constant speed v. In the above notation, |v| = v. What is the centripetalacceleration of this object? You may know from a physics class that the answer is v2/rwhere r is the radius. This follows from the above quite easily. First, what is the curvatureof a circle of radius r? A parameterization of such a curve is

R(t) = (r cos t,r sin t)

1To osculate means to kiss. Thus this plane could be called the kissing plane. However, that does not soundformal enough so we call it the osculating plane.