282 CHAPTER 15. MOTION ON A SPACE CURVE

and so

N =dTds

/

∣∣∣∣dTds

∣∣∣∣=−((cos t)i+(sin t)j)

The binormal:

B =1√

a2 +b2

∣∣∣∣∣∣∣i j k

−asin t acos t b−cos t −sin t 0

∣∣∣∣∣∣∣=1√

a2 +b2((bsin t) i−bcos tj+ak)

Now the curvature κ (t) =∣∣ dT

ds

∣∣=√( acos ta2+b2

)2+(

asin ta2+b2

)2= a

a2+b2 . Note the curvature

is constant in this example. The final task is to find the torsion. Recall that B′ = τN wherethe derivative on B is taken with respect to arc length. Therefore, remembering that t is afunction of s,

B′ (s) =1√

a2 +b2((bcos t) i+ (bsin t)j)

dtds

=1

a2 +b2 ((bcos t) i+ (bsin t)j)

= τ (−(cos t)i− (sin t)j) = τN

and it follows −b/(a2 +b2

)= τ .

An important application of the usefulness of these ideas involves the decompositionof the acceleration in terms of these vectors of an object moving over a space curve.

Corollary 15.2.6 Let R(t) be a space curve and denote by v (t) the velocity, v (t) =R′ (t),let v(t)≡ |v (t)| denote the speed, and let a(t) denote the acceleration. Then v = vT anda= dv

dt T +κv2N.

Proof: T = dRds = dR

dtdtds = v dt

ds . Also, s =∫ t

0 v(r) dr and so dsdt = v which implies

dtds =

1v . Therefore, T = v/v which implies v = vT as claimed.

Now the acceleration is just the derivative of the velocity and so by the Serrat Frenetformulas,

a=dvdt

T + vdTdt

=dvdt

T + vdTds

v =dvdt

T + v2κN

Note how this decomposes the acceleration into a component tangent to the curve and one

which is normal to it. Also note that from the above, v∣∣T ′∣∣ T ′(t)|T ′| = v2κN and so |T

′|v = κ

and N = T ′(t)|T ′| . ■

15.3 Exercises1. Find a parametrization for the intersection of the planes 2x+ y+ 3z = −2 and 3x−

2y+ z =−4.

2. Find a parametrization for the intersection of the plane 3x + y + z = −3 and thecircular cylinder x2 + y2 = 1.

3. Find a parametrization for the intersection of the plane 4x + 2y+ 3z = 2 and theelliptic cylinder x2 +4z2 = 9.