784 CHAPTER 13 VECTOR CALCULUS
If φ (u) > 0 for all u ∈ J , then r and R are said to differ by an orientationpreserving change of parameter. In this case, r and R take on exactly the same values
in exactly the same order. In other words, they produce exactly the same oriented curve
(Exercise 44).
If, on the other hand, φ (u) < 0 for all u ∈ J . then the change in parameter is said
to be orientation-reversing. In this case r and R still take on exactly the same values but
in opposite order. The paths are the same, but they are traversed in opposite directions.
You have already seen one example of this:
z
R(u) = r(a + b − u).
y
Example 5 The parametrization
( a, 0, 0)
x
r(t) = a cos t i + a sin t j + bt k.
Figure 13.3.13
t ∈ [0, 2π]
z
gives one “spiral” of the circular helix with the orientation indicated by the arrows
(Figure 13.3.13). If we let φ(u) = π u, u ∈ [0, 2], then φ maps the interval J = [0, 2]
onto the interval I = [0, 2π], and φ (u) = π > 0. Thus,
R(u) = r(φ(u)) = a cos (πu) i + a sin (π u) j + bπu k,
u ∈ [0, 2]
is precisely the same curve with the same orientation.
On the other hand, if we let ψ(u) = (2 − u)π, u ∈ [0, 2], then ψ also maps the
interval J = [0, 2] onto the interval I = [0, 2π], but ψ (u) = −π < 0. Thus,
y
R(u) = r(ψ(u)) = a cos (2 − u)π i + a sin (2 − u)π j + b(2 − u)π k
( a, 0, 0)
x
Figure 13.3.14
produces the same curve but with the opposite orientation. See Figure 13.3.14.
EXERCISES 13.3
Find the tangent vector r (t) at the indicated point and
parametrize the tangent line.
1. r(t) = cos π t i + sin π t j + t k
−t
2. r(t) = e i + e j − ln t k
t
3. r(t) = a + t b + t c
2
at t = 2.
at t = 1.
at t = −1.
4. r(t) = (t + 1) i + (t + 1) j + (t 3 + 1) k
2
5. r(t) = 2t 2 i + (1 − t) j + (3 + 2t 2 ) k
6. r(t) = 3t a + b − t 2 c
at P(1, 1, 1).
at P(2, 0, 5).
at t = 2.
7. r(t) = 2 cos t i + 3 sin t j + t k;
t = π/4.
8. r(t) = t sin t i + t cos t j + 2t k;
t = π/2.
9. Show that r(t) = at i + bt 2 j parametrizes a parabola. Find
an equation in x and y for this parabola.
10. Show that r(t) = 12 a(eωt + e−ωt ) i + 12 a(eωt − t −ωt ) j
parametrizes the right branch (x > 0) of the hyperbola
x 2 − y 2 = a2 .
11. Find (a) the points on the curve r(t) = t i+(1+t 2 ) j at which
r(t) and r (t) are perpendicular; (b) the points at which they
have the same direction; (c) the points at which they have
opposite directions.
12. Find the curve given that r (t) = αr(t) for all real t and
r(0) = i + 2j + 3k.
13. Suppose that r (t) and r(t) are parallel for all t. Show that,
if r (t) is never 0, then the tangent line at each point passes
through the origin.
c In Exercises 14–16, the given curves intersect at the indicated
point. Find the angle of intersection. Express your answer in
radians rounded to the nearest hundredth, and in degrees rounded
to the nearest tenth.
14. r1 (t) = t i + t 2 j + t 3 k,
r2 (u) = sin 2u i + u cos u j + u k; P(0, 0, 0).
15. r1 (t) = (et − 1) i + 2 sin t j + ln (t + 1) k,
r2 (u) = (u + 1) i + (u2 − 1) j + (u3 + 1) k; P(0, 0, 0).
13.3 CURVES 16. r1 (t) = e−t i + cos t j + (t 2 + 4) k,
r2 (u) = (2 + u) i + u4 j + 4u2 k; P(1, 1, 4).
17. Find the point at which the curves
(b) Find the unit tangents at P(1, 2, 0), first taking t = t1 ,
then taking t = t2 .
33. Find the point(s) at which the twisted cubic
r1 (t) = et i + 2 sin (t + 12 π) j + (t 2 − 2) k,
r2 (u) = u i + 2 j + (u2 − 3) k
intersect and find the angle of intersection.
18. Consider the vector function f (t) = t i + f (t) j formed from
a differentiable real-valued function f . The vector function
f parametrizes the graph of f .
(a) Parametrize the tangent line at P(t0 , f (t0 )).
(b) Show that the parametrization obtained in part (a) can
be reduced to the usual equation for the tangent line:
y − f (t0 ) = f (t0 )(x − t0 )
y = f (t0 )
f (t0 ) = 0;
if
if f (t0 ) = 0.
19. Define a vector function r on the interval [0, 2π ] that satisfies the initial condition r(0) = ai and, as t increases to 2π,
traces out the ellipse b2 x2 + a2 y2 = a2 b2 :
(a) Once in a counterclockwise manner.
(b) Once in a clockwise manner.
(c) Twice in a counterclockwise manner.
(d) Three times in a clockwise manner.
20. Repeat Exercise 19 given that r(0) = b j.
In Exercise 21–26, sketch the plane curve determined by the
given vector-valued function r and indicate the orientation. Find
r (t) and draw the position vector and the tangent vector for the
indicated value of t, placing the tangent vector at the tip of the
position vector.
21. r(t) = 14 t 4 i + t 2 j; t = 2.
22. r(t) = 2t i + (t 2 + 1) j;
23. r(t) = e i + e
2t
−4t
j;
t = 4.
t = 0.
24. r(t) = sin t i − 2 cos t j;
t = π/3.
25. r(t) = 2 cos t i + 3 sin t j;
26. r(t) = sec t i + tan t j,
t = π/6.
|t| < π/2;
t = π/4.
Find a vector parametrization for the curve.
27. y2 = x − 1,
y ≥ 1.
28. r = 1 − cos θ,
29. r = sin 3θ,
30. y = x ,
4
3
θ ∈ [0, 2π ].
θ ∈ [0, π].
(Polar coordinates)
(Polar coordinates)
y ≤ 0.
31. Find an equation in x and y for the curve r(t) = t 3 i + t 2 j.
Draw the curve. Does the curve have a tangent vector at the
origin? If so, what is the unit tangent vector?
32. (a) Show that the curve
r(t) = (t 2 − t + 1) i + (t 3 − t + 2) j + ( sin πt) k
intersects itself at P(1, 2, 0) by finding numbers t1 < t2
for which P is the tip of both r(t1 ) and r(t2 ).
785
r(t) = t i + t 2 j + t 3 k
intersects the plane 4x + 2y + z = 24. What is the angle of
intersection between the curve and the normal to the plane?
34. (a) Find the unit tangent and the principal normal at an
arbitrary point of the ellipse
r(t) = a cos t i + b sin t j.
(b) Write vector equations for the tangent line and the
normal line at the tip of r( 14 π).
Find the unit tangent vector, the principal normal vector, and
an equation in x, y, z for the osculating plane at the point on the
curve corresponding to the indicated value of t.
35. r(t) = i + 2t j + t 2 k;
36. r(t) = t i + t j + 2t k;
2
2
t = 1.
t = 1.
37. r(t) = cos 2t i + sin 2t j + t k
at t = 14 π.
38. r(t) = t i + 2t j + t 2 k
at t = 2.
39. r(t) = t i + t 2 j + t 3 k
at t = 1.
40. r(t) = cos 3t i + t j − sin 3t k
at t = 13 π.
41. r(t) = et sin t i + et cos t j + et k;
t = 0.
42. r(t) = ( cos t + t sin t) i + ( sin t − t cos t) j + 2k;
t = 14 π.
43. Let r = r(t), t ∈ [a, b] and set
R(u) = r (a + b − u),
u ∈ [a, b].
Show that this change of parameter changes the sign of the
unit tangent but does not alter the principal normal.
HINT: Let P be the tip of R(u) = r(a+b−u). At that point,
R produces a unit tangent T1 (u) and a principal normal
N1 (u). At that same point, r produces a unit tangent T(a +
b − u) and a principal normal N(a + b − u).
44. Show that two vector functions that differ by a orientationpreserving change of parameter take on exactly the same
values in exactly the same order. That is, set
r = r(t),
t ∈ I.
Assume that φ maps an interval J onto the interval I and
that φ (u) > 0 for all u ∈ J . Set
R(u) = r (φ(u)),
and show that R and r take on exactly the same values in
exactly the same order.
45. Show that the unit tangent vector, the principal normal vector, and the osculating plane are left invariant
(left unchanged) by every orientation-preserving change of
parameter.
786 CHAPTER 13 VECTOR CALCULUS
In Exercises 46 and 47, let r be the vector-valued function
defined by
r(t) = 2 cos t i + 2 sin t j + 4t k
for
0 ≤ t ≤ 2π.
The graph of r is one revolution on a circular helix, starting at
the point (2, 0, 0) and ending at the point (2, 0, 8π).
√
46. Let ϕ(u) = u2 for 0 ≤ u ≤ 2π, and let
R(u) = r[ϕ(u)] = 2 cos u2 i + 2 sin u2 j + 4u2 k.
(a) Show that ϕ determines
√ a orientation-preserving change
of parameter on [0, 2π].
(b) Show that the unit tangent and principal normal vectors
for r at the point t = 14 π are the same as the unit tangent
√
and principal normal vectors for R at u = 12 π.
√
47. Let ψ(v) = 2π − v2 for 0 ≤ v ≤ 2π, and let
R(v) = r[ψ(v)] = 2 cos (2π − v2 ) i +
2 sin (2π − v2 ) j + 4(2π − v2 ) k.
(a) Show that ψ determines
√ a orientation-reversing change
of parameter on [0, 2π].
(b) Show that the principal normal vector for r at t =√π/4
is the same as the principal normal for R at v = 12 7π ,
and show that the unit tangent vector for r at π/4 √
is the
negative of the unit tangent vector for R at v = 12 7π.
√
√
c 48. Let r(t) = 2 cos t i + 2 sin t j + t k, 0 ≤ t ≤ 2π.
(a) Find scalar parametric equations for the tangent line to
the curve at the point (1, 1, π/4).
(b) Use a CAS to draw the curve and the tangent line
together. Experiment with the t-interval to find a good
illustration of the curve and the tangent line.
(c) Are there points on the curve where the tangent line is
parallel to the xy-plane? If so, find them.
√
√
c 49. Let r(t) = 2 cos t i + 2 sin t j + sin 5t k, 0 ≤ t ≤ 2π.
(a) Find scalar parametric equations
√ for the tangent line to
the curve at the point (1, 1, − 2/2).
(b) Use a CAS to draw the curve and the tangent line
together. Experiment with the t-interval to find a good
illustration of the curve and the tangent line.
(c) Are there points on the curve where the tangent line is
parallel to the xy-plane? If so, find them
√
√
c 50. Let r(t) = 2 cos t i + 2 sin t j + ln t k,
0 ≤ t ≤ 2π.
(a) Find the scalar parametric equations for the tangent line
to the curve at the point (1, 1, ln (π/4).
(b) Use a CAS to draw the curve and the tangent line
together. Experiment with the t-interval to find a good
illustration of the curve and the tangent line.
13.4 ARC LENGTH
In Section 9.8 we considered arc length in an intuitive manner and decided that the
length of the path C traced out by a pair of continuously differentiable functions
x = x(t),
y = y(t),
t ∈ [a, b]
is given by the formula
L(C) =
b
[x (t)]2 + [ y (t)]2 dt.
a
Applied to a path C in space traced out by
x = x(t),
y = y(t),
z = z(t),
t ∈ [a, b]
the formula becomes
L(C) =
b
[x (t)]2 + [y (t)]2 + [z (t)]2 dt.
a
In vector notation, both formulas can be written
L(C) =
b
r (t) dt.
a
We will prove this result in this form, but first we give a precise definition of arc length.