Filetype PDF | Posted on 26 Jan 2023 | 2 years ago
Authentication
digital resources
161x Filetype PDF File size 0.07 MB Source: crubweb.uncoma.edu.ar
Tom M. Apostol
CALCULUS
VOLUME 1
One-Variable Calculus, with an
Introduction to Linear Algebra
SECOND EDITION
John Wiley & Sons, Inc.
New York l Santa Barbara l London l Sydney l Toronto
536 Calculus of vector-valuedfunctions
18. If 0 < 6 < 4a, let r(f) = a(t - sin t)i + a(1 - COS t)j + b sin ht k. Show that the length of
the path traced out from t = 0 to t = 2~ is 8&(k), where E(k) has the meaning given in
Exercise 17 and k2 = 1 - (b/4~)~.
19. A particle moves with position vector
r(t) = tA + t2B + 2(5t)3/3 A x B ,
where A and B are two fixed unit vectors making an angle of ~13 radians with each other.
Compute the speed of the particle at time t and find how long it takes for it to move a distance
of 12 units of arc length from the initial position r(0).
20. (a) When a circle rolls (without slipping) along a straight line, a point on the circumference
traces out a curve called a cycloid. If the fixed line is the x-axis and if the tracing point (x, y)
is originally at the origin, show that when the circle rolls through an angle 0 we have
x = a(0 - sin 0) , y = a(1 - COS e) >
where a is the radius of the circlle. These serve as parametric equations for the cycloid.
(b) Referring to part (a), show that dy/dx = cet $0 and deduce that the tangent line of the
cycloid at (x, y) makes an angle + (7~ - 0) with the x-axis. Make a sketch and show that the
tangent line passes through the highest point on the circle.
21. Let C be a curve described by two equivalent functions X and Y, where Y(t) = X[u(t)] for
c 5 t 5 d. If the function u which defines the change of parameter has a continuous deriv-
ative in [c, d] prove that
‘u(d) IIX’(u)Ij du =j”” 11 Y’(t)i/ dt ,
iu(c) c
and deduce that the arc length of C is invariant under such a change of parameter.
22. Consider the plane curve whose vector equation is r(t) = ti +f(t)j, where
rr
f(t) = f COS z if t#O, f(0) = 0 .
0
Consider the following partition of the interval [0, 11:
1 1 1
2n >2n _ 1 9*. . >3 > 2 >1 .
p=i 0,’ ~ -- 1
Show that the corresponding inscribed polygon T(P) has length
and deduce that this curve is nonrectifiable.
14.14 Curvature of a curve
For a straight line the unit tangent vector T does not change its direction, and hence
T’ = 0. If the curve is not a straight line, the derivative T’ measures the tendency of the
tangent to change its direction. The rate of change of the unit tangent with respect to arc
Curvature of a curve 537
length is called the curvature vector of the curve. We denote this by dT/ds, where s repre-
sents arc length. The chain rule, used in conjunction with the relation s’(t) = v(t), tells us
that the curvature vector dT/ds is related to the “time” derivative T’ by the equation
dT dt dT
-c--c-
ds ds dt &> T’(t) = & T’(t).
Since T’(t) = II T’(t) // N(t), we obtain
dT IITYOII Nctj
(14.19) -=-
ds 40 ’
which shows that the curvature vector has the same direction as the principal normal N(t).
The scalar factor which multiplies N(t) in (14.19) is a nonnegative number called the
curvature of the curve at t and it is denoted by K(t) (K is the Greek letter kappa). Thus the
curvature K(t), defined to be the length of the curvature vector, is given by the following
formula :
(14.20) K(t) - 11 T’(t)11 .
40
EXAMPLE 1. Curcature of a circle. For a circle of radius a, given by r(t) = a COS t i +
a sin tj, we have v(t) = -a sin t i + a COS t j, v(t) = a, T(t) = -sin t i + COS tj, and
T’(t) = -COS t i - sin t j. Hence we have 11 T’(t)11 = 1 SO K(t) = I/a. This shows that a
circle has constant curvature. The reciprocal of the curvature is the radius of the circle.
When K(t) # 0, its reciprocal is called the radius of curvature and is denoted by p(t)
(p is the Greek letter rho). That circle in the osculating plane with radius p(t) and tenter
at the tip of the curvature vector is called the osculating circle. It cari be shown that the
osculating circle is the limiting position of circles passing through three nearby points on
the curve as two of the points approach the third. Because of this property, the osculating
circle is often called the circle that “best fits the curve” at each of its points.
EXAMPLE 2. Curvature of a plane curve. For a plane curve, we have seen that 11 T’(t) Ij =
Iu’(t)l, where u(t) is the angle the tangent vector makes with the positive x-axis, as shown
in Figure 14.11. From the chain rule, we have cr’(t) = dcc/dt = (dcc/ds)(ds/dt) = v(t)dcc/ds,
SO Equation (14.20) implies
In other words, the curvature of a plane curve is the absolute value of the rate of change of
cx with respect to arc length. It measures the change of direction per unit distance along the
curve.
EXAMPLE 3. Plane curves of constant curvature. If dulds is a nonzero constant, say
da/ds = a, then tc = as + b, where b is a constant. Hence, if we use the arc length s as
Calcur’us of vector-valuedfunctions
538
a parameter, we have T = COS (as + b)i + sin (as + ZI)~. Integrating, we find that r =
(lia) sin (as + b)i - (l/a) COS (as + 6)j + A, where A is a constant vector. Therefore
11~ - A 11 = I/lal, SO the curve is a circle (or an arc of a circle) with tenter at A and radius
l/lal. This proves that a plane curve of constant curvature K # 0 is a circle (or an arc of a
circle) with radius l/~.
Now we prove a theorem which relates the curvature to the velocity and acceleration.
THEOREM 14.14. For any motion with velocity v(t), speed v(t), acceleration u(t), and
curvature K(t), we have
(14.21) u(t) -= v’(t)T(t) + K(t)v’(t)N(t).
This formula, in turn, implies
(14.22) K(t) = Ila x V(al
u3(t)
Proof. TO prove (14.21), we rewrite (14.20) in the form I/ T’(t) 11 = K(t)V(t), which gives us
T’(t) = K(t)V(f)N(t). Substituting this expression for T’(t) in Equation (14.8), we obtain
(14.21).
TO prove (14.22), we form the cross product u(t) x v(t), using (14.21) for u(t) and the
formula v(t) = v(t)T(t) for the velocity. This gives us
(14.23) a x v = v’vT x T + KV~N x T = KV~N x T
since T x T = 0. If we take the length of each member of (14.23) and note that
Il N x Tlj = lINl/ II TII sin &T = 1 ,
we obtain Ila x vlj = KV~, which proves (14.22).
In practice it is fairly easy to compute the vectors u and a (by differentiating the position
vector Y); hence Equation (14.22) provides a useful method for computing the curvature.
This method is usually simpler than determining the curvature from its definition.
For a straight line we have u x v = 0, SO the curvature is everywhere zero. A curve
with a small curvature at a point has a large radius of curvature there and hence does not
differ much from a straight line in the immediate vicinity of the point. Thus the curvature
is a measure of the tendency of a curve to deviate from a straight line.
14.15 Exercises
1. Refer to the curves described in Exercises 1 through 6 of Section 14.9 and in each case determine
the curvature K(I) for the value of t indicated.
2. A helix is described by the position function r(t) = a COS ot i + a sin otj + bwtk. Prove that
it has constant curvature K = a/(~? + @).
no reviews yet
Please Login to review.
