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CHAPTER 34
34.80 Prove the converse of Problem 34.79: If the speed is constant, then the velocity and acceleration vectors are
perpendicular.
This is a special case of Problem 34.59 if we substitute R'(0 for R(r).
34.81
Let T(f) be the unit tangent vector to a curve R(t). Show that, wherever T(t) * 0, the principal unit normal
vector N = TV |T | is perpendicular to T.
The argument of Problem 34.59 applies to any differentiable vector function. Therefore, T • T' = 0, which
in turn implies T-N = 0 wherever N is defined (i.e., wherever |T'| >0).
34.82 Show that the principal unit normal vector N(f) (Problem 34.81) points in the direction in which T(f) is turning as t
increases.
Since N(t) has the same direction as T', we need prove the result only for T'. T'(0 =
For small Af, AT/Af has approximately the same direction as T'(0> and, when
and
AT/Af have the same direction. If we place T(f + At) and T(f) with their tails at the origin, AT is the vector
from the head of T(f) to the head of 1(t + Af). Figure 34-16 shows that AT points in the direction in which T is
turning (to the right in this case).
Fig. 34-16
34.83 If R(t) = (r cos t, r sin t), where t represents time, show that the principal unit normal vector N(f) has the same
direction as the acceleration vector.
R'(<) = K-sin t, cos t) and |R'(Ol = '• So, T(f) = R'/|R'| = (-sin t, cost). Hence, T'(0 = (~cos/,
-sinf) and |T'(Ol = 1. Thus, N(r) =T'(0 = -(cos r.sin t) = - - R. In this case, by Problem 34.74, N(f)
has the same direction as the acceleration vector. Note that N(f) is pointing in the direction in which T(f) is
turning, that is, "inside" the curve.
34.84 Compute N(f) for the curve R(t) = (2 cos t
2 , 2 sin f
2 ).
Thus,
So, |R'(OI=4|<|.
R'(0 = (~4f sin t
2 ,4t cos r
2 ) = 4r(-sin t
2 , cos t
2 ).
Hence,
So,
and, therefore, N(f) =
T(f)/|T(f)| = — (cos t
2 , sin t
2 ). Again, N(f) has the same direction as the acceleration vector, that is, it points
toward the center of the circle traced out by R(t).
34.85 Show that when arc length s is chosen as the curve parameter, the principal unit normal vector has the simple
expression N(s) = R"(s) I \R"(s)\.
By Problem 34.47, T(s) = R'(s), and so T'(s) = R"(s) and N(s) =T'(s)/|T(s)| = R"(s) /|R"(*)| [except where R"(s) = 0].
34.86
Show that, for the cycloid R(() = a(t - sin t, 1 - cos t), where / is the time, the acceleration vector is not parallel
to the principal unit normal vector.
CHAPTER 34
34.80 Prove the converse of Problem 34.79: If the speed is constant, then the velocity and acceleration vectors are
perpendicular.
This is a special case of Problem 34.59 if we substitute R'(0 for R(r).
34.81
Let T(f) be the unit tangent vector to a curve R(t). Show that, wherever T(t) * 0, the principal unit normal
vector N = TV |T | is perpendicular to T.
The argument of Problem 34.59 applies to any differentiable vector function. Therefore, T • T' = 0, which
in turn implies T-N = 0 wherever N is defined (i.e., wherever |T'| >0).
34.82 Show that the principal unit normal vector N(f) (Problem 34.81) points in the direction in which T(f) is turning as t
increases.
Since N(t) has the same direction as T', we need prove the result only for T'. T'(0 =
For small Af, AT/Af has approximately the same direction as T'(0> and, when
and
AT/Af have the same direction. If we place T(f + At) and T(f) with their tails at the origin, AT is the vector
from the head of T(f) to the head of 1(t + Af). Figure 34-16 shows that AT points in the direction in which T is
turning (to the right in this case).
Fig. 34-16
34.83 If R(t) = (r cos t, r sin t), where t represents time, show that the principal unit normal vector N(f) has the same
direction as the acceleration vector.
R'(<) = K-sin t, cos t) and |R'(Ol = '• So, T(f) = R'/|R'| = (-sin t, cost). Hence, T'(0 = (~cos/,
-sinf) and |T'(Ol = 1. Thus, N(r) =T'(0 = -(cos r.sin t) = - - R. In this case, by Problem 34.74, N(f)
has the same direction as the acceleration vector. Note that N(f) is pointing in the direction in which T(f) is
turning, that is, "inside" the curve.
34.84 Compute N(f) for the curve R(t) = (2 cos t
2 , 2 sin f
2 ).
Thus,
So, |R'(OI=4|<|.
R'(0 = (~4f sin t
2 ,4t cos r
2 ) = 4r(-sin t
2 , cos t
2 ).
Hence,
So,
and, therefore, N(f) =
T(f)/|T(f)| = — (cos t
2 , sin t
2 ). Again, N(f) has the same direction as the acceleration vector, that is, it points
toward the center of the circle traced out by R(t).
34.85 Show that when arc length s is chosen as the curve parameter, the principal unit normal vector has the simple
expression N(s) = R"(s) I \R"(s)\.
By Problem 34.47, T(s) = R'(s), and so T'(s) = R"(s) and N(s) =T'(s)/|T(s)| = R"(s) /|R"(*)| [except where R"(s) = 0].
34.86
Show that, for the cycloid R(() = a(t - sin t, 1 - cos t), where / is the time, the acceleration vector is not parallel
to the principal unit normal vector.