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352 CHAPTER 11 Vector Differential Calculus

for t > 0. It is usually more convenient to compute such quantities as the unit tangent and the
curvature in terms of the parameter t used to define C, rather than attempting to solve for t in
terms of the arc length s.

Given a position vector F(t) for a curve C, we have a unit tangent at any point where the
component functions are differentiable and their derivatives are not all zero. We claim that,
in terms of s, the vector
1

N(s) = T (s).
κ(s)
is a unit normal vector (orthogonal to the tangent) to C.

First, N(s) is a unit vector because κ(s) = T (s) ,so
1

N(s) = T (s) = 1.

T (s)
We claim also that N(s) is orthogonal to the tangent vector T(s). To see this, recall that T(s) is a
unit vector, so
2
T(s) = T(s) · T(s) = 1.
Differentiate this equation to get

T (s) · T(s) + T(s) · T (s) = 2T(s) · T (s) = 0.

Therefore, T(s) is orthogonal to T (s).But N(s) is a scalar multiple of T (s), hence it is in the

same direction as T (s). Therefore, T(s) is orthogonal to N(s).
At any point where F is twice differentiable, we may now place a unit tangent and a unit
normal vector, as in Figure 11.6. With these in hand, we claim that we can write the acceleration
in terms of tangential and normal components:
a(t) = a T T(t) + a N N(t)
z

T
N

y

x

FIGURE 11.6 Unit tangent and normal vectors to
acurve.

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October 15, 2010 16:11 THM/NEIL Page-352 27410_11_ch11_p343-366