Unformatted text preview: 5E14(pp 884893) 1/18/06 11:47 AM Page 884 CHAPTER 14 The calculus of vectorvalued functions is used
in Section 14.4 to prove
Kepler’s laws. These
describe the motion of
the planets about the Sun
and also apply to the orbit
of a satellite about the
Earth, such as the Hubble
Space Telescope. V ector Functions 5E14(pp 884893) 1/18/06 11:47 AM Page 885 The functions that we have been using so far have been realvalued functions. We now study functions whose values are
vectors because such functions are needed to describe curves
and surfaces in space. We will also use vectorvalued functions
to describe the motion of objects through space. In particular,
we will use them to derive Kepler’s laws of planetary motion.  14.1 Vector Functions and Space Curves
In general, a function is a rule that assigns to each element in the domain an element in the
range. A vectorvalued function, or vector function, is simply a function whose domain
is a set of real numbers and whose range is a set of vectors. We are most interested in vector functions r whose values are threedimensional vectors. This means that for every number t in the domain of r there is a unique vector in V3 denoted by r t . If f t , t t , and h t
are the components of the vector r t , then f , t, and h are realvalued functions called the
component functions of r and we can write
rt f t ,t t ,h t fti tt j ht k We use the letter t to denote the independent variable because it represents time in most
applications of vector functions.
EXAMPLE 1 If t 3, ln 3 rt t , st then the component functions are
ft t3 tt ln 3 t ht st By our usual convention, the domain of r consists of all values of t for which the expression for r t is deﬁned. The expressions t 3, ln 3 t , and st are all deﬁned when
3 t 0 and t 0. Therefore, the domain of r is the interval 0, 3 .
The limit of a vector function r is deﬁned by taking the limits of its component functions as follows.
1
 If lim t l a r t
L, this deﬁnition is equivalent to saying that the length and direction of the
vector r t approach the length and direction of
the vector L. If r t f t , t t , h t , then
lim r t
tla lim f t , lim t t , lim h t
tla tla tla provided the limits of the component functions exist.
Equivalently, we could have used an  deﬁnition (see Exercise 43). Limits of vector
functions obey the same rules as limits of realvalued functions (see Exercise 41). 885 5E14(pp 884893) 886 ❙❙❙❙ 1/18/06 11:47 AM Page 886 CHAPTER 14 VECTOR FUNCTIONS EXAMPLE 2 Find lim r t , where r t t3 i 1 tl0 sin t
k.
t te t j SOLUTION According to Deﬁnition 1, the limit of r is the vector whose components are the
limits of the component functions of r : lim r t t3 lim 1 tl0 tl0 i k i lim te t j tl0 lim
tl0 sin t
k
t (by Equation 3.5.2) A vector function r is continuous at a if
lim r t ra tla z P { f(t), g(t), h(t)} In view of Deﬁnition 1, we see that r is continuous at a if and only if its component functions f , t, and h are continuous at a.
There is a close connection between continuous vector functions and space curves.
Suppose that f , t, and h are continuous realvalued functions on an interval I . Then the set
C of all points x, y, z in space, where C x 2
0 r(t)=k f(t), g(t), h(t)l x y F IGURE 1 C is traced out by the tip of a moving
position vector r(t). ft y z tt ht and t varies throughout the interval I , is called a space curve. The equations in (2) are
called parametric equations of C and t is called a parameter. We can think of C as being
traced out by a moving particle whose position at time t is f t , t t , h t . If we now conf t , t t , h t , then r t is the position vector of the
sider the vector function r t
point P f t , t t , h t on C. Thus, any continuous vector function r deﬁnes a space curve
C that is traced out by the tip of the moving vector r t , as shown in Figure 1.
EXAMPLE 3 Describe the curve deﬁned by the vector function rt
Visual 14.1A shows several curves being
traced out by position vectors, including
those in Figures 1 and 2. 1 t, 2 5t, 1 6t SOLUTION The corresponding parametric equations are x 1 t y 2 z 5t 1 6t which we recognize from Equations 13.5.2 as parametric equations of a line passing
through the point 1, 2, 1 and parallel to the vector 1, 5, 6 . Alternatively, we could
1, 2, 1 and
observe that the function can be written as r r0 t v, where r0
v
1, 5, 6 , and this is the vector equation of a line as given by Equation 13.5.1.
Plane curves can also be represented in vector notation. For instance, the curve given
by the parametric equations x t 2 2 t and y t 1 (see Example 1 in Section 11.1)
could also be described by the vector equation
rt
where i 1, 0 and j t2 2 t, t 1 t2 2t i 0, 1 . EXAMPLE 4 Sketch the curve whose vector equation is rt cos t i sin t j tk t 1j 5E14(pp 884893) 1/18/06 11:47 AM Page 887 SECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES ❙❙❙❙ 887 S OLUTION The parametric equations for this curve are x cos t y z sin t t Since x 2 y 2 cos 2t sin 2t 1, the curve must lie on the circular cylinder
x 2 y 2 1. The point x, y, z lies directly above the point x, y, 0 , which moves
counterclockwise around the circle x 2 y 2 1 in the x yplane. (See Example 2 in
Section 11.1.) Since z t, the curve spirals upward around the cylinder as t increases.
The curve, shown in Figure 2, is called a helix.
z π ”0, 1, 2 ’ F IGURE 2 x y (1, 0, 0) The corkscrew shape of the helix in Example 4 is familiar from its occurrence in coiled
springs. It also occurs in the model of DNA (deoxyribonucleic acid, the genetic material
of living cells). In 1953 James Watson and Francis Crick showed that the structure of the
DNA molecule is that of two linked, parallel helices that are intertwined as in Figure 3.
In Examples 3 and 4 we were given vector equations of curves and asked for a geometric description or sketch. In the next two examples we are given a geometric description of a curve and are asked to ﬁnd parametric equations for the curve.
EXAMPLE 5 Find a vector equation and parametric equations for the line segment that
joins the point P 1, 3, 2 to the point Q 2, 1, 3 .
FIGURE 3  Figure 4 shows the line segment PQ in
Example 5. SOLUTION In Section 13.5 we found a vector equation for the line segment that joins the
tip of the vector r 0 to the tip of the vector r 1: rt 1 t r0 tr1 0 t 1 z (See Equation 13.5.4.) Here we take r 0
1, 3, 2 and r 1
vector equation of the line segment from P to Q: Q(2, _1, 3) rt 1 t 1, 3, 1 t, 3 2 t 2, 4 t, 2 1, 3 2, 1, 3 to obtain a 0 t t 1 1 or P(1, 3, _2)
FIGURE 4 rt y x 5t 0 The corresponding parametric equations are
x 1 t y 3 4t z 2 5t 0 t 1 5E14(pp 884893) 888 ❙❙❙❙ 1/18/06 11:47 AM Page 888 CHAPTER 14 VECTOR FUNCTIONS EXAMPLE 6 Find a vector function that represents the curve of intersection of the cylinder x2 y2 z 1 and the plane y 2. SOLUTION Figure 5 shows how the plane and the cylinder intersect, and Figure 6 shows the
curve of intersection C, which is an ellipse.
z z y+z=2 (0, _1, 3) (_1, 0, 2) C
(1, 0, 2) (0, 1, 1) ≈+¥=1
0
y x y x FIGURE 5 FIGURE 6 The projection of C onto the xyplane is the circle x 2
from Example 2 in Section 11.1 that we can write
x cos t y sin t 0 y2 t 1, z 0. So we know 2 From the equation of the plane, we have
z 2 y 2 sin t So we can write parametric equations for C as
x cos t y z sin t 2 2 sin t k sin t 0 t 2 The corresponding vector equation is
rt cos t i sin t j 0 t 2 This equation is called a parametrization of the curve C. The arrows in Figure 6 indicate
the direction in which C is traced as the parameter t increases. Using Computers to Draw Space Curves
Space curves are inherently more difﬁcult to draw by hand than plane curves; for an accurate representation we need to use technology. For instance, Figure 7 shows a computergenerated graph of the curve with parametric equations
x 4 sin 20 t cos t y 4 sin 20 t sin t z cos 20 t It’s called a toroidal spiral because it lies on a torus. Another interesting curve, the tre 5E14(pp 884893) 1/18/06 11:47 AM Page 889 ❙❙❙❙ S ECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES 889 foil knot, with equations
x 2 cos 1.5t cos t y 2 cos 1.5t sin t z sin 1.5t is graphed in Figure 8. It wouldn’t be easy to plot either of these curves by hand.
z z x y y x FIGURE 7 A toroidal spiral FIGURE 8 A trefoil knot Even when a computer is used to draw a space curve, optical illusions make it difﬁcult
to get a good impression of what the curve really looks like. (This is especially true in
Figure 8. See Exercise 42.) The next example shows how to cope with this problem.
t, t 2, t 3 . EXAMPLE 7 Use a computer to sketch the curve with vector equation r t This curve is called a twisted cubic.
SOLUTION We start by using the computer to plot the curve with parametric equations x t, y t 2, z t 3 for 2 t 2. The result is shown in Figure 9(a), but it’s hard to
see the true nature of the curve from that graph alone. Most threedimensional computer
graphing programs allow the user to enclose a curve or surface in a box instead of displaying the coordinate axes. When we look at the same curve in a box in Figure 9(b), we
have a much clearer picture of the curve. We can see that it climbs from a lower corner
of the box to the upper corner nearest us, and it twists as it climbs. z
6 _2 6 6
z0 x 2 _6 4 z0 _6
0 2 _2 y y2 0x (a) 0 2 4 _6 (b)
8
4
z0 z0 1 _4 _4 2
3 _2 4 _8 0x 2
y 0x 8 _1 1 4 2 (c) _2 0 y2 4 (d)
FIGURE 9 Views of the twisted cubic 2 1 0
x (e) _1 _2 _8 0 1 2
y (f ) 3 4 5E14(pp 884893) 890 ❙❙❙❙ 1/18/06 11:47 AM Page 890 CHAPTER 14 VECTOR FUNCTIONS In Visual 14.1B you can rotate the box
in Figure 9 to see the curve from any
viewpoint. z x y We get an even better idea of the curve when we view it from different vantage
points. Part (c) shows the result of rotating the box to give another viewpoint. Parts (d),
(e), and (f) show the views we get when we look directly at a face of the box. In particular, part (d) shows the view from directly above the box. It is the projection of the
curve on the xyplane, namely, the parabola y x 2. Part (e) shows the projection on the
x zplane, the cubic curve z x 3. It’s now obvious why the given curve is called a
twisted cubic.
Another method of visualizing a space curve is to draw it on a surface. For instance, the
twisted cubic in Example 7 lies on the parabolic cylinder y x 2. (Eliminate the parameter from the ﬁrst two parametric equations, x t and y t 2.) Figure 10 shows both the
cylinder and the twisted cubic, and we see that the curve moves upward from the origin
along the surface of the cylinder. We also used this method in Example 4 to visualize the
helix lying on the circular cylinder (see Figure 2).
A third method for visualizing the twisted cubic is to realize that it also lies on the cylinder z x 3. So it can be viewed as the curve of intersection of the cylinders y x 2 and
z x 3. (See Figure 11.) FIGURE 10
8
4
z Visual 14.1C shows how curves arise
as intersections of surfaces. 0
_4
_8
_1 FIGURE 11  Some computer algebra systems provide us
with a clearer picture of a space curve by enclosing it in a tube. Such a plot enables us to see
whether one part of a curve passes in front of or
behind another part of the curve. For example,
Figure 13 shows the curve of Figure 12(b) as rendered by the tubeplot command in Maple. B x 0 1 0 2 4
y We have seen that an interesting space curve, the helix, occurs in the model of DNA.
Another notable example of a space curve in science is the trajectory of a positively
charged particle in orthogonally oriented electric and magnetic ﬁelds E and B. Depending
on the initial velocity given the particle at the origin, the path of the particle is either a
space curve whose projection on the horizontal plane is the cycloid we studied in Section
11.1 [Figure 12(a)] or a curve whose projection is the trochoid investigated in Exercise 38
in Section 11.1 [Figure 12(b)]. B E E t
(a) r(t) = ktsin t, 1cos t, t l t
3 3 (b) r(t) = kt 2 sin t, 1 2 cos t, t l FIGURE 12 Motion of a charged particle in orthogonally oriented electric and magnetic fields FIGURE 13 5E14(pp 884893) 1/18/06 11:47 AM Page 891 SECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES ❙❙❙❙ 891 For further details concerning the physics involved and animations of the trajectories of
the particles, see the following web sites:
■ ■ 1–2 Exercises 1. r t t 2, st 2. r t t
t ■ 3–6  cos t, y sin t, z cos t, y sin t, z ln t ■ ■ ■ ■ ■ cos 10 t, y e z sin 10t, z I
■ 4. lim et
t 1 s1
, 5. lim st
6. lim
tl t z II ■ ■ , 3 ■ x y x tan t
k
t 1
j
1 t III 1 ln t
t arctan t , e 2t, ■ 1 t
t
t2 3i tl1 z z y ■ ■ ■ ■ ■ ■ ■ x ■ x  7. r t t4 1, t 8. r t 9. r t t, cos 2 t, sin 2 t 10. r t 11. r t y IV Sketch the curve with the given vector equation. Indicate
with an arrow the direction in which t increases. 1 z V t 3, t 2
t, 3t, y
z VI t sin t, 3, cos t 12. r t ti 13. r t t2 i 14. r t sin t i tj x cos t k t4 j ■ ■ 17. P 1, 1, 2 , ■ ■ y x sin t j
■ s2 cos t k ■ ■ ■ ■ ■ ■ ■ ■ 15–18  Find a vector equation and parametric equations for the
line segment that joins P to Q. 15. P 0, 0, 0 , y t6 k
■ ■ e cos t, sin t, t ln t tl0 ■ t sin 5t 24. x t2 k ln 9 ■ t e 23. x t sin t j ■ t 22. x Find the limit. tl0 7–14 1, s5 2
i
2
■ 3. lim ■ www.physics.ucla.edu/plasmaexp/Beam/ Find the domain of the vector function.  ■ www.phy.ntnu.edu.tw/java/emField/emField.html ■  14.1 lompado.uah.edu/Links/CrossedFields.html 16. P 1, 0, 1 , Q 1, 2, 3 18. P Q 4, 1, 7
■ ■ ■ ■ 19–24 ■ ■  Match the parametric equations with the graphs
(labeled I–VI). Give reasons for your choices. 19. x cos 4 t, y t, t, y t 2, z e 21. x t, y 11 t2 , ■ sin 4 t
t z t2 ■ ■ ■ ■ ■ ■ ■ t cos t,
y 2, and use this x2 26. Show that the curve with parametric equations x sin t,
cos t, z sin 2t is the curve of intersection of the surfaces
x 2 and x 2 y 2 1. Use this fact to help sketch the curve. y
z
■ ; 27–30  Use a computer to graph the curve with the given vector
equation. Make sure you choose a parameter domain and viewpoints that reveal the true nature of the curve. 27. r t z 20. x ■ y t sin t, z t lies on the cone z 2
fact to help sketch the curve. 1, 2
■ ■ 25. Show that the curve with parametric equations x Q 2, 3, 1 2, 4, 0 , Q 6,
■ ■ sin t, cos t, t 2 29. r t t 2, st 30. r t
■ ■ 1, s5 t4 28. r t t2 1, t, t 2 t sin t, sin 2 t, sin 3 t
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E14(pp 884893) 892 ❙❙❙❙ 1/18/06 11:47 AM Page 892 CHAPTER 14 VECTOR FUNCTIONS trajectories of two particles are given by the vector functions ; 31. Graph the curve with parametric equations
x
1 cos 16 t cos t, y
1 cos 16 t sin t,
z 1 cos 16 t. Explain the appearance of the graph by
showing that it lies on a cone. for t x s1 r2 t 4t 3, t 2, 5t 6 0. Do the particles collide? s1 0.25 cos 2 10 t sin t z t, t 2, t 3 r1 t 0.25 cos 2 10 t cos t y 0.5 cos 10 t r2 t 1 2 t, 1 6 t, 1 14 t Do the particles collide? Do their paths intersect?
41. Suppose u and v are vector functions that possess limits as Explain the appearance of the graph by showing that it lies on a
sphere. t l a and let c be a constant. Prove the following properties of
limits.
(a) lim u t
vt
lim u t
lim v t
tla t 2,
, z 1 t 3 passes through the points (1, 4, 0) and
y 1 3t
(9, 8, 28) but not through the point (4, 7, 6). 33. Show that the curve with parametric equations x tla (b) lim c u t
tla (c) lim u t
tla Find a vector function that represents the curve of intersection of the two surfaces. tla c lim u t
tla vt lim u t
tla lim v t
tla  34. The cylinder x
35. The cone z 2 y ■ ■ 2 sx 2 36. The paraboloid z
■ 12, t 2 40. Two particles travel along the space curves ; 32. Graph the curve with parametric equations 34–36 t 2, 7t r1 t ■ 4 and the surface z
y 2 and the plane z 4x
■ 2 1 ■ ■ ■ x ■ 2 2 ; 38. Try to sketch by hand the curve of intersection of the
parabolic cylinder y x 2 and the top half of the ellipsoid
x 2 4y 2 4 z 2 16. Then ﬁnd parametric equations for
this curve and use these equations and a computer to graph
the curve.
39. If two objects travel through space along two different curves, it’s often important to know whether they will collide. (Will a
missile hit its moving target? Will two aircraft collide?) The
curves might intersect, but we need to know whether the
objects are in the same position at the same time. Suppose the x 2 lim v t tla cos 1.5t cos t
z ■ cylinder x
y
4 and the parabolic cylinder z x . Then
ﬁnd parametric equations for this curve and use these equations
and a computer to graph the curve.  14.2 lim u t tla it doesn’t reveal the whole story. Use the parametric equations
2 ; 37. Try to sketch by hand the curve of intersection of the circular
2 vt 42. The view of the trefoil knot shown in Figure 8 is accurate, but y y and the parabolic cylinder y
■ tla xy 2 ■ (d) lim u t y 2 cos 1.5t sin t sin 1.5t to sketch the curve by hand as viewed from above, with gaps
indicating where the curve passes over itself. Start by showing
that the projection of the curve onto the xyplane has polar
coordinates r 2 cos 1.5t and
t, so r varies between 1
and 3. Then show that z has maximum and minimum values
when the projection is halfway between r 1 and r 3.
; When you have ﬁnished your sketch, use a computer to draw
the curve with viewpoint directly above and compare with your
sketch. Then use the computer to draw the curve from several
other viewpoints. You can get a better impression of the curve
if you plot a tube with radius 0.2 around the curve. (Use the
tubeplot command in Maple.)
43. Show that lim t l a r t there is a number
0
.
ta b if and only if for every
0
whenever
0 such that r t
b Derivatives and Integrals of Vector Functions
Later in this chapter we are going to use vector functions to describe the motion of planets and other objects through space. Here we prepare the way by developing the calculus
of vector functions. Derivatives
The derivative r of a vector function r is deﬁned in much the same way as for realvalued functions: 5E14(pp 884893) 1/18/06 11:47 AM Page 893 ❙❙❙❙ SECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS dr
dt 1 rt lim rt h
h hl0 893 rt if this limit exists. The geometric signiﬁcance of this deﬁnition is shown in Figure 1. If the
l
points P and Q have position vectors r t and r t h , then PQ represents the vector
rt h
r t , which can therefore be regarded as a secant vector. If h 0, the scalar
multiple 1 h r t h
r t has the same direction as r t h
r t . As h l 0, it
appears that this vector approaches a vector that lies on the tangent line. For this reason,
the vector r t is called the tangent vector to the curve deﬁned by r at the point P, provided that r t exists and r t
0. The tangent line to C at P is deﬁned to be the line
through P parallel to the tangent vector r t . We will also have occasion to consider the
unit tangent vector, which is
rt
rt Tt
z Visual 14.2 shows an animation of
Figure 1. z r (t+h)r (t) r ª(t) Q P P Q r (t) r (t)
r (t+h) r (t+h) C C
0 0
y x FIGURE 1 r (t+h)r (t)
h y x (a) The secant vector (b) The tangent vector The following theorem gives us a convenient method for computing the derivative of a
vector function r; just differentiate each component of r.
f t ,t t ,h t
2 Theorem If r t
are differentiable functions, then
rt fti f t ,t t ,h t tt j f ti h t k, where f , t, and h ttj htk Proof rt lim 1
rt
t lim 1
t tl0 tl0 lim ft tl0 lim tl0 ft t ft rt
t ,t t t
t ft t
t ft f t ,t t ,h t t ,h t
, t t tt tt t , lim tl0 tt t f t ,t t ,h t
, ht ht
t tt
t t , lim tl0 ht t ht
t 5E14(pp 894903) 894 ❙❙❙❙ 1/18/06 12:40 PM Page 894 CHAPTER 14 VECTOR FUNCTIONS EXAMPLE 1 (a) Find the derivative of r t
1 t 3 i te t j
(b) Find the unit tangent vector at the point where t sin 2 t k.
0. SOLUTION (a) According to Theorem 2, we differentiate each component of r :
rt
(b) Since r 0 i and r 0 3t 2 i
j j 2k
s1 4 EXAMPLE 2 For the curve r t
st i
r 1 and the tangent vector r 1 . y
2 2 cos 2 t k 2 k, the unit tangent vector at the point 1, 0, 0 is r0
r0 T0 t e tj 1 2 1
j
s5 2
k
s5 t j, ﬁnd r t and sketch the position vector SOLUTION We have
(1, 1) r (1)
0 r ª(1) 1
i
2 st rt
x 1 j and 1
i
2 r1 j The curve is a plane curve and elimination of the parameter from the equations x st,
y 2 t gives y 2 x 2, x 0. In Figure 2 we draw the position vector r 1
ij
starting at the origin and the tangent vector r 1 starting at the corresponding point 1, 1 . FIGURE 2 EXAMPLE 3 Find parametric equations for the tangent line to the helix with parametric equations
 The helix and the tangent line in Example 3
are shown in Figure 3. x
at the point 0, 1, y rt 4 _0.5 FIGURE 3 y0 0.5 1 2 _2
0x t 2. 8
z z sin t SOLUTION The vector equation of the helix is r t 12 0
_1 2 cos t 2 cos t, sin t, t , so 2 sin t, cos t, 1 The parameter value corresponding to the point 0, 1, 2 is t
2, so the tangent
vector there is r
2
2, 0, 1 . The tangent line is the line through 0, 1, 2
parallel to the vector
2, 0, 1 , so by Equations 13.5.2 its parametric equations are
x 2t y 1 z 2 t Just as for realvalued functions, the second derivative of a vector function r is the
derivative of r , that is, r
r . For instance, the second derivative of the function in
Example 3 is
rt 2 cos t, sin t, 0 A curve given by a vector function r t on an interval I is called smooth if r is continuous and r t
0 (except possibly at any endpoints of I ). For instance, the helix in
Example 3 is smooth because r t is never 0.
EXAMPLE 4 Determine whether the semicubical parabola r t 1 t 3, t 2 is smooth. 5E14(pp 894903) 1/18/06 12:40 PM Page 895 S ECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS y 895 S OLUTION Since 3 t 2, 2 t rt
cusp
0 ❙❙❙❙ 1 FIGURE 4 The curve r(t)=k1+t #, t @ l
is not smooth. x we have r 0
0, 0
0 and, therefore, the curve is not smooth. The point that corresponds to t 0 is (1, 0), and we see from the graph in Figure 4 that there is a sharp
corner, called a cusp, at (1, 0). Any curve with this type of behavior—an abrupt change
in direction—is not smooth.
A curve, such as the semicubical parabola, that is made up of a ﬁnite number of smooth
pieces is called piecewise smooth. Differentiation Rules
The next theorem shows that the differentiation formulas for realvalued functions have
their counterparts for vectorvalued functions.
3 Theorem Suppose u and v are differentiable vector functions, c is a scalar, and
f is a realvalued function. Then
d
1.
ut
vt
ut
vt
dt
d
cu t
cu t
2.
dt
d
f tut
f tut
ftu t
3.
dt
d
ut vt
u t vt
ut v t
4.
dt
d
ut
vt
ut
vt
ut
vt
5.
dt
d
uft
f tu ft
(Chain Rule)
6.
dt This theorem can be proved either directly from Deﬁnition 1 or by using Theorem 2 and
the corresponding differentiation formulas for realvalued functions. The proofs of Formulas
1, 2, 3, 5, and 6 are left as exercises.
Proof of Formula 4 Let ut f1 t , f2 t , f3 t vt t1 t , t2 t , t3 t
3 Then ut vt f1 t t1 t f2 t t2 t f3 t t3 t fi t ti t
i1 so the ordinary Product Rule gives
d
ut
dt vt d
dt 3 3 fi t ti t
i1 i1 d
fi t ti t
dt 3 f i t ti t fi t t i t i1
3 3 f i t ti t
i1 ut fi t t i t
i1 vt ut vt 5E14(pp 894903) 896 ❙❙❙❙ 1/18/06 12:40 PM Page 896 CHAPTER 14 VECTOR FUNCTIONS EXAMPLE 5 Show that if r t c (a constant), then r t is orthogonal to r t for all t . SOLUTION Since rt rt 2 rt c2 and c 2 is a constant, Formula 4 of Theorem 3 gives
d
rt
dt 0 rt rt rt rt rt 2r t
rt 0, which says that r t is orthogonal to r t .
Thus, r t r t
Geometrically, this result says that if a curve lies on a sphere with center the origin,
then the tangent vector r t is always perpendicular to the position vector r t . Integrals
The deﬁnite integral of a continuous vector function r t can be deﬁned in much the same
way as for realvalued functions except that the integral is a vector. But then we can
express the integral of r in terms of the integrals of its component functions f, t, and h as
follows. (We use the notation of Chapter 5.) y b a n r t dt r t*
i lim nl t i1
n n f t*
i lim nl n t t*
i ti i1 h t*
i tj i1 tk i1 and so y b a r t dt y b a y f t dt i b a y t t dt j b a h t dt k This means that we can evaluate an integral of a vector function by integrating each component function.
We can extend the Fundamental Theorem of Calculus to continuous vector functions as
follows: y b a r t dt Rt
b
a Rb where R is an antiderivative of r, that is, R t
indeﬁnite integrals (antiderivatives).
EXAMPLE 6 If r t 2 cos t i yrt dt sin t j y 2 cos t dt
2 sin t i Ra r t . We use the notation 2 t k, then
i cos t j y sin t dt
t2 k y 2 t dt j k C where C is a vector constant of integration, and y 0 2 r t dt [2 sin t i cos t j t2 k 0 2 2 2i j 4 k xr t dt for 5E14(pp 894903) 1/18/06 12:40 PM Page 897 ❙❙❙❙ S ECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS  14.2 897 Exercises 1. The ﬁgure shows a curve C given by a vector function r t . (a) Draw the vectors r 4.5
(b) Draw the vectors
r 4.5 r4 r 4 and r 4.2 and 0.5 r4. r 4.2 t 13. r t 2 ei j ln 1 a 16. r t ta ■ ■ 3t k
c cos 3t k b sin t j
t2 c tb
b ■ k 3 at cos 3 t i 15. r t (c) Write expressions for r 4 and the unit tangent vector T(4).
(d) Draw the vector T(4). t2 j s1 14. r t r4
0.2 tc ■ ■ ■ ■ ■ ■ ■ ■ ■ 17–20  Find the unit tangent vector T t at the point with the
given value of the parameter t. y R C sin 1t i 12. r t 6 t 5, 4 t 3, 2 t , t 17. r t r(4.5) 2 18. r t P tj cos t i 3t j 20. r t Q r(4.2) 4 st i 19. r t 1 2 sin t i ■ ■ ■ 1 t k, t 2 sin 2 t k, 2 cos t j
■ 1 ■ t 0 tan t k, t
■ ■ ■ 4
■ ■ ■ ■ r(4)
0 23. x t2 25. x e 26. x r1 t 5, y 24. x 0.1 ln t, ■ Explain why these vectors are so close to each other in
length and direction. 4. r t 1 t, st , t 5. r t 1 ti 6. r t t ei e 7. r t et i e 3 t j, 8. r t 2 sin t i ■ ■ 9–16  ■ j, t
t ■ ■ t 2, 1
i t, st j e 4t k r t. cos t, t 3; (1, 1, 1) t2 1, z y y e 2 st,
■ t 1; 1, 1, 1 t z e t; 1, 0, 1 sin t,
t 2; z ■ 0, 2, 1 ■ ■ ■ ■ ■ ■ ■ s2 cos t, z
3e , z cos t, y
■ ■ s2 sin t ; 2t ■ 3e ■ 2t 1, 3, 3 ; ■ 4, 1, 1 ■ ■ ■ ■ ■ ■ t 3, t 4, t 5
cos 3t, sin 3t (b) r t t3 t, t 4, t 5 30. (a) Find the point of intersection of the tangent lines to the 0
0 ■ t ■ (a) r t
(c) r t 1 3 cos t j, , ﬁnd T 0 , r 0 , and r t 29. Determine whether the curve is smooth. 1 t ; 3
■ curve r t
sin t, 2 sin t, cos t at the points where
t 0 and t 0.5.
(b) Illustrate by graphing the curve and both tangent lines.
t, t 2, t 3 and r2 t
sin t, sin 2 t, t intersect at the origin. Find their angle of intersection correct to the
nearest degree. 31. The curves r1 t
■ ■ ■ ■ Find the derivative of the vector function. 9. r t
11. r t t t 4, z
1, y t, y 28. x 4 t 2 j, t ■ 27. x
■ t r t.  Find parametric equations for the tangent line to the
curve with the given parametric equations at the speciﬁed point.
Illustrate by graphing both the curve and the tangent line on a common screen.  cos t, sin t , , te 2t ; 27–28 (a) Sketch the plane curve with the given vector equation.
(b) Find r t .
(c) Sketch the position vector r t and the tangent vector r t for
the given value of t.
3. r t e ,e 2t 23–26  Find parametric equations for the tangent line to the
curve with the given parametric equations at the speciﬁed point. function r t
t 2, t , 0 t 2, and draw the vectors
r(1), r(1.1), and r(1.1) r(1).
(b) Draw the vector r 1 starting at (1, 1) and compare it with
the vector 3–8 2t 22. If r t 2. (a) Make a large sketch of the curve described by the vector r 1.1 t, t 2, t 3 , ﬁnd r t , T 1 , r t , and r t 21. If r t x 1 10. r t cos 3t, t, sin 3t ■ t, 1 t, 3 t 2 and
r2 s
3 s, s 2, s intersect? Find their angle of intersection correct to the nearest degree. 32. At what point do the curves r1 t
2 5E14(pp 894903) 898 ❙❙❙❙ 33–38 33.
34. y Page 898 1 43. Prove Formula 5 of Theorem 3. Evaluate the integral.  16t 3 i 9t 2 j 44. Prove Formula 6 of Theorem 3. 25 t 4 k dt i 2 t 2 j 3 t 3 k and v t
ﬁnd d dt u t v t . 45. If u t 4 1 t2 1 0
2 2t j y 36. y 37. y et i 2t j 38. y cos ti 4 te t j st i ■ 3 sin t cos t j ■ ■ 39. Find r t if r t t2 i 40. Find r t if r t sin t i
2 k. r0 i j d dt 2 sin t cos t k dt sin t k , 47. Show that if r is a vector function such that r exists, then d
rt
dt rt 49. If r t t k dt
■ ■ 4t 3 j ■ ■ t 2 k and r 0 cos t j ■ ■ 0, show that [Hint: ■ j. 2 rt rt rt rt d
ut
dt 48. Find an expression for vt wt . d
rt
dt
rt 1
rt rt r t. 50. If a curve has the property that the position vector r t is always perpendicular to the tangent vector r t , show that
the curve lies on a sphere with center the origin. 2 t k and 51. If u t rt rt 42. Prove Formula 3 of Theorem 3. r t , show that ut 41. Prove Formula 1 of Theorem 3.  14.3 cos t j vt . ut 1
k dt
t2 tj ti 46. If u and v are the vector functions in Exercise 45, ﬁnd ln t k dt
sin ■ k dt
2 3 sin t cos t i 0 1 t2 1 2 35. ■ 12:41 PM CHAPTER 14 VECTOR FUNCTIONS 0 y 1/18/06 rt rt rt Arc Length and Curvature
In Section 11.2 we deﬁned the length of a plane curve with parametric equations x f t ,
y t t , a t b, as the limit of lengths of inscribed polygons and, for the case where
f and t are continuous, we arrived at the formula
1 z L y b a 2 sf t 2 tt dt y dx
dt b a 2 dy
dt 2 dt The length of a space curve is deﬁned in exactly the same way (see Figure 1). Suppose
that the curve has the vector equation r t
f t , t t , h t , a t b, or, equivalently,
the parametric equations x f t , y t t , z h t , where f , t , and h are continuous.
If the curve is traversed exactly once as t increases from a to b, then it can be shown that
its length is 0
y 2 L y b a x y F IGURE 1 b a The length of a space curve is the limit
of lengths of inscribed polygons. sf t
dx
dt 2 tt
2 2 dy
dt ht
2 2 dt dz
dt 2 dt Notice that both of the arc length formulas (1) and (2) can be put into the more compact form
3 L y b a r t dt 5E14(pp 894903) 1/18/06 12:41 PM Page 899 S ECTION 14.3 ARC LENGTH AND CURVATURE because, for plane curves r t f ti whereas, for space curves r t
rt  Figure 2 shows the arc of the helix whose
length is computed in Example 1. f ti sf t tt j tt 2 tt 2 h t k, fti
ttj 2 2 ttj htk sf t ht 2 EXAMPLE 1 Find the length of the arc of the circular helix with vector equation z rt cos t i t k from the point 1, 0, 0 to the point 1, 0, 2 . sin t j SOLUTION Since r t sin t i cos t j rt (1, 0, 2π) s k, we have sin t 2 cos 2 t 1 s2 The arc from 1, 0, 0 to 1, 0, 2 is described by the parameter interval 0
and so, from Formula 3, we have (1, 0, 0)
x 899 t t j, fti rt ❙❙❙❙ y y L FIGURE 2 2 0 r t dt y 2 0 s2 dt t 2 2 s2 A single curve C can be represented by more than one vector function. For instance, the
twisted cubic
r1 t 4 t, t 2, t 3 1 t 0 u 2 could also be represented by the function  Piecewisesmooth curves were introduced on
page 895.
z s(t)
C
r(a)
0
x FIGURE 3 y ln 2 where the connection between the parameters t and u is given by t e u. We say that
Equations 4 and 5 are parametrizations of the curve C. If we were to use Equation 3 to
compute the length of C using Equations 4 and 5, we would get the same answer. In general, it can be shown that when Equation 3 is used to compute the length of any piecewisesmooth curve, the arc length is independent of the parametrization that is used.
Now we suppose that C is a piecewisesmooth curve given by a vector function
rt
f t i t t j h t k, a t b, and C is traversed exactly once as t increases
from a to b. We deﬁne its arc length function s by
6 r(t) e u, e 2u, e 3u r2 u 5 st y t a ru du y t a dx
du 2 dy
du 2 dz
du 2 du Thus, s t is the length of the part of C between r a and r t . (See Figure 3.) If we differentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus, we
obtain
ds
rt
7
dt
It is often useful to parametrize a curve with respect to arc length because arc length
arises naturally from the shape of the curve and does not depend on a particular coordinate
system. If a curve r t is already given in terms of a parameter t and s t is the arc length
function given by Equation 6, then we may be able to solve for t as a function of s: t t s .
Then the curve can be reparametrized in terms of s by substituting for t : r r t s . Thus,
if s 3 for instance, r t 3 is the position vector of the point 3 units of length along the
curve from its starting point. 5E14(pp 894903) 900 ❙❙❙❙ 1/18/06 12:41 PM Page 900 CHAPTER 14 VECTOR FUNCTIONS E XAMPLE 2 Reparametrize the helix r t cos t i sin t j t k with respect to arc
length measured from 1, 0, 0 in the direction of increasing t . SOLUTION The initial point 1, 0, 0 corresponds to the parameter value t
Example 1 we have ds
dt
and so s Therefore, t y st t rt s2 r u du 0 0. From y t 0 s2 du s2 t s s2 and the required reparametrization is obtained by substituting for t :
rts cos(s s2 ) i sin(s s2 ) j (s s2 ) k Curvature
If C is a smooth curve deﬁned by the vector function r, then r t
tangent vector T t is given by
z Tt 0
x C y FIGURE 4 Unit tangent vectors at equally spaced
points on C
Visual 14.3A shows animated unit tangent vectors, like those in Figure 4, for a
variety of plane curves and space curves. 0. Recall that the unit rt
rt and indicates the direction of the curve. From Figure 4 you can see that T t changes direction very slowly when C is fairly straight, but it changes direction more quickly when C
bends or twists more sharply.
The curvature of C at a given point is a measure of how quickly the curve changes direction at that point. Speciﬁcally, we deﬁne it to be the magnitude of the rate of change of the
unit tangent vector with respect to arc length. (We use arc length so that the curvature will
be independent of the parametrization.)
8 Definition The curvature of a curve is dT
ds
where T is the unit tangent vector.
The curvature is easier to compute if it is expressed in terms of the parameter t instead
of s, so we use the Chain Rule (Theorem 14.2.3, Formula 6) to write
dT
dt
But ds dt 9 rt d T ds
ds dt dT
ds and d T dt
ds dt from Equation 7, so t Tt
rt EXAMPLE 3 Show that the curvature of a circle of radius a is 1 a. 5E14(pp 894903) 1/18/06 12:41 PM Page 901 S ECTION 14.3 ARC LENGTH AND CURVATURE ❙❙❙❙ 901 SOLUTION We can take the circle to have center the origin, and then a parametrization is rt
Therefore rt a cos t i
a cos t j a sin t i so a sin t j
and rt
rt Tt and sin t i Tt This gives T t cos t i rt a cos t j sin t j 1, so using Equation 9, we have
Tt
rt t 1
a The result of Example 3 shows that small circles have large curvature and large circles
have small curvature, in accordance with our intuition. We can see directly from the deﬁnition of curvature that the curvature of a straight line is always 0 because the tangent vector is constant.
Although Formula 9 can be used in all cases to compute the curvature, the formula
given by the following theorem is often more convenient to apply.
10 Theorem The curvature of the curve given by the vector function r is rt t Proof Since T r r and r rt
3 rt ds dt, we have
r ds
T
dt rT so the Product Rule (Theorem 14.2.3, Formula 3) gives
d 2s
T
dt 2 r
Using the fact that T T ds
T
dt 0 (see Example 2 in Section 13.4), we have
r ds
dt r 2 T T Now T t
1 for all t , so T and T are orthogonal by Example 5 in Section 14.2.
Therefore, by Theorem 13.4.6,
r
Thus and r 2 ds
dt T
T T 2 ds
dt T r
r
ds dt 2
T
r r r
r r r
r T 3 2 ds
dt 2 T 5E14(pp 894903) 902 ❙❙❙❙ 1/18/06 12:41 PM Page 902 CHAPTER 14 VECTOR FUNCTIONS t, t 2, t 3 at a general point E XAMPLE 4 Find the curvature of the twisted cubic r t and at 0, 0, 0 .
SOLUTION We ﬁrst compute the required ingredients: 1, 2 t, 3 t 2 rt
rt rt rt 4t 2 s1
i
1
0 rt 0, 2, 6 t 9t 4 j
k
2t 3t 2
2 6t s36 t 4 rt rt 6t 2 i 36 t 2 4 6t j 2k 2 s9 t 4 9t 2 1 Theorem 10 then gives
rt t rt At the origin the curvature is 2 s1 9 t 2 9 t 4
1 4t 2 9t 4 3 2 rt
3 2. 0 For the special case of a plane curve with equation y f x , we can choose x as the
parameter and write r x
x i f x j. Then r x
i f x j and r x
f x j.
Since i
j
k and j
j
0, we have r x
rx
f x k. We also have
rx
f x 2 and so, by Theorem 10,
s1 x 11 1 fx
fx 2 32 y
2 EXAMPLE 5 Find the curvature of the parabola y y=≈ and 2, 4 .
SOLUTION Since y
y=k(x) 0 FIGURE 5 The parabola y=≈ and its
curvature function x 2 at the points 0, 0 , 1, 1 , 1 2 x and y
x 2, Formula 11 gives 1 y
y 2 32 1 2
4x 2 32 x The curvature at 0, 0 is 0
2. At 1, 1 it is 1
2 5 3 2 0.18. At 2, 4 it is
32
2
2 17
0.03. Observe from the expression for x or the graph of in Figure 5 that x l 0 as x l
. This corresponds to the fact that the parabola appears
to become ﬂatter as x l
. The Normal and Binormal Vectors
At a given point on a smooth space curve r t , there are many vectors that are orthogonal
to the unit tangent vector T t . We single out one by observing that, since T t
1 for
all t, we have T t T t
0 by Example 5 in Section 14.2, so T t is orthogonal to 5E14(pp 894903) 1/18/06 12:42 PM Page 903 S ECTION 14.3 ARC LENGTH AND CURVATURE  We can think of the normal vector as indicating the direction in which the curve is turning
at each point. ❙❙❙❙ 903 T t . Note that T t is itself not a unit vector. But if r is also smooth, we can deﬁne the
principal unit normal vector N t (or simply unit normal) as
Tt
Tt Nt T(t)
B(t)
N(t) The vector B t
Tt
N t is called the binormal vector. It is perpendicular to both T
and N and is also a unit vector. (See Figure 6.)
EXAMPLE 6 Find the unit normal and binormal vectors for the circular helix FIGURE 6 rt  Figure 7 illustrates Example 6 by showing
the vectors T, N, and B at two locations on the
helix. In general, the vectors T, N, and B, starting at the various points on a curve, form a set of
orthogonal vectors, called the TNB frame, that
moves along the curve as t varies. This TNB
frame plays an important role in the branch of
mathematics known as differential geometry and
in its applications to the motion of spacecraft. cos t i sin t j tk SOLUTION We ﬁrst compute the ingredients needed for the unit normal vector: rt sin t i cos t j rt
rt Tt rt 1
s2 sin t i cos t i 1
s2 Tt k sin t j s2 cos t j k
1
s2 Tt z T Tt
Tt Nt
B N
T B cos t i Bt Tt i
sin t
cos t 1
s2 Nt y FIGURE 7 Visual 14.3B shows how the TNB frame
moves along several curves. cos t, sin t, 0 This shows that the normal vector at a point on the helix is horizontal and points toward
the zaxis. The binormal vector is N x sin t j j
cos t
sin t k
1
0 1
sin t,
s2 cos t, 1 The plane determined by the normal and binormal vectors N and B at a point P on a
curve C is called the normal plane of C at P. It consists of all lines that are orthogonal
to the tangent vector T. The plane determined by the vectors T and N is called the osculating plane of C at P. The name comes from the Latin osculum, meaning “kiss.” It is the
plane that comes closest to containing the part of the curve near P. (For a plane curve, the
osculating plane is simply the plane that contains the curve.)
The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lies
on the concave side of C (toward which N points), and has radius
1 (the reciprocal
of the curvature) is called the osculating circle (or the circle of curvature) of C at P. It is
the circle that best describes how C behaves near P; it shares the same tangent, normal,
and curvature at P.
EXAMPLE 7 Find the equations of the normal plane and osculating plane of the helix in Example 6 at the point P 0, 1, 2. SOLUTION The normal plane at P has normal vector r 1, 0, 1 , so an equation 2 is
1x 0 0y 1 1z 2 0 or z x 2 5E14(pp 904913) ❙❙❙❙ 904 1/18/06 12:33 PM Page 904 CHAPTER 14 VECTOR FUNCTIONS  Figure 8 shows the helix and the osculating
plane in Example 7.
z The osculating plane at P contains the vectors T and N, so its normal vector is
T N B. From Example 6 we have
1
sin t,
s2 Bt z=_x+π
2 cos t, 1 B 1
1
, 0,
s2
s2 2 A simpler normal vector is 1, 0, 1 , so an equation of the osculating plane is
P x 1x y FIGURE 8 0 0y 1z 1 0 2 z or x x 2 at the origin. EXAMPLE 8 Find and graph the osculating circle of the parabola y SOLUTION From Example 5 the curvature of the parabola at the origin is 0
2. So the
1
1
radius of the osculating circle at the origin is 1
2 and its center is (0, 2 ). Its equation
is therefore
2
1
x2 (y 1)
2
4 y y=≈ osculating
circle 2 For the graph in Figure 9 we use parametric equations of this circle:
1
2 x 1
2 0 y 1
2 1
2 sin t We summarize here the formulas for unit tangent, unit normal and binormal vectors,
and curvature. x 1 cos t FIGURE 9
Visual 14.3C shows how the osculating
circle changes as a point moves along a
curve.  14.3
1–6  2 t , sin t 3. r t
4. r t t2 i 5. r t i 6. r t 12 t i tj t k, 0 0 t
t t rt Nt rt
rt 3 ■ 7. Use Simpson’s Rule with n arc of the twisted cubic x
the point 2, 4, 8 .  Reparametrize the curve with respect to arc length measured from the point where t 0 in the direction of increasing t. 10 9. r t t ■ 1 3 sin t i 3t j e 2 t cos 2 t i 11. r t e ■ t 2t i 10. r t 1 1 3t 2 k, 0 ■ t t sin t , 0 ln t k, 1
3 8t 3 2 j
■ 10 e t k, 2t j
2 ■ Tt
rt Tt 9–11 t cos t, cos t
et j s2 t i ■ Bt Exercises 2 sin t, 5t, 2 cos t , 2. r t Tt
Tt Nt
dT
ds Find the length of the curve. 1. r t ■ rt
rt Tt ■ ■ 4t k e 2 t sin 2 t k 2j
4t j ■ 5 3 cos t k ■ ■ ■ ■ ■ ■ ■ ■ 1
■ ■ ■ ■ ■ 10 to estimate the length of the
t, y t 2, z t 3 from the origin to ; 8. Graph the curve with parametric equations x cos t,
y sin 3 t, z sin t. Find the total length of this curve correct
to four decimal places. 12. Reparametrize the curve rt 2
t2 1 1i 2t
t2 1 j with respect to arc length measured from the point (1, 0) in the
direction of increasing t. Express the reparametrization in its
simplest form. What can you conclude about the curve? 5E14(pp 904913) 1/18/06 12:33 PM Page 905 ❙❙❙❙ S ECTION 14.3 ARC LENGTH AND CURVATURE 13–16  905 (a) Find the unit tangent and unit normal vectors T t and N t .
(b) Use Formula 9 to ﬁnd the curvature. 32–33  Two graphs, a and b, are shown. One is a curve y
fx
and the other is the graph of its curvature function y
x . Identify each curve and explain your choices. 13. r t 32. 2 sin t, 5 t, 2 cos t
2 14. r t t , sin t 15. r t s2 t, e t, e ■ ■ 17–19 ■ t ■ t sin t , t ■ ■ ■ ■ ■ t2 i 18. r t ti 19. r t 3t i ■ ■ b ■ tj t2 k 1 4 sin t j
■ ■ ■ ■ ■ ■ ■ ■ t 24. y ■ ■ ■ ■ 4t 3 2 1. ■ metric curve x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ t t is
x y yx
x2 y2 3 2 ■ where the dots indicate derivatives with respect to t.
37–38 ex 27. y ln x f t,y 4x 5 2 26–27  At what point does the curve have maximum curvature?
What happens to the curvature as x l ? 26. y ■ 36. Use Theorem 10 to show that the curvature of a plane para 25. y
■ ■ t 3 sin t, 1 3 cos t, t is shown in
2
2
Figure 12(b) in Section 14.1. Where do you think the curvature
is largest? Use a CAS to ﬁnd and graph the curvature function.
For which values of t is the curvature largest? t2 z cos x
■ ■ 35. The graph of r t Use Formula 11 to ﬁnd the curvature. x3 ■ C AS t, t 2, t 3 at the point (1, 1, 1). and ﬁnd the curvature at the point 1, 4,
 ■ sin 3 t, sin 2 t, sin 3 t . At how
many points on the curve does it appear that the curvature
has a local or absolute maximum?
(b) Use a CAS to ﬁnd and graph the curvature function. Does
this graph conﬁrm your conclusion from part (a)? e t cos t, e t sin t, t at the y ■ x 34. (a) Graph the curve r t ■ ; 22. Graph the curve with parametric equations
x ■ C AS 4 cos t k ■ 21. Find the curvature of r t ■ b
x point (1, 0, 0). 23. y a ■ tk 20. Find the curvature of r t 23–25 y a t 2, 2 t, ln t 16. r t
■ 0 Use Theorem 10 to ﬁnd the curvature.  17. r t ■ t cos t, cos t 33.
y ■ ■ ■  e t cos t, 37. x
38. x 28. Find an equation of a parabola that has curvature 4 at the ■ Use the formula in Exercise 36 to ﬁnd the curvature. t 3, 1
■ e t sin t y
y ■ ■ t2 t
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ origin.
29. (a) Is the curvature of the curve C shown in the ﬁgure greater 39–40 at P or at Q ? Explain.
(b) Estimate the curvature at P and at Q by sketching the
osculating circles at those points. 39. r t
40. r t
■ y 41. x 0 ■ x 1 ■ ■ 1, 0, 1 ■ ■ ■ ■ ■ ■ 31. y
■ ■ ■ ■ x
■ t,z
■ 3 t;
■ 2 cos 3 t ; 0, , 2 1, 1, 1
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1
2 x2
at the points 0, 0 and (1, 2 ). Graph both osculating circles and
the parabola on the same screen. ; 44. Find equations of the osculating circles of the parabola y
1 4
■ ■ t, z 9x 2 4y 2 36 at the points 2, 0 and 0, 3 . Use a graphing
calculator or computer to graph the ellipse and both osculating
circles on the same screen.  Use a graphing calculator or computer to graph both the
curve and its curvature function x on the same screen. Is the
graph of what you would expect?
x t, y
■ 2 ; 43. Find equations of the osculating circles of the ellipse ; 30–31 xe (1, 2 , 1)
3 e t, e t sin t, e t cos t ,
■ 2 sin 3 t, y 42. x Q ■ t 2, 2 t 3, t ,
3 41–42  Find equations of the normal plane and osculating plane
of the curve at the given point. 1 ■ ■ Find the vectors T, N, and B at the given point. P
C 30. y  5E14(pp 904913) ❙❙❙❙ 906 1/18/06 12:33 PM CHAPTER 14 VECTOR FUNCTIONS t 3, y 3 t, z t 4 is the normal
6 y 8 z 1? 45. At what point on the curve x plane parallel to the plane 6 x
CAS Page 906 (b) r r (c) r 47. Show that the curvature s 2 N s 3 B r
r 2 t, 1 t 2, 1 t 3 .
2
3 curve r t d ds ,
where is the angle between T and i ; that is, is the angle of
inclination of the tangent line. (This shows that the deﬁnition
of curvature is consistent with the deﬁnition for plane curves
given in Exercise 69 in Section 11.2.) 54. Find the curvature and torsion of the curve x y sinh t, t at the point 0, 1, 0 . Figure 3 on page 887). The radius of each helix is about
10 angstroms (1 Å 10 8 cm). Each helix rises about 34 Å
during each complete turn, and there are about 2.9 10 8 complete turns. Estimate the length of each helix. (b) Show that d B ds is perpendicular to T.
(c) Deduce from parts (a) and (b) that d B ds
s N for
some number s called the torsion of the curve. (The
torsion measures the degree of twisting of a curve.)
(d) Show that for a plane curve the torsion is s
0. 56. Let’s consider the problem of designing a railroad track to make a smooth transition between sections of straight track.
Existing track along the negative xaxis is to be joined
smoothly to a track along the line y 1 for x 1.
(a) Find a polynomial P P x of degree 5 such that the function F deﬁned by 50. The following formulas, called the FrenetSerret formulas, are of fundamental importance in differential geometry:
1. d T ds
N
2. d N ds
T
B
3. d B ds
N
(Formula 1 comes from Exercise 47 and Formula 3 comes
from Exercise 49.) Use the fact that N B T to deduce
Formula 2 from Formulas 1 and 3. Fx 51. Use the FrenetSerret formulas to prove each of the following. (Primes denote derivatives with respect to t. Start as in the
proof of Theorem 10.)
(a) r
sT
s 2N cosh t, z 55. The DNA molecule has the shape of a double helix (see 49. (a) Show that d B ds is perpendicular to B. ; 0
Px
1 if x
if 0
if x 0
x
1 1 is continuous and has continuous slope and continuous
curvature.
(b) Use a graphing calculator or computer to draw the graph
of F . Motion in Space: Velocity and Acceleration
r (t+h)r (t)
h
r ª(t)
Q z P
r (t) r (t+h) In this section we show how the ideas of tangent and normal vectors and curvature can be
used in physics to study the motion of an object, including its velocity and acceleration,
along a space curve. In particular, we follow in the footsteps of Newton by using these
methods to derive Kepler’s First Law of planetary motion.
Suppose a particle moves through space so that its position vector at time t is r t .
Notice from Figure 1 that, for small values of h, the vector C
1 O FIGURE 1 3 ss 53. Use the formula in Exercise 51(d) to ﬁnd the torsion of the N 48. Show that the curvature of a plane curve is x r
r T a cos t, a sin t, b t , where
a and b are positive constants, has constant curvature and constant torsion. [Use the result of Exercise 51(d).] vectors by the equation  14.4 3 52. Show that the circular helix r t is related to the tangent and normal
dT
ds r (d) lating plane is parallel to the plane x y z 1? (Note: You
will need a CAS for differentiating, for simplifying, and for
computing a cross product.) B s 2 s 46. Is there a point on the curve in Exercise 45 where the oscu 3 s rt h
h rt y approximates the direction of the particle moving along the curve r t . Its magnitude measures the size of the displacement vector per unit time. The vector (1) gives the average 5E14(pp 904913) 1/18/06 12:33 PM Page 907 S ECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ❙ ❙❙❙ 907 velocity over a time interval of length h and its limit is the velocity vector v t at time t : vt 2 lim rt h
h hl0 rt rt Thus, the velocity vector is also the tangent vector and points in the direction of the tangent line.
The speed of the particle at time t is the magnitude of the velocity vector, that is, v t .
This is appropriate because, from (2) and from Equation 14.3.7, we have
vt rt ds
dt rate of change of distance with respect to time As in the case of onedimensional motion, the acceleration of the particle is deﬁned as the
derivative of the velocity:
at vt rt EXAMPLE 1 The position vector of an object moving in a plane is given by rt
t 3 i t 2 j. Find its velocity, speed, and acceleration when t
geometrically. y 1 and illustrate SOLUTION The velocity and acceleration at time t are
v (1) vt
(1, 1) rt 3t 2 i 2t j at rt 6t i 2j a (1)
x 0 and the speed is
FIGURE 2 s 3t 2 vt Visual 14.4 shows animated velocity and
acceleration vectors for objects moving
along various curves. When t z 2 s9 t 4 4t 2 3i 2j a1 6i 2j v1 EXAMPLE 2 Find the velocity, acceleration, and speed of a particle with position vector t 2, e t, te t . SOLUTION v (1) vt 1
y FIGURE 3 rt 2 t, e t, 1 t et at x s13 These velocity and acceleration vectors are shown in Figure 2. rt a (1) 2t 1, we have
v1  Figure 3 shows the path of the particle in
Example 2 with the velocity and acceleration
vectors when t 1. 2 vt 2, e t, 2 t et s4 t 2 e 2t t 2e 2t vt 1 The vector integrals that were introduced in Section 14.2 can be used to ﬁnd position
vectors when velocity or acceleration vectors are known, as in the following example. 5E14(pp 904913) 908 ❙❙❙❙ 1/18/06 12:33 PM Page 908 CHAPTER 14 VECTOR FUNCTIONS EXAMPLE 3 A moving particle starts at an initial position r 0
1, 0, 0 with initial
velocity v 0
i j k. Its acceleration is a t
4 t i 6 t j k. Find its velocity
and position at time t .
SOLUTION Since a t v t , we have yat vt 2t 2 i y dt 4t i 3t 2 j 6t j tk k dt C To determine the value of the constant vector C, we use the fact that v 0
The preceding equation gives v 0
C, so C i j k and
2t 2 i vt 3t 2 j  The expression for r t that we obtained in
Example 3 was used to plot the path of the particle in Figure 4 for 0 t 3. Since v t yvt z4 0
0 (2t3
3
5
y 10 1j t 1k 1i 3t 2 1j t k. 1 k dt j k dt 2t 2 y 6 (1, 0, 0) 3t 2 j r t , we have
rt 2 i 1i 2t 2 tk i t) i tj (1t2
2 t3 t3 tj t) k D 0
15 20 20 x Putting t 0, we ﬁnd that D (2t3
3 rt F IGURE 4 r0
t i, so
1) i (1t2
2 t) k In general, vector integrals allow us to recover velocity when acceleration is known and
position when velocity is known:
vt v t0 y t t0 a u du rt r t0 y t t0 v u du If the force that acts on a particle is known, then the acceleration can be found from
Newton’s Second Law of Motion. The vector version of this law states that if, at any time
t, a force F t acts on an object of mass m producing an acceleration a t , then
 The angular speed of the object moving with
d dt, where is the angle
position P is
shown in Figure 5. Ft ma t EXAMPLE 4 An object with mass m that moves in a circular path with constant angular
speed has position vector r t
a cos t i a sin t j. Find the force acting on the
object and show that it is directed toward the origin. y P SOLUTION vt rt a sin t i at ¨
0 vt a a cos t j x
2 cos t i a 2 sin t j Therefore, Newton’s Second Law gives the force as
FIGURE 5 Ft ma t m 2 a cos t i a sin t j 5E14(pp 904913) 1/18/06 12:33 PM Page 909 S ECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ❙❙❙❙ 909 Notice that F t
m 2 r t . This shows that the force acts in the direction opposite to
the radius vector r t and therefore points toward the origin (see Figure 5). Such a force
is called a centripetal (centerseeking) force.
y EXAMPLE 5 A projectile is ﬁred with angle of elevation and initial velocity v0. (See
Figure 6.) Assuming that air resistance is negligible and the only external force is due to
gravity, ﬁnd the position function r t of the projectile. What value of maximizes the
range (the horizontal distance traveled)? v¸ a
0 x d SOLUTION We set up the axes so that the projectile starts at the origin. Since the force due
to gravity acts downward, we have F FIGURE 6 where t ma mt j 9.8 m s2 . Thus a a
Since v t tj a, we have
vt where C tt j C v0 . Therefore v0 rt vt tt j v0 t v0 D Integrating again, we obtain
1
2 rt
But D tt 2 j 0, so the position vector of the projectile is given by r0 1
2 rt 3 If we write v0 tt 2 j t v0 v0 (the initial speed of the projectile), then v0 v0 cos i v0 sin j and Equation 3 becomes
rt v0 cos [ v0 sin ti 1
2 t tt 2 ] j The parametric equations of the trajectory are therefore
 If you eliminate t from Equations 4, you will
see that y is a quadratic function of x. So the
path of the projectile is part of a parabola. x 4 v0 cos t y v0 sin t 1
2 tt 2 The horizontal distance d is the value of x when y 0. Setting y
or t
2v0 sin
t. The latter value of t then gives
d x v0 cos 2v0 sin
t Clearly, d has its maximum value when sin 2 v 2 2 sin
0 cos 0, we obtain t v 2 sin 2
0 t
1, that is, t
4. 0 5E14(pp 904913) 910 ❙❙❙❙ 1/18/06 12:33 PM Page 910 CHAPTER 14 VECTOR FUNCTIONS E XAMPLE 6 A projectile is ﬁred with muzzle speed 150 m s and angle of elevation 45
from a position 10 m above ground level. Where does the projectile hit the ground, and
with what speed?
SOLUTION If we place the origin at ground level, then the initial position of the projectile
is (0, 10) and so we need to adjust Equations 4 by adding 10 to the expression for y.
With v 0 150 m s,
45 , and t 9.8 m s2, we have x 150 cos y 10 4t 75 s2 t 150 sin 1
2 4t 9.8 t 2 10 Impact occurs when y 0, that is, 4.9 t 2 75 s2 t 10
equation (and using only the positive value of t), we get
t 75 s2 s11,250
9.8 4.9 t 2 75 s2 t 196 0. Solving this quadratic 21.74 Then x 75 s2 21.74
2306 , so the projectile hits the ground about 2306 m away.
The velocity of the projectile is
vt rt (75 s2 75 s2 i 9.8 t) j So its speed at impact is
v 21.74 s(75 s2 )2 (75 s2 9.8 21.74)2 151 m s Tangential and Normal Components of Acceleration
When we study the motion of a particle, it is often useful to resolve the acceleration into
two components, one in the direction of the tangent and the other in the direction of the
normal. If we write v
v for the speed of the particle, then
rt
rt Tt
and so v vt
vt v
v vT If we differentiate both sides of this equation with respect to t , we get
a 5 v vT vT If we use the expression for the curvature given by Equation 14.3.9, then we have
6 T
r T so v T The unit normal vector was deﬁned in the preceding section as N
T TN vN and Equation 5 becomes
7 a vT v2 N v T T , so (6) gives 5E14(pp 904913) 1/18/06 12:33 PM Page 911 SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ❙❙❙❙ 911 Writing a T and a N for the tangential and normal components of acceleration, we have aT a T
a aTT aN N and aN where N aT 8 v v2 aN FIGURE 7 This resolution is illustrated in Figure 7.
Let’s look at what Formula 7 says. The ﬁrst thing to notice is that the binormal vector
B is absent. No matter how an object moves through space, its acceleration always lies in
the plane of T and N (the osculating plane). (Recall that T gives the direction of motion
and N points in the direction the curve is turning.) Next we notice that the tangential component of acceleration is v , the rate of change of speed, and the normal component of
acceleration is v 2, the curvature times the square of the speed. This makes sense if we
think of a passenger in a car—a sharp turn in a road means a large value of the curvature
, so the component of the acceleration perpendicular to the motion is large and the passenger is thrown against a car door. High speed around the turn has the same effect; in fact,
if you double your speed, aN is increased by a factor of 4.
Although we have expressions for the tangential and normal components of acceleration in Equations 8, it’s desirable to have expressions that depend only on r, r , and r . To
this end we take the dot product of v v T with a as given by Equation 7:
va vT
vv T v2 N vT v3 T T N vv (since T T 1 and T N 0) Therefore
aT 9 va v rt v rt
rt Using the formula for curvature given by Theorem 14.3.10, we have
10 aN rt v2 rt
rt 3 rt rt 2 EXAMPLE 7 A particle moves with position function r t t 2, t 2, t 3 . Find the tangential and normal components of acceleration.
rt t2 i t2 j rt 2t i 2t j rt 2i rt SOLUTION s8t 2 2j t3 k
3t 2 k
6t k 9t 4 Therefore, Equation 9 gives the tangential component as
aT rt rt
rt rt
rt 8 t 18 t 3
s8 t 2 9 t 4 5E14(pp 904913) 912 ❙❙❙❙ 1/18/06 12:33 PM Page 912 CHAPTER 14 VECTOR FUNCTIONS rt Since ijk
2 t 2 t 3t 2
2 2 6t rt 6t 2 i 6t 2 j Equation 10 gives the normal component as
aN rt rt
rt 6 s2 t 2
s8 t 2 9 t 4 Kepler ’s Laws of Planetary Motion
We now describe one of the great accomplishments of calculus by showing how the material of this chapter can be used to prove Kepler’s laws of planetary motion. After 20 years
of studying the astronomical observations of the Danish astronomer Tycho Brahe, the
German mathematician and astronomer Johannes Kepler (1571–1630) formulated the following three laws.
Kepler’s Laws
1. A planet revolves around the Sun in an elliptical orbit with the Sun at one focus.
2. The line joining the Sun to a planet sweeps out equal areas in equal times.
3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
In his book Principia Mathematica of 1687, Sir Isaac Newton was able to show that
these three laws are consequences of two of his own laws, the Second Law of Motion and
the Law of Universal Gravitation. In what follows we prove Kepler’s First Law. The remaining laws are left as exercises (with hints).
Since the gravitational force of the Sun on a planet is so much larger than the forces
exerted by other celestial bodies, we can safely ignore all bodies in the universe except the
Sun and one planet revolving about it. We use a coordinate system with the Sun at the origin and we let r r t be the position vector of the planet. (Equally well, r could be the
position vector of the Moon or a satellite moving around the Earth or a comet moving
around a star.) The velocity vector is v r and the acceleration vector is a r . We use
the following laws of Newton:
Second Law of Motion: F
Law of Gravitation: F ma
GMm
r
r3 GMm
u
r2 where F is the gravitational force on the planet, m and M are the masses of the planet and
r , and u
1 r r is the unit vector in the
the Sun, G is the gravitational constant, r
direction of r.
We ﬁrst show that the planet moves in one plane. By equating the expressions for F in
Newton’s two laws, we ﬁnd that
a GM
r
r3 5E14(pp 904913) 1/18/06 12:33 PM Page 913 SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION and so a is parallel to r. It follows that r
to write
d
r
dt v a ❙❙❙❙ 913 0. We use Formula 5 in Theorem 14.2.3 r v r v v v r a Therefore r v 0 0 0 h where h is a constant vector. (We may assume that h 0; that is, r and v are not parallel.)
This means that the vector r r t is perpendicular to h for all values of t , so the planet
always lies in the plane through the origin perpendicular to h. Thus, the orbit of the planet
is a plane curve.
To prove Kepler’s First Law we rewrite the vector h as follows:
h r v r ru ru 2 r ru ru r2 u u ru ru u rr u u Then
a GM
u
r2 h r2u u GM u u u
But u u
that u u and so FIGURE 8 v u (by Theorem 13.4.8, Property 6) 1, it follows from Example 5 in Section 14.2 GM u h v v 11 ¨ y r
x h u h a h GM u Integrating both sides of this equation, we get h
c u uu u 2 1 and, since u t
0. Therefore
a z GM u v
u h GM u c where c is a constant vector.
At this point it is convenient to choose the coordinate axes so that the standard basis
vector k points in the direction of the vector h. Then the planet moves in the xyplane.
Since both v h and u are perpendicular to h, Equation 11 shows that c lies in the
xyplane. This means that we can choose the x and yaxes so that the vector i lies in the
direction of c, as shown in Figure 8.
If is the angle between c and r, then r, are polar coordinates of the planet. From
Equation 11 we have
r v h r GM u GMr u u c GM r u
r c cos rc
GMr rc cos 5E14(pp 914919) 914 ❙❙❙❙ 1/18/06 12:21 PM Page 914 CHAPTER 14 VECTOR FUNCTIONS where c c . Then
r where e v h vh
e cos r v h hh h 2 h2 h . So
r Writing d 1r
GM 1 c GM . But
r where h rvh
GM c cos h 2 GM
1 e cos eh 2 c
1 e cos h 2 c, we obtain the equation
r 12 1 ed
e cos Comparing with Theorem 11.6.6, we see that Equation 12 is the polar equation of a conic
section with focus at the origin and eccentricity e. We know that the orbit of a planet is a
closed curve and so the conic must be an ellipse.
This completes the derivation of Kepler’s First Law. We will guide you through the
derivation of the Second and Third Laws in the Applied Project on page 916. The proofs
of these three laws show that the methods of this chapter provide a powerful tool for
describing some of the laws of nature.  14.4 Exercises 1. The table gives coordinates of a particle moving through space along a smooth curve.
(a) Find the average velocities over the time intervals [0, 1],
[0.5, 1], [1, 2], and [1, 1.5].
(b) Estimate the velocity and speed of the particle at t 1. (d) Draw an approximation to the vector v(2) and estimate the
speed of the particle at t 2.
y r(2.4)
t x y 0
0.5
1.0
1.5
2.0 2.7
3.5
4.5
5.9
7.3 9.8
7.2
6.0
6.4
7.8 2 3.7
3.3
3.0
2.8
2.7 2. The ﬁgure shows the path of a particle that moves with position vector r t at time t.
(a) Draw a vector that represents the average velocity of the
particle over the time interval 2 t 2.4.
(b) Draw a vector that represents the average velocity over the
time interval 1.5 t 2.
(c) Write an expression for the velocity vector v(2). r(2) 1 z r(1.5) 0 2 1 x 3–8  Find the velocity, acceleration, and speed of a particle with
the given position function. Sketch the path of the particle and
draw the velocity and acceleration vectors for the speciﬁed value
of t. 3. r t t2 4. r t 2 5. r t t ei 1, t , t 1 t, 4 st , t
t e j, t 1
0 5E14(pp 914919) 1/18/06 12:21 PM Page 915 S ECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION 6. r t sin t i 2 cos t j, 7. r t sin t i tj ; 8. r t
■ t2 j ti ■ ■ t t 3 k, t ■ ■ t that can be used to hit a target 800 m away. 0 28. A batter hits a baseball 3 ft above the ground toward the center 1
■ ■ ■ ■ ■ ■ ﬁeld fence, which is 10 ft high and 400 ft from home plate. The
ball leaves the bat with speed 115 ft s at an angle 50 above
the horizontal. Is it a home run? (In other words, does the ball
clear the fence?) ■ 9–14  Find the velocity, acceleration, and speed of a particle with
the given position function. t2 9. r t
10. r t 1, t 3, t 2
et j s2 t i
2 12. r t ti 13. r t t e cos t i 14. r t t sin t i ■ 1 ; 29. Water traveling along a straight portion of a river normally
ﬂows fastest in the middle, and the speed slows to almost zero
at the banks. Consider a long stretch of river ﬂowing north,
with parallel banks 40 m apart. If the maximum water speed
is 3 m s, we can use a quadratic function as a basic model for
the rate of water ﬂow x units from the west bank:
3
x.
fx
400 x 40
(a) A boat proceeds at a constant speed of 5 m s from a point
A on the west bank while maintaining a heading perpendicular to the bank. How far down the river on the opposite
bank will the boat touch shore? Graph the path of the boat.
(b) Suppose we would like to pilot the boat to land at the point
B on the east bank directly opposite A. If we maintain a
constant speed of 5 m s and a constant heading, ﬁnd the
angle at which the boat should head. Then graph the actual
path the boat follows. Does the path seem realistic? 2 cos t, 3 t, 2 sin t 11. r t ■ e tk ln t j ■ tk
sin t j tk
t2 k t cos t j
■ ■ ■ ■ ■ ■ ■ ■ ■ 15–16  Find the velocity and position vectors of a particle that
has the given acceleration and the given initial velocity and
position. 15. a t k, ■ ■ 17–18 i 10 k, 16. a t v0 v0 ■ ■ j, r0 i
■ 0 j
■ k, r 0
■ 2i ■ 3j ■ ■ ■ ■ 30. Another reasonable model for the water speed of the river in  Exercise 29 is a sine function: f x
3 sin x 40 . If a boater
would like to cross the river from A to B with constant heading
and a constant speed of 5 m s, determine the angle at which
the boat should head. (a) Find the position vector of a particle that has the given acceleration and the speciﬁed initial velocity and position.
; (b) Use a computer to graph the path of the particle.
17. a t
18. a t
■ ■ i 2j ti
■ 2 t k, 2 tj v0 0, cos 2 t k, v0 ■ ■ ■ ■ r0
i
■ i k k, r0
■ j
■ ■ ■ 19. The position function of a particle is given by rt 915 27. A gun has muzzle speed 150 m s. Find two angles of elevation 6 cos t k, ❙❙❙❙ t 2, 5 t, t 2 16 t . When is the speed a minimum? 20. What force is required so that a particle of mass m has the position function r t t3 i t2 j t 3 k? 21. A force with magnitude 20 N acts directly upward from the xyplane on an object with mass 4 kg. The object starts at the
origin with initial velocity v 0
i j. Find its position function and its speed at time t.
22. Show that if a particle moves with constant speed, then the velocity and acceleration vectors are orthogonal. 31–36  Find the tangential and normal components of the acceleration vector. 31. r t 3t t3 i 32. r t 1 ti 33. r t 3t 2 j
2t j sin t j cos t i t2 tk 2 34. r t ti 35. r t t ei s2 t j e tk 36. r t ti cos 2t j sin 2t k ■ ■ ■ tj ■ 3t k ■ ■ ■ ■ ■ ﬁgure to estimate the tangential and normal components of a.
y 24. Rework Exercise 23 if the projectile is ﬁred from a position a 200 m above the ground.
25. A ball is thrown at an angle of 45 to the ground. If the ball lands 90 m away, what was the initial speed of the ball?
26. A gun is ﬁred with angle of elevation 30 . What is the muzzle speed if the maximum height of the shell is 500 m? ■ ■ 37. The magnitude of the acceleration vector a is 10 cm s2. Use the 23. A projectile is ﬁred with an initial speed of 500 m s and angle of elevation 30 . Find (a) the range of the projectile, (b) the
maximum height reached, and (c) the speed at impact. ■ 0 x 5E14(pp 914919) 916 ❙❙❙❙ 1/18/06 12:21 PM Page 916 CHAPTER 14 VECTOR FUNCTIONS 38. If a particle with mass m moves with position vector r t , then 40. A rocket burning its onboard fuel while moving through space its angular momentum is deﬁned as L t
mr t
v t and
its torque as t
mr t
a t . Show that L t
t.
Deduce that if t
0 for all t, then L t is constant. (This is
the law of conservation of angular momentum.) has velocity v t and mass m t at time t. If the exhaust gases
escape with velocity ve relative to the rocket, it can be deduced
from Newton’s Second Law of Motion that
m 39. The position function of a spaceship is rt 3 ti 2 ln t j 7 4
t2 1 dv
dt dm
ve
dt m0
ve.
mt
(b) For the rocket to accelerate in a straight line from rest to
twice the speed of its own exhaust gases, what fraction of
its initial mass would the rocket have to burn as fuel? k (a) Show that v t and the coordinates of a space station are 6, 4, 9 . The captain
wants the spaceship to coast into the space station. When
should the engines be turned off? v0 ln APPLIED PROJECT
Kepler’s Laws
Johannes Kepler stated the following three laws of planetary motion on the basis of masses of
data on the positions of the planets at various times.
Kepler’s Laws
1. A planet revolves around the Sun in an elliptical orbit with the Sun at one focus.
2. The line joining the Sun to a planet sweeps out equal areas in equal times.
3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
Kepler formulated these laws because they ﬁtted the astronomical data. He wasn’t able to see
why they were true or how they related to each other. But Sir Isaac Newton, in his Principia
Mathematica of 1687, showed how to deduce Kepler’s three laws from two of Newton’s own
laws, the Second Law of Motion and the Law of Universal Gravitation. In Section 14.4 we
proved Kepler’s First Law using the calculus of vector functions. In this project we guide you
through the proofs of Kepler’s Second and Third Laws and explore some of their consequences.
1. Use the following steps to prove Kepler’s Second Law. The notation is the same as in the proof of the First Law in Section 14.4. In particular, use polar coordinates so that
r
r cos i
r sin j.
(a) Show that h r2 d
k.
dt (b) Deduce that r 2 d
dt h. (c) If A A t is the area swept out by the radius vector r
as in the ﬁgure, show that y r (t)
A(t) dA
dt r (t¸) 1
2 r2 r t in the time interval t0 , t d
dt (d) Deduce that
0 x dA
dt 1
2 h constant This says that the rate at which A is swept out is constant and proves Kepler’s Second
Law. 5E14(pp 914919) 1/18/06 12:21 PM Page 917 C HAPTER 14 REVIEW ❙❙❙❙ 917 2. Let T be the period of a planet about the Sun; that is, T is the time required for it to travel once around its elliptical orbit. Suppose that the lengths of the major and minor axes of the
ellipse are 2a and 2b.
(a) Use part (d) of Problem 1 to show that T
2 (b) Show that h
GM 2 a b h. 2 ed b
.
a (c) Use parts (a) and (b) to show that T 2 423
a.
GM This proves Kepler’s Third Law. [Notice that the proportionality constant 4
independent of the planet.] 2 GM is 3. The period of the Earth’s orbit is approximately 365.25 days. Use this fact and Kepler’s Third Law to ﬁnd the length of the major axis of the Earth’s orbit. You will need the mass of
the Sun, M 1.99 10 30 kg, and the gravitational constant, G 6.67 10 11 N m 2 kg2 .
4. It’s possible to place a satellite into orbit about the Earth so that it remains ﬁxed above a given location on the equator. Compute the altitude that is needed for such a satellite. The
Earth’s mass is 5.98 10 24 kg; its radius is 6.37 10 6 m. (This orbit is called the Clarke
Geosynchronous Orbit after Arthur C. Clarke, who ﬁrst proposed the idea in 1948. The ﬁrst
such satellite, Syncom II, was launched in July 1963.)  14 Review ■ CONCEPT CHECK 1. What is a vector function? How do you ﬁnd its derivative and its integral?
2. What is the connection between vector functions and space curves? ■ 6. (a) What is the deﬁnition of curvature? (b) Write a formula for curvature in terms of r t and T t .
(c) Write a formula for curvature in terms of r t and r t .
(d) Write a formula for the curvature of a plane curve with
equation y f x . 3. (a) What is a smooth curve? (b) How do you ﬁnd the tangent vector to a smooth curve at a
point? How do you ﬁnd the tangent line? The unit tangent
vector?
4. If u and v are differentiable vector functions, c is a scalar, and f is a realvalued function, write the rules for differentiating
the following vector functions.
(a) u t
(b) c u t
(c) f t u t
vt
(d) u t v t
(e) u t
(f) u f t
vt 7. (a) Write formulas for the unit normal and binormal vectors of a smooth space curve r t .
(b) What is the normal plane of a curve at a point? What is the
osculating plane? What is the osculating circle?
8. (a) How do you ﬁnd the velocity, speed, and acceleration of a particle that moves along a space curve?
(b) Write the acceleration in terms of its tangential and normal
components. 5. How do you ﬁnd the length of a space curve given by a vector function r t ? 9. State Kepler’s Laws. 5E14(pp 914919) 918 ❙❙❙❙ 1/18/06 12:21 PM Page 918 CHAPTER 14 VECTOR FUNCTIONS ■ TRUEFALSE QUIZ 6. If r t is a differentiable vector function, then Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. t3 i 1. The curve with vector equation r t 2t 3 j d
rt
dt 3t 3 k is a line.
t, t 3, t 5 is smooth. 3. The curve with vector equation r t cos t, t 2, t 4 is smooth. 8. The binormal vector is B t 10. Different parametrizations of the same curve result in identical vt tangent vectors at a given point on the curve. ■ EXERCISES 1. (a) Sketch the curve with vector function ti cos t j sin t k 2. Let r t t t , t , t , ﬁnd (a) the unit tangent vector, (b) the unit normal vector, and (c) the curvature. 0 1. 13. Find the curvature of the curve y 16 and the plane x z 5. x t 2, y t 4, z t 3 at the point 1, 1, 1 . Graph the curve
and the tangent line on a common screen.
t2 i t cos t j sin t k, evaluate x01 r t dt. 2 t 3, y 2 t 1,
z ln t. Find (a) the point where C intersects the xzplane,
(b) parametric equations of the tangent line at 1, 1, 0 , and
(c) an equation of the normal plane to C at 1, 1, 0 . 6. Let C be the curve with equations x 7. Use Simpson’s Rule with n 4 to estimate the length of the
arc of the curve with equations x st, y 4 t, z t 2 1
from 1, 4, 2 to 2, 1, 17 . 8. Find the length of the curve r t 0 t 4 sin t at the x 4 at the point 1, 1 . ; 14. Find an equation of the osculating circle of the curve ; 4. Find parametric equations for the tangent line to the curve 5. If r t 3 cos t, y points 3, 0 and 0, 4 . 3. Find a vector function that represents the curve of intersection y2 1312
3
2 12. Find the curvature of the ellipse x 1 t, ln t
s2 t, e
(a) Find the domain of r.
(b) Find lim t l 0 r t .
(c) Find r t .
of the cylinder x 2 ■ 11. For the curve given by r t (b) Find r t and r t .
t Tt. tangent vector, normal vector, and curvature as C at that
point. 5. If u t and v t are differentiable vector functions, then rt Nt 9. The osculating circle of a curve C at a point has the same tiating each component function. ut d T dt . curvature is 4. The derivative of a vector function is obtained by differen vt rt 7. If T t is the unit tangent vector of a smooth curve, then the 2. The curve with vector equation r t d
ut
dt ■ y x 4 x 2 at the origin. Graph both the curve and its osculating circle.
15. Find an equation of the osculating plane of the curve x sin 2 t, y t, z cos 2 t at the point 0, , 1 . 16. The ﬁgure shows the curve C traced by a particle with position vector r t at time t.
(a) Draw a vector that represents the average velocity of the
particle over the time interval 3 t 3.2.
(b) Write an expression for the velocity v(3).
(c) Write an expression for the unit tangent vector T(3) and
draw it.
y 2 t 3 2, cos 2 t, sin 2 t , 1. C
1 9. The helix r1 t cos t i sin t j t k intersects the curve
r2 t
1 t i t 2 j t 3 k at the point 1, 0, 0 . Find the
angle of intersection of these curves.
e t i e t sin t j e t cos t k
with respect to arc length measured from the point 1, 0, 1 in
the direction of increasing t . r(3)
r(3.2) 10. Reparametrize the curve r t 0 1 x 5E14(pp 914919) 1/18/06 12:21 PM Page 919 C HAPTER 14 REVIEW 17. A particle moves with position function rt 18. A particle starts at the origin with initial velocity i j 3 k.
6 t k. Find its position 12 t 2 j 6t i 20. Find the tangential and normal components of the acceleration vector of a particle with position function
t2 k 2t j 21. A disk of radius 1 is rotating in the counterclockwise direction cos t i sin t j (a) Show that the velocity v of the particle is
v cos t i sin t j t vd where vd R t is the velocity of a point on the edge of
the disk.
(b) Show that the acceleration a of the particle is
a 2 vd t ad where a d R t is the acceleration of a point on the rim
of the disk. The extra term 2 vd is called the Coriolis acceleration; it is the result of the interaction of the rotation of
the disk and the motion of the particle. One can obtain a
physical demonstration of this acceleration by walking
toward the edge of a moving merrygoround. 1
s1
s2 Fx at a constant angular speed . A particle starts at the center of
the disk and moves toward the edge along a ﬁxed radius so that
its position at time t, t 0, is given by r t
t R t , where
Rt e t sin t j road tracks, it’s important to realize that the acceleration of the
train should be continuous so that the reactive force exerted by
the train on the track is also continuous. Because of the formulas for the components of acceleration in Section 14.4, this will
be the case if the curvature varies continuously.
(a) A logical candidate for a transfer curve to join existing
tracks given by y 1 for x 0 and y s2 x for
x 1 s2 might be the function f x
s1 x 2,
, whose graph is the arc of the circle shown
0 x 1 s2
in the ﬁgure. It looks reasonable at ﬁrst glance. Show that
the function at an initial speed of 43 ft s. It leaves his hand 7 ft above the
ground.
(a) Where is the shot 2 seconds later?
(b) How high does the shot go?
(c) Where does the shot land? ti e t cos t i 22. In designing transfer curves to connect sections of straight rail 19. An athlete throws a shot at an angle of 45 to the horizontal rt 919 (c) Determine the Coriolis acceleration of a particle that moves
on a rotating disk according to the equation rt
t ln t i t j e t k. Find the velocity, speed, and
acceleration of the particle.
Its acceleration is a t
function. ❙❙❙❙ ; if x
if 0
if x x2
x 0
x 1 s2
1 s2 is continuous and has continuous slope, but does not have
continuous curvature. Therefore, f is not an appropriate
transfer curve.
(b) Find a ﬁfthdegree polynomial to serve as a transfer curve
between the following straight line segments: y 0 for
x 0 and y x for x 1. Could this be done with a
fourthdegree polynomial? Use a graphing calculator or
computer to sketch the graph of the “connected” function
and check to see that it looks like the one in the ﬁgure.
y y
1 y=x y=F(x) y=0
0 1
2
œ„ x 0 transfer curve
1 x 5E14(pp 920921) 1/17/06 11:42 AM PROBLEMS
PLUS Page 920 1. A particle P moves with constant angular speed around a circle whose center is at the origin
and whose radius is R. The particle is said to be in uniform circular motion. Assume that the
motion is counterclockwise and that the particle is at the point R, 0 when t 0. The position
vector at time t 0 is r t
R cos t i R sin t j.
(a) Find the velocity vector v and show that v r 0. Conclude that v is tangent to the circle
and points in the direction of the motion.
(b) Show that the speed v of the particle is the constant R. The period T of the particle is
the time required for one complete revolution. Conclude that
2R
v T y v vt r x 2 (c) Find the acceleration vector a. Show that it is proportional to r and that it points toward
the origin. An acceleration with this property is called a centripetal acceleration. Show
that the magnitude of the acceleration vector is a
R 2.
(d) Suppose that the particle has mass m. Show that the magnitude of the force F that is
required to produce this motion, called a centripetal force, is
F mv
R 2 2. A circular curve of radius R on a highway is banked at an angle
FIGURE FOR PROBLEM 1 F ¨ FIGURE FOR PROBLEM 2 so that a car can safely
traverse the curve without skidding when there is no friction between the road and the tires.
The loss of friction could occur, for example, if the road is covered with a ﬁlm of water or ice.
The rated speed vR of the curve is the maximum speed that a car can attain without skidding.
Suppose a car of mass m is traversing the curve at the rated speed vR. Two forces are acting on
the car: the vertical force, m t, due to the weight of the car, and a force F exerted by, and
normal to, the road. (See the ﬁgure.)
The vertical component of F balances the weight of the car, so that F cos
m t. The
horizontal component of F produces a centripetal force on the car so that, by Newton’s Second Law and part (d) of Problem 1,
F sin mg 2
mvR
R 2
(a) Show that vR Rt tan .
(b) Find the rated speed of a circular curve with radius 400 ft that is banked at an angle of 12 .
(c) Suppose the design engineers want to keep the banking at 12 , but wish to increase the
rated speed by 50%. What should the radius of the curve be? 3. A projectile is ﬁred from the origin with angle of elevation and initial speed v0. Assuming
that air resistance is negligible and that the only force acting on the projectile is gravity, t,
we showed in Example 5 in Section 14.4 that the position vector of the projectile is
v0 cos t i
[ v0 sin t 1 t t 2 ] j . We also showed that the maximum horizontal
rt
2
distance of the projectile is achieved when
45 and in this case the range is R v 2 t.
0
(a) At what angle should the projectile be ﬁred to achieve maximum height and what is the
maximum height?
(b) Fix the initial speed v0 and consider the parabola x 2 2 Ry R 2 0, whose graph is
shown in the ﬁgure. Show that the projectile can hit any target inside or on the boundary
y _R 920 0 y Rx 0 D x 5E14(pp 920921) 1/17/06 11:42 AM Page 921 of the region bounded by the parabola and the xaxis, and that it can’t hit any target outside this region.
(c) Suppose that the gun is elevated to an angle of inclination in order to aim at a target that
is suspended at a height h directly over a point D units downrange. The target is released
at the instant the gun is ﬁred. Show that the projectile always hits the target, regardless of
the value v0, provided the projectile does not hit the ground “before” D.
y 4. (a) A projectile is ﬁred from the origin down an inclined plane that makes an angle
v¸
a
x ¨ FIGURE FOR PROBLEM 4 with the
horizontal. The angle of elevation of the gun and the initial speed of the projectile are
and v0, respectively. Find the position vector of the projectile and the parametric equations
of the path of the projectile as functions of the time t. (Ignore air resistance.)
(b) Show that the angle of elevation that will maximize the downhill range is the angle
halfway between the plane and the vertical.
(c) Suppose the projectile is ﬁred up an inclined plane whose angle of inclination is . Show
that, in order to maximize the (uphill) range, the projectile should be ﬁred in the direction
halfway between the plane and the vertical.
(d) In a paper presented in 1686, Edmond Halley summarized the laws of gravity and projectile
motion and applied them to gunnery. One problem he posed involved ﬁring a projectile to
hit a target a distance R up an inclined plane. Show that the angle at which the projectile
should be ﬁred to hit the target but use the least amount of energy is the same as the angle
in part (c). (Use the fact that the energy needed to ﬁre the projectile is proportional to
the square of the initial speed, so minimizing the energy is equivalent to minimizing the
initial speed.) 5. A projectile of mass m is ﬁred from the origin at an angle of elevation . In addition to grav ity, assume that air resistance provides a force that is proportional to the velocity and that
opposes the motion. Then, by Newton’s Second Law, the total force acting on the projectile
satisﬁes the equation
1 d 2R
dt 2 m mt j k dR
dt where R is the position vector and k 0 is the constant of proportionality.
(a) Show that Equation 1 can be integrated to obtain the equation
k
R
m dR
dt v0 tt j dR
0.
dt
(b) Multiply both sides of the equation in part (a) by e k m t and show that the lefthand side of
the resulting equation is the derivative of the product e k m t R t . Then integrate to ﬁnd an
expression for the position vector R t .
where v0 v0 6. Find the curvature of the curve with parametric equations x y t 0 sin ( 1
2 )d 2 y y t 0 cos ( 1
2 )d 2 7. A ball rolls off a table with a speed of 2 ft s. The table is 3.5 ft high.
3.5 ft FIGURE FOR PROBLEM 7 ¨¨ (a) Determine the point at which the ball hits the ﬂoor and ﬁnd its speed at the instant of
impact.
(b) Find the angle between the path of the ball and the vertical line drawn through the point
of impact. (See the ﬁgure.)
(c) Suppose the ball rebounds from the ﬂoor at the same angle with which it hits the ﬂoor, but
loses 20% of its speed due to energy absorbed by the ball on impact. Where does the ball
strike the ﬂoor on the second bounce?
8. A cable has radius r and length L and is wound around a spool with radius R without over lapping. What is the shortest length along the spool that is covered by the cable? 921 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.
 Fall '09
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