PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION
285
and
Thus N(f) has the direction of (sin t, cos t- 1). However, the acceleration vector R"(f) = a(sin t, cos t).
34.87
At a given point on a curve, does the unit tangent vector depend on the direction in which the curve is being swept
out?
We let S(t) = R(-t). Then S(t) traces the same curve as R(f), but in the reverse direction. S'(t) =
-R'(-0 and |S'(0| = |R'(~Ol- Let T*(f) be the unit tangent vector for the curve S(r). Then T*(r) =
S'(0/|S'(Ol = ~
R '(~0/I
R '(-OI = -T(-f)- Hence, at each point, the direction of the unit tangent vector has
been reversed.
34.88 At a given point on a curve, does the principal unit normal vector depend on the direction in which the curve is
being swept out?
Use the same notation as in Problem 34.87. Let N*(/) be the principal unit normal vector on the reversed
curve S(0- Since T*(t) =-T(-t), -r T*(f) = T(-t), by the chain rule. Hence,
Thus, the principal unit vector is not changed by reversing the direction along a curve.
34.89 Let be the angle between the velocity vector and the positive ;t-axis. Show that \dT/dd>\ = 1.
Since T is a unit vector in the same direction as the velocity vector,
dT/d = (-sin, cos) and \dT/d\ = 1.
T = (cos <£, sin).
Hence,
34.90
Show that there is a scalar function /(r) such that N'(0 = f(t)T(t), and find a formula for /(/).
Parameterize the curve as R(<) = (t, t
3 ).
Then
So, T'(0) = 0 and, therefore, N(0) is not denned.
34.91 Show that there is a scalar function/(r) such that N'(0 =/(0T(0> and nnd a formula for/(f).
Since N(t) is a unit vector, Problem 34.59 shows that N'(0 -I N(f). Since T(0 1 N(r), N'(0 and T(t)
must be parallel (remember that we are in two dimensions). Hence, there is a scalar function /(r) such that
T'
N'(r)=/(r)T(0- Since N(()-T(0 = 0, the product rule yields N-T' + N'-T = 0, N'-T= - 7^71 •!" =
-|T'|. Hence, - |T'| = N''-T =/(/)T• T =f(t),
since T-T = |T|
2 = 1. When t = s = arc length, the
scalar / measures the curvature of R($); see Problem 34.105.
34.92 Show that
By the product rule,
T()t)=1/v R'(t).
285
and
Thus N(f) has the direction of (sin t, cos t- 1). However, the acceleration vector R"(f) = a(sin t, cos t).
34.87
At a given point on a curve, does the unit tangent vector depend on the direction in which the curve is being swept
out?
We let S(t) = R(-t). Then S(t) traces the same curve as R(f), but in the reverse direction. S'(t) =
-R'(-0 and |S'(0| = |R'(~Ol- Let T*(f) be the unit tangent vector for the curve S(r). Then T*(r) =
S'(0/|S'(Ol = ~
R '(~0/I
R '(-OI = -T(-f)- Hence, at each point, the direction of the unit tangent vector has
been reversed.
34.88 At a given point on a curve, does the principal unit normal vector depend on the direction in which the curve is
being swept out?
Use the same notation as in Problem 34.87. Let N*(/) be the principal unit normal vector on the reversed
curve S(0- Since T*(t) =-T(-t), -r T*(f) = T(-t), by the chain rule. Hence,
Thus, the principal unit vector is not changed by reversing the direction along a curve.
34.89 Let be the angle between the velocity vector and the positive ;t-axis. Show that \dT/dd>\ = 1.
Since T is a unit vector in the same direction as the velocity vector,
dT/d
T = (cos <£, sin
Hence,
34.90
Show that there is a scalar function /(r) such that N'(0 = f(t)T(t), and find a formula for /(/).
Parameterize the curve as R(<) = (t, t
3 ).
Then
So, T'(0) = 0 and, therefore, N(0) is not denned.
34.91 Show that there is a scalar function/(r) such that N'(0 =/(0T(0> and nnd a formula for/(f).
Since N(t) is a unit vector, Problem 34.59 shows that N'(0 -I N(f). Since T(0 1 N(r), N'(0 and T(t)
must be parallel (remember that we are in two dimensions). Hence, there is a scalar function /(r) such that
T'
N'(r)=/(r)T(0- Since N(()-T(0 = 0, the product rule yields N-T' + N'-T = 0, N'-T= - 7^71 •!" =
-|T'|. Hence, - |T'| = N''-T =/(/)T• T =f(t),
since T-T = |T|
2 = 1. When t = s = arc length, the
scalar / measures the curvature of R($); see Problem 34.105.
34.92 Show that
By the product rule,
T()t)=1/v R'(t).