Engineering Math

10.1 Space Curves

A fly buzzing around the room, a person riding a roller coaster, and a satellite orbiting the earth all have something in common. They are moving over some sort of curve in three dimensions.

Denote by R

the position vector of the point on the curve which occurs at time t. Assume that R^′,R^′′ exist and are continuous. Thus R^′ = v, the velocity and R^′′ = a is defined as the acceleration.

Proof: It follows from the definition that

= 1. Therefore, T ⋅ T = 1 and so, upon differentiating both sides,

T′ ⋅T + T ⋅T′ = 2T ′ ⋅T = 0.

Therefore, T^′ is perpendicular to T. Let N

≡ T^′. Note that if = 0, you could let N be any unit vector. Then letting κ be defined such that ≡ κ, it follows

′ ′ T (t) = |T (t)|N (t) = κ(t)|v (t)|N (t). ■

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

Proof:

a = dv-= -d(R ′) = d-(|v |T ) = d-|v-|T + |v|T′ = d|v-|T + |v|2 κN. dt dt dt dt dt

This proves the first part.

For the second part,

2 |a| = (aTT + aN N)⋅(aTT + aN N)

= a2TT ⋅T + 2aNaT T⋅N +a2N N ⋅N = a2T + a2N

because T ⋅ N = 0. ■

From 10.1 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 angle between N and T is 90^∘,

2 |a-×-v| |v-×-a| |a × v| = κ|v| |v|,κ = |v|3 = |v |3 (10.2)

(10.2)

Finally, it is good to point out that the curvature is a property of the curve itself, and does not depend on the parametrization of the curve. If the curve is given by two different vector valued functions R

and R, then from the formula above for the curvature,

′ ||dT dτ|| ||dT|| κ (t) = |T--(t)|= |ddτR-ddtτ|= |ddτR| ≡ κ(τ). |v(t)| |dτ-dt| |dτ|

From this, it is possible to give an important formula from physics. Suppose an object orbits a point at constant speed v. In the above notation,

= v. What is the centripetal acceleration of this object? You may know from a physics class that the answer is v²∕r where r is the radius. This follows from the above quite easily. First, what is the curvature of a circle of radius r? A parameterization of such a curve is

R (t) = (rcost,rsin t)

Thus using 10.2 and this parametrization,

|| || || i j k || v× a = ||− rsin t rcost 0 ||= kr2 |− rcost − rsin t 0 |

Thus

κ = r2 = 1 r3 r

Since v is constant, it follows from 10.1 that

1 1 a = - |v|2N = -v2N r r

First of all, v

= and so the speed is given by = . Therefore,

(∘ --------------) a = -d sin2 (t)+ 1+ 4t2 = ∘sin-(t)cos(t)+-4t-. T dt (2+ 4t2 − cos2t)

It remains to find a_N. To do this, you can find the curvature first if you like.

a(t) = R′′(t) = (− cost,0,2).

Then

∘ --------------------------------- 4+ (− 2sin(t)+ 2(cos(t))t)2 + cos2(t) κ = |(− c(o∘st,0,2)×-(−-sin-t,1),32t)|=-------(∘---------------)3-------- sin2(t)+ 1 + 4t2 sin2(t)+ 1+ 4t2

Then a_N = κ

²

You can observe the formula a_N² + a_T² = ² holds. Indeed a_N² + a_T² =