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**Unformatted text preview: **5E-14(pp 884-893)
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CHAPTER 14
The calculus of vectorvalued functions is used in Section 14.4 to prove Keplers 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
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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 vector-valued functions to describe the motion of objects through space. In particular, we will use them to derive Keplers 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 vector-valued 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 three-dimensional 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 real-valued 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
rt then the component functions are ft t3 tt
t 3, ln 3
t , st
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 dened. The expressions t 3, ln 3 t , and st are all dened 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 dened by taking the limits of its component functions as follows.
1
|||| If lim t l a r t L, this denition 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 - denition (see Exercise 43). Limits of vector functions obey the same rules as limits of real-valued functions (see Exercise 41).
885
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CHAPTER 14 VECTOR FUNCTIONS
EXAMPLE 2 Find lim r t , where r t
tl0
1
t3 i
te t j
sin t k. t
SOLUTION According to Denition 1, the limit of r is the vector whose components are the limits of the component functions of r :
lim r t
tl0
tl0
lim 1 k
t3
i
tl0
lim te
t
j
lim
tl0
sin t k t
i
(by Equation 3.5.2)
A vector function r is continuous at a if lim r t
tla
ra
z
P { f(t), g(t), h(t)} C
In view of Denition 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 real-valued functions on an interval I . Then the set C of all points x, y, z in space, where
2
x
ft
y
tt
z
ht
0 x
r(t)=k f(t), g(t), h(t)l
y
F IGURE 1
C is traced out by the tip of a moving position vector r(t).
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 denes 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 dened 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
5t
z
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 t 1j
0, 1 .
EXAMPLE 4 Sketch the curve whose vector equation is
rt
cos t i
sin t j
tk
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S OLUTION The parametric equations for this curve are
x
cos t
y
sin t
z
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 y-plane. (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
0, 1, 2 (1, 0, 0) y
F IGURE 2
x
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 nd 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 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:
|||| Figure 4 shows the line segment PQ in Example 5. z
rt
1
t r0
tr1
0
t
1 2, 1, 3 to obtain a
Q(2, _1, 3)
(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: rt or 1 t 1, 3, 2 t 2, 1, 3 0
t
1
x
y
rt
1
t, 3
4 t,
2
5t
0
t
1
P(1, 3, _2) FIGURE 4
The corresponding parametric equations are x 1 t y 3 4t z 2 5t 0 t 1
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CHAPTER 14 VECTOR FUNCTIONS
EXAMPLE 6 Find a vector function that represents the curve of intersection of the cylinder
x2
y2
1 and the plane y
z
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 (0, _1, 3)
y+z=2
(_1, 0, 2)
C
(1, 0, 2)
(0, 1, 1)
+=1
0 x y x y
FIGURE 5
FIGURE 6
The projection of C onto the xy-plane is the circle x 2 from Example 2 in Section 11.1 that we can write x cos t y sin t 0 t
y2
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 sin t z 2 sin t 0 t 2
The corresponding vector equation is rt cos t i sin t j 2 sin t k 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 difcult 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
Its called a toroidal spiral because it lies on a torus. Another interesting curve, the tre-
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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 wouldnt be easy to plot either of these curves by hand.
z z
y x
x
y
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 difcult 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.
EXAMPLE 7 Use a computer to sketch the curve with vector equation r t
t, t 2, t 3 .
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 its hard to see the true nature of the curve from that graph alone. Most three-dimensional 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
_2
6 z0 _6 0 y y2 4 2 0x
6 z0 _2 _6 0 y2 4 2 0x
x
2 4
_2
(a)
_2 _1 0x 1 2 0 1 2 y 3 4 8 4 z0 _4 _8
(b)
8 4 z0 _4 _8
(c)
2
1
0 x
_1
_2
0
1
2 y
3
4
(d) FIGURE 9 Views of the twisted cubic
(e)
(f )
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CHAPTER 14 VECTOR FUNCTIONS
In Visual 14.1B you can rotate the box in Figure 9 to see the curve from any viewpoint.
z
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 xy-plane, namely, the parabola y x 2. Part (e) shows the projection on the x z-plane, the cubic curve z x 3. Its 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 rst 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.)
8 4
x
y
FIGURE 10
Visual 14.1C shows how curves arise as intersections of surfaces.
z
0 _4 _8 _1 0 2 4 y
FIGURE 11
x
1
0
|||| 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.
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 elds 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
B
E
E
t (a) r(t) = kt-sin t, 1-cos t, t l FIGURE 12
3
t (b) r(t) = kt- 2 sin t, 1- 2 cos t, t l FIGURE 13
3
Motion of a charged particle in orthogonally oriented electric and magnetic fields
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For further details concerning the physics involved and animations of the trajectories of the particles, see the following web sites:
lompado.uah.edu/Links/CrossedFields.html www.phy.ntnu.edu.tw/java/emField/emField.html www.physics.ucla.edu/plasma-exp/Beam/
|||| 14.1
12
||||
Exercises
22. x 23. x
Find the domain of the vector function. t 2, st t t
e
t
cos 10 t, y y
y sin t, sin t,
e z z
t
sin 10t, sin 5t ln t
II
z
e
t
1. r t 2. r t
1, s5 sin t j
t ln 9
cos t, cos t,
z
2 i 2
t2 k
24. x
I
z
36
||||
Find the limit. cos t, sin t, t ln t et t 1 s1 , 3i t t2 ln t t
3. lim
tl0
4. lim
tl0
t t
1 1 j 1
,
3 1 t tan t k t
x III z y
x IV z
y
5. lim st
tl1
6. lim
tl
arctan t , e 2t,
y x x V z VI z y
Sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increases.
||||
714
7. r t 9. r t 11. r t 12. r t 13. r t 14. r t
t4
1, t
8. r t 10. r t
t 3, t 2 1 t, 3t, t
t, cos 2 t, sin 2 t sin t, 3, cos t ti t2 i sin t i
tj t4 j
cos t k t6 k
x
y x y
sin t j
s2 cos t k
25. Show that the curve with parametric equations x
1518 |||| Find a vector equation and parametric equations for the line segment that joins P to Q.
y t sin t, z t lies on the cone z 2 fact to help sketch the curve. y z
x2
t cos t, y 2, and use this
26. Show that the curve with parametric equations x
15. P 0, 0, 0 , 17. P 1,
Q 1, 2, 3 Q 4, 1, 7
16. P 1, 0, 1 , 18. P
Q 2, 3, 1 1, 2
1, 2 ,
2, 4, 0 , Q 6,
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.
; 2730
1924
|||| Match the parametric equations with the graphs (labeled IVI). Give reasons for your choices.
|||| 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.
19. x 20. x 21. x
cos 4 t, t, t, y y
y t 2, 11
t, z
z e t2 ,
t
27. r t
sin t, cos t, t 2 t 2, st
28. r t
t4
t2
1, t, t 2
sin 4 t
29. r t 30. r t
1, s5
t
z
t2
sin t, sin 2 t, sin 3 t
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; 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.
trajectories of two particles are given by the vector functions r1 t for t t 2, 7t 12, t 2 r2 t 4t 3, t 2, 5t 6
0. Do the particles collide?
; 32. Graph the curve with parametric equations
x y z s1 s1 0.25 cos 2 10 t cos t 0.25 cos 2 10 t sin t
40. Two particles travel along the space curves
r1 t
t, t 2, t 3
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
0.5 cos 10 t
Explain the appearance of the graph by showing that it lies on a sphere.
33. Show that the curve with parametric equations x
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 tla 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).
||||
(b) lim c u t
tla
c lim u t
tla
(c) lim u t
tla
vt vt
lim u t
tla
lim v t
tla
Find a vector function that represents the curve of intersection of the two surfaces.
34. The cylinder x 35. The cone z
2
3436
(d) lim u t
tla
lim u t
tla
lim v t
tla
y
2
4 and the surface z y 2 and the plane z 1
xy y x
2
42. The view of the trefoil knot shown in Figure 8 is accurate, but
sx 2
it doesnt reveal the whole story. Use the parametric equations x 2 cos 1.5t cos t z y sin 1.5t 2 cos 1.5t sin t
36. The paraboloid z
4x
2
y and the parabolic cylinder y
2
; 37. Try to sketch by hand the curve of intersection of the circular
cylinder x y 4 and the parabolic cylinder z x . Then nd parametric equations for this curve and use these equations and a computer to graph the curve.
2 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 nd 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,
its 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
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 xy-plane 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 nished 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
|||| 14.2
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 dened in much the same way as for realvalued functions:
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1
dr dt
rt
hl0
lim
rt
h h
rt
if this limit exists. The geometric signicance of this denition 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 dened by r at the point P, provided that r t exists and r t 0. The tangent line to C at P is dened 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 Tt
Visual 14.2 shows an animation of Figure 1. z
rt rt
z
r (t+h)-r (t) h
r (t+h)-r (t) P r (t) r (t+h) Q
r (t) P r (t) C
Q r (t+h)
C
0 x y x
0 y
FIGURE 1
(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
Proof
fti f ti
tt j ttj
h t k, where f , t, and h htk
f t ,t t ,h t
rt
lim lim lim
tl0
1 rt t 1 t ft ft ft t t t t
t
rt t ,t t ft ft , tt tt t ,h t t t t t
tl0
tl0
t tt , ht tt
f t ,t t ,h t t t , lim ht t t ht
tl0
ht
tl0
lim
tl0
, lim
f t ,t t ,h t
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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
SOLUTION
sin 2 t k. 0.
(a) According to Theorem 2, we differentiate each component of r : rt (b) Since r 0 i and r 0 T0 3t 2 i j r0 r0 1 t e tj 2 cos 2 t k
2 k, the unit tangent vector at the point 1, 0, 0 is j 2k s1 4 2 1 j s5 2 k s5
y 2 (1, 1)
EXAMPLE 2 For the curve r t st i r 1 and the tangent vector r 1 .
SOLUTION We have r (1) r (1)
1 x
t j, nd r t and sketch the position vector
rt
1 i 2 st
j
and
r1
1 i 2
j
0
FIGURE 2
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 .
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, 2.
2 cos t
y
sin t
z
t
12 8 z 4 0 _1 _2 0x
SOLUTION The vector equation of the helix is r t
2 cos t, sin t, t , so
rt
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 t
_0.5
y0
0.5
1
2
FIGURE 3
2
Just as for real-valued 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.
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S ECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
895
y
S OLUTION Since
rt
cusp
0 1 x
3 t 2, 2 t
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 behavioran abrupt change in directionis not smooth. A curve, such as the semicubical parabola, that is made up of a nite number of smooth pieces is called piecewise smooth.
FIGURE 4
The curve r(t)=k1+t #, t @ l is not smooth.
Differentiation Rules
The next theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions.
3 Theorem Suppose u and v are differentiable vector functions, c is a scalar, and f is a real-valued 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 Denition 1 or by using Theorem 2 and the corresponding differentiation formulas for real-valued functions. The proofs of Formulas 1, 2, 3, 5, and 6 are left as exercises.
Proof of Formula 4 Let
ut Then ut vt
f1 t , f2 t , f3 t f1 t t1 t f2 t t2 t
vt
t1 t , t2 t , t3 t
3
f3 t t3 t
i1
fi t ti t
so the ordinary Product Rule gives d ut dt vt d dt
3 3 3
fi t ti t
i1 i1
d fi t ti t dt
f i t ti t
i1 3
fi t t i t
3
f i t ti t
i1 i1
fi t t i t ut vt
ut
vt
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CHAPTER 14 VECTOR FUNCTIONS
EXAMPLE 5 Show that if r t
SOLUTION Since
c (a constant), then r t is orthogonal to r t for all t . rt rt rt
2
c2
and c 2 is a constant, Formula 4 of Theorem 3 gives 0 d rt dt 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 denite integral of a continuous vector function r t can be dened in much the same way as for real-valued 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
n
a
r t dt
nl
lim
i1
r t* i
n
t
n n
nl
lim
i1
f t* i
ti
i1
t t* i
tj
i1
h t* i
tk
and so
y
b
a
r t dt
y
b
a
f t dt i
y
b
a
t t dt j
y
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
Ra
where R is an antiderivative of r, that is, R t indenite integrals (antiderivatives).
EXAMPLE 6 If r t
r t . We use the notation
xr t
dt for
2 cos t i dt
sin t j
2 t k, then i
yrt
y 2 cos t dt
2 sin t i
y sin t dt
t2 k C
j
y 2 t dt
k
cos t j
where C is a vector constant of integration, and
y
2
0
r t dt
[2 sin t i
cos t j
t2 k
]
2
2
0
2i
j
4
k
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S ECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
897
|||| 14.2
Exercises
12. r t 13. r t 14. r t 15. r t 16. r t
1. The gure shows a curve C given by a vector function r t .
sin 1t i ei
t
2
s1 j ln 1 t2 c tc
t2 j 3t k
3
k c cos 3t k
(a) Draw the vectors r 4.5 (b) Draw the vectors r 4.5 0.5 r4
r 4 and r 4.2 r 4.2 0.2
r4. r4
at cos 3 t i a ta
b sin t j
and
tb b
(c) Write expressions for r 4 and the unit tangent vector T(4). (d) Draw the vector T(4).
y
C
r(4.5) 1 r(4.2)
R
1720 |||| Find the unit tangent vector T t at the point with the given value of the parameter t.
17. r t Q 18. r t 19. r t P 20. r t
6 t 5, 4 t 3, 2 t , t 4 st i cos t i 2 sin t i
1 1 t 0 4
tj 3t j
2
t k, t
2 sin 2 t k,
2 cos t j
tan t k, t
r(4) 0 1 x
21. If r t 22. If r t
t, t 2, t 3 , nd r t , T 1 , r t , and r t e ,e
2t 2t
r t. r t.
, te
2t
, nd T 0 , r 0 , and r t
2. (a) Make a large sketch of the curve described by the vector
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 r 1.1 0.1 Explain why these vectors are so close to each other in length and direction.
38
||||
2326 |||| Find parametric equations for the tangent line to the curve with the given parametric equations at the specied point.
23. x 24. x 25. x 26. x
t 5, y t2 e
t
t 4, z 1, y t2 y 2 st,
t 3; (1, 1, 1) 1, z e
t
t z
1; e t; 0, 2, 1
1, 1, 1 1, 0, 1
r1
cos t, y
sin t, t 2;
ln t,
z
; 2728
(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 4. r t 5. r t 6. r t 7. r t 8. r t
|||| Find parametric equations for the tangent line to the curve with the given parametric equations at the specied point. Illustrate by graphing both the curve and the tangent line on a common screen.
27. x 28. x
t, y
s2 cos t, z 3e , z
s2 sin t ; 3e
4, 1, 1
cos t, y
2t
2t
;
1, 3, 3
cos t, sin t , 1 1 ei et i
t
t 1
4
29. Determine whether the curve is smooth.
t, st , t ti e
t
t 2 j, t j, t t 0 0
1
(a) r t (c) r t
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
e 3 t j,
2 sin t i
3 cos t j,
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.
31. The curves r1 t
916
||||
Find the derivative of the vector function. t 2, 1 i j t, st e 4t k
10. r t
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. 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.
2
9. r t 11. r t
cos 3t, t, sin 3t
32. At what point do the curves r1 t
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898
CHAPTER 14 VECTOR FUNCTIONS
3338
||||
Evaluate the integral. 9t 2 j j 25 t 4 k dt 2t 1 t2 k dt
2
43. Prove Formula 5 of Theorem 3. 44. Prove Formula 6 of Theorem 3. 45. If u t
33. 34. 35. 36. 37. 38.
y y y y y y
1
0 1
16t 3 i 4 1
2
i 2 t 2 j 3 t 3 k and v t nd d dt u t v t . d dt ut vt .
ti
cos t j
sin t k ,
0
t2
2
46. If u and v are the vector functions in Exercise 45, nd
0 4
3 sin t cos t i st i te t j 2t j ti
3 sin t cos t j 1 k dt t2
2 sin t cos t k dt
47. Show that if r is a vector function such that r exists, then
1
d rt dt
48. Find an expression for
rt d ut dt d rt dt rt
rt vt 1 rt
rt wt . rt r t.
et i cos
ln t k dt sin
tj
t k dt
49. If r t
0, show that rt
2
[Hint:
rt
]
39. Find r t if r t 40. Find r t if r t
t2 i sin t i 2 k.
4t 3 j
t 2 k and r 0 2 t k and
j.
50. If a curve has the property that the position vector r t is
cos t j
r0
i
j
always perpendicular to the tangent vector r t , show that the curve lies on a sphere with center the origin.
51. If u t
41. Prove Formula 1 of Theorem 3. 42. Prove Formula 3 of Theorem 3.
rt
rt ut
r t , show that rt rt rt
|||| 14.3
Arc Length and Curvature
In Section 11.2 we dened 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
L
y
b
a
sf t
2
tt
2
dt
y
b
a
dx dt
2
dy dt
2
dt
z
The length of a space curve is dened 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
2
0 y x
L
y
b
a b
sf t dx dt
2
tt
2
2
ht
2
2
dt
2
F IGURE 1
y
The length of a space curve is the limit of lengths of inscribed polygons.
a
dy dt
dz dt
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
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S ECTION 14.3 ARC LENGTH AND CURVATURE
899
because, for plane curves r t rt whereas, for space curves r t
|||| Figure 2 shows the arc of the helix whose length is computed in Example 1. z
fti f ti fti ttj
t t j, ttj tt j htk sf t h t k, sf t
2 2
tt
2
rt
f ti
tt
2
ht
2
EXAMPLE 1 Find the length of the arc of the circular helix with vector equation
rt
cos t i
sin t j
t k from the point 1, 0, 0 to the point 1, 0, 2 . sin t i rt cos t j s sin t k, we have
2
SOLUTION Since r t
(1, 0, 2)
cos 2 t
1
s2 t 2
(1, 0, 0) x y
The arc from 1, 0, 0 to 1, 0, 2 is described by the parameter interval 0 and so, from Formula 3, we have L
FIGURE 2
y
2
0
r t dt
y
2
0
s2 dt
2 s2
A single curve C can be represented by more than one vector function. For instance, the twisted cubic
4
r1 t
t, t 2, t 3
1
t
2
could also be represented by the function
5
r2 u
e u, e 2u, e 3u
0
u
ln 2
|||| Piecewise-smooth curves were introduced on page 895. z
s(t) C r(t) r(a)
0 x y
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 piecewise-smooth 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 dene its arc length function s by
6
st
y
t
a
ru
du
y
t
a
dx du
2
dy du
2
dz du
2
du
FIGURE 3
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.
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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 . 0. From
SOLUTION The initial point 1, 0, 0 corresponds to the parameter value t Example 1 we have
ds dt and so Therefore, t s st
rt r u du
s2
y
t
0
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 dened by the vector function r, then r t tangent vector T t is given by
z
0. Recall that the unit
Tt
rt rt
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.
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. Specically, we dene 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 rt d T ds ds dt and dT ds d T dt ds dt
from Equation 7, so Tt rt
9
t
EXAMPLE 3 Show that the curvature of a circle of radius a is 1 a.
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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 so and This gives T t rt a sin t i Tt Tt
a cos t i a cos t j rt rt
a sin t j and sin t i rt cos t j a
cos t i
sin t j
1, so using Equation 9, we have t Tt rt 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 denition 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
t
rt rt
rt
3
Proof Since T
r
r and r r
ds dt, we have rT ds T dt
so the Product Rule (Theorem 14.2.3, Formula 3) gives r Using the fact that T T d 2s T dt 2 ds T dt
0 (see Example 2 in Section 13.4), we have r r ds dt
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 r ds dt T
2
T
T r r ds dt 2 T r
ds dt
2
T r r
T r
2
ds dt
2
T
and
r r
r
3
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902
CHAPTER 14 VECTOR FUNCTIONS
E XAMPLE 4 Find the curvature of the twisted cubic r t
t, t 2, t 3 at a general point
and at 0, 0, 0 .
SOLUTION We rst compute the required ingredients:
rt rt
1, 2 t, 3 t 2 s1 i 1 0 4t 2 j k 2t 3t 2 2 6t 36 t 2 9t 4
rt
0, 2, 6 t
rt
rt
6t 2 i
6t j
2k
rt Theorem 10 then gives t
rt
s36 t 4
4
2 s9 t 4
9t 2
1
rt rt 0
rt
3
2 s1 9 t 2 9 t 4 1 4t 2 9t 4 3 2
At the origin the curvature is
2.
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 fx fx
11
x
1
2 32
y 2
y=
EXAMPLE 5 Find the curvature of the parabola y
x 2 at the points 0, 0 , 1, 1 ,
and 2, 4 .
SOLUTION Since y y=k(x)
2 x and y x
2, Formula 11 gives y y 2 4x 2
1
2 32
1
32
0
1
x
FIGURE 5
The parabola y= and its curvature function
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 atter 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
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S ECTION 14.3 ARC LENGTH AND CURVATURE
903
|||| We can think of the normal vector as indicating the direction in which the curve is turning at each point.
T t . Note that T t is itself not a unit vector. But if r is also smooth, we can dene the principal unit normal vector N t (or simply unit normal) as Nt Tt Tt
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
|||| 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. z
rt
cos t i
sin t j
tk
SOLUTION We rst compute the ingredients needed for the unit normal vector:
rt Tt
sin t i rt rt 1 s2 Tt Tt
cos t j 1 s2
k sin t i
rt cos t j k
s2
Tt
cos t i
sin t j
Tt
1 s2 sin t, 0
T B N B N T
Nt
cos t i
sin t j
cos t,
This shows that the normal vector at a point on the helix is horizontal and points toward the z-axis. The binormal vector is 1 s2 i sin t cos t j cos t sin t k 1 0 1 sin t, s2
Bt
y x
Tt
Nt
cos t, 1
FIGURE 7
Visual 14.3B shows how the TNB frame moves along several curves.
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, is 1x 0 0y
2. 2 0 or 1, 0, 1 , so an equation z x
SOLUTION The normal plane at P has normal vector r
1
1z
2
2
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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 Bt 1 sin t, s2 cos t, 1 B 1 1 , 0, s2 s2
z=_x+ 2
2
A simpler normal vector is 1, 0, 1 , so an equation of the osculating plane is
x
P
y
1x
0
0y
1
1z
2
0
or
z
x
2
FIGURE 8
EXAMPLE 8 Find and graph the osculating circle of the parabola y
x 2 at the origin.
y
osculating circle
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
For the graph in Figure 9 we use parametric equations of this circle:
1 2
x
1 x
1 2
cos t
y
1 2
1 2
sin t
0
We summarize here the formulas for unit tangent, unit normal and binormal vectors, and curvature. Tt rt rt dT ds Nt Tt rt Tt Tt rt rt Bt rt
3
FIGURE 9
Visual 14.3C shows how the osculating circle changes as a point moves along a curve.
Tt
Nt
|||| 14.3
16
||||
Exercises
911
Find the length of the curve. 2 sin t, 5t, 2 cos t , t , sin t s2 t i t2 i i
1. r t 2. r t 3. r t 4. r t 5. r t 6. r t
10
t
10 t
|||| Reparametrize the curve with respect to arc length measured from the point where t 0 in the direction of increasing t.
2
t cos t, cos t et j e t k, ln t k, 1 t k, 0
3
t sin t , 0 t t 1 t
9. r t 10. r t 11. r t
2t i
1
3t j 2j 4t j
5
4t k
0
1 e
e 2 t cos 2 t i 3 sin t i
e 2 t sin 2 t k 3 cos t k
2t j tj
2
t
12 t i
8t 3 2 j
3t 2 k, 0
1
12. Reparametrize the curve
7. Use Simpsons Rule with n
arc of the twisted cubic x the point 2, 4, 8 .
10 to estimate the length of the t, y t 2, z t 3 from the origin to
rt
2 t2 1
1i
2t t2 1
j
; 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.
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?
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S ECTION 14.3 ARC LENGTH AND CURVATURE
905
1316
||||
(a) Find the unit tangent and unit normal vectors T t and N t . (b) Use Formula 9 to nd the curvature.
13. r t 14. r t 15. r t
3233 |||| 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.
2 sin t, 5 t, 2 cos t t , sin t s2 t, e t, e
32.
33.
y y
2
t cos t, cos t
t
t sin t , t
16. r t
0
a
t 2, 2 t, ln t
a b
x
b
x
1719
||||
Use Theorem 10 to nd the curvature. t2 i ti 3t i
17. r t 18. r t 19. r t
tk tj
1
t2 k 4 cos t k
4 sin t j
C AS
34. (a) Graph the curve r t
20. Find the curvature of r t
e t cos t, e t sin t, t at the t, t 2, t 3 at the point (1, 1, 1).
C AS
point (1, 0, 0).
21. Find the curvature of r t
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 nd and graph the curvature function. Does this graph conrm your conclusion from part (a)? 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 nd and graph the curvature function. For which values of t is the curvature largest? metric curve x f t,y t t is x y yx x2 y2 3 2 where the dots indicate derivatives with respect to t.
35. The graph of r t
; 22. Graph the curve with parametric equations
x t y 4t 3 2 z t2 1. and nd the curvature at the point 1, 4,
2325
||||
36. Use Theorem 10 to show that the curvature of a plane para-
Use Formula 11 to nd the curvature.
24. y
23. y
x3
cos x
25. y
4x 5 2
2627 |||| At what point does the curve have maximum curvature? What happens to the curvature as x l ?
3738
||||
Use the formula in Exercise 36 to nd the curvature. e t cos t, 1 t 3,
26. y
ln x
27. y
ex
37. x 38. x
y y
e t sin t t
28. Find an equation of a parabola that has curvature 4 at the
t2
origin.
29. (a) Is the curvature of the curve C shown in the gure greater
3940
||||
Find the vectors T, N, and B at the given point. t 2, 2 t 3, t , 3
at P or at Q ? Explain. (b) Estimate the curvature at P and at Q by sketching the osculating circles at those points.
y
39. r t 40. r t
(1, 2 , 1) 3
1, 0, 1
e t, e t sin t, e t cos t ,
P C
4142 |||| Find equations of the normal plane and osculating plane of the curve at the given point.
1
41. x Q 42. x
2 sin 3 t, y t, y
t, z t;
2 cos 3 t ; 1, 1, 1
0, ,
2
t,z
2
3
0
1
x
; 43. Find equations of the osculating circles of the ellipse
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.
; 3031
|||| 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?
30. y
xe
x
31. y
x
4
; 44. Find equations of the osculating circles of the parabola y 1
x2 at the points 0, 0 and (1, 2 ). Graph both osculating circles and the parabola on the same screen.
1 2
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CHAPTER 14 VECTOR FUNCTIONS
45. At what point on the curve x
plane parallel to the plane 6 x
CAS
t 3, y 3 t, z t 4 is the normal 6 y 8 z 1?
(b) r (c) r (d)
r s r r r
s
2
3
B
3
s
T
3 ss
s
2
N
s
3
B
46. Is there a point on the curve in Exercise 45 where the oscu-
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.)
47. Show that the curvature
r r
2
52. Show that the circular helix r t
is related to the tangent and normal dT ds
vectors by the equation N
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).] t, 1 t 2, 1 t 3 . 2 3 sinh t,
53. Use the formula in Exercise 51(d) to nd the torsion of the
curve r t
48. Show that the curvature of a plane curve is
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 denition of curvature is consistent with the denition for plane curves given in Exercise 69 in Section 11.2.) (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.
54. Find the curvature and torsion of the curve x
y
cosh t, z
t at the point 0, 1, 0 .
55. The DNA molecule has the shape of a double helix (see
49. (a) Show that d B ds is perpendicular to B.
Figure 3 on page 887). The radius of each helix is about 10 angstroms (1 10 8 cm). Each helix rises about 34 during each complete turn, and there are about 2.9 10 8 complete turns. Estimate the length of each helix.
56. Lets consider the problem of designing a railroad track to
50. The following formulas, called the Frenet-Serret 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.
51. Use the Frenet-Serret formulas to prove each of the following.
make a smooth transition between sections of straight track. Existing track along the negative x-axis 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 dened by 0 Px 1 if x if 0 if x 0 x 1
Fx
1
(Primes denote derivatives with respect to t. Start as in the proof of Theorem 10.) (a) r sT s 2N
;
is continuous and has continuous slope and continuous curvature. (b) Use a graphing calculator or computer to draw the graph of F .
|||| 14.4
z
Motion in Space: Velocity and Acceleration
r (t+h)-r (t) h r (t) Q 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 Keplers 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
1
C
O x y
rt
h h
rt
FIGURE 1
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
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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 :
2
vt
hl0
lim
rt
h h
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 one-dimensional motion, the acceleration of the particle is dened as the derivative of the velocity: at vt rt
EXAMPLE 1 The position vector of an object moving in a plane is given by
y
rt t 3 i t 2 j. Find its velocity, speed, and acceleration when t geometrically.
SOLUTION The velocity and acceleration at time t are v (1)
1 and illustrate
vt
(1, 1) 0
rt rt
3t 2 i 6t i
2t j 2j
a (1)
at
x
and the speed is
FIGURE 2
Visual 14.4 shows animated velocity and acceleration vectors for objects moving along various curves.
vt When t 1, we have v1 3i 2j
s 3t 2
2
2t
2
s9 t 4
4t 2
a1
6i
2j
v1
s13
|||| Figure 3 shows the path of the particle in Example 2 with the velocity and acceleration vectors when t 1. z
These velocity and acceleration vectors are shown in Figure 2.
EXAMPLE 2 Find the velocity, acceleration, and speed of a particle with position vector
a (1) v (1)
rt
SOLUTION
t 2, e t, te t . vt at rt vt s4 t 2 2 t, e t, 1 2, e t, 2 e 2t 1 t et t et t 2e 2t
1 y x
vt
FIGURE 3
The vector integrals that were introduced in Section 14.2 can be used to nd position vectors when velocity or acceleration vectors are known, as in the following example.
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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 vt
yat
2t 2 i
dt
y
4t i tk
6t j C
k dt
3t 2 j
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 vt 2t 2 i 2t 2
|||| 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.
i
j
k.
3t 2 j 1i
tk 3t 2
i 1j
j t
k 1k
Since v t
r t , we have rt
yvt y
dt 1i t) i r0 t t3 i, so 1) i t3 tj 3t 2 tj 1j t 1 k dt t) k D
6 z4 2 0 0 5 y (1, 0, 0) 10 15 20 20 0 x
2t 2
(2t3 3
Putting t 0, we nd that D rt
(1t2 2
F IGURE 4
(2t3 3
(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 Newtons 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. y
Ft
ma t
P
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.
SOLUTION
x
0
vt at
rt vt
a sin t i a
2
a cos t j a
2
cos t i
sin t j
Therefore, Newtons Second Law gives the force as
FIGURE 5
Ft
ma t
m
2
a cos t i
a sin t j
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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 (center-seeking) force.
y
v
EXAMPLE 5 A projectile is red 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, nd the position function r t of the projectile. What value of maximizes the range (the horizontal distance traveled)?
x
a
0
d FIGURE 6
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 where t a 9.8 m s2 . Thus
ma
mt j
a Since v t a, we have vt where C v0 v0 . Therefore rt Integrating again, we obtain rt But D
3
1 2
tj
tt j
C
vt
tt j
v0
tt 2 j
t v0
D
r0
0, so the position vector of the projectile is given by rt
1 2
tt 2 j
t v0
If we write v0
v0 (the initial speed of the projectile), then
v0 and Equation 3 becomes rt
v0 cos
v0 cos
i
v0 sin
j
ti
[ v0 sin
t
1 2
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.
4
x
v0 cos
t
y
v0 sin
t
1 2
tt 2 0, we obtain t 0
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
v 2 2 sin 0
cos t
v 2 sin 2 0
t 4.
Clearly, d has its maximum value when sin 2
1, that is,
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CHAPTER 14 VECTOR FUNCTIONS
E XAMPLE 6 A projectile is red 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 y
150 cos 10
4t
75 s2 t 4t
1 2
150 sin
9.8 t 2
10
75 s2 t
4.9 t 2
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 9.8 196
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 So its speed at impact is v 21.74 rt 75 s2 i
(75 s2
9.8 t) j
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 Tt and so rt rt v
vT
vt vt
v
v
If we differentiate both sides of this equation with respect to t , we get
5
a
v
vT
vT
If we use the expression for the curvature given by Equation 14.3.9, then we have
6
T r
T
v
so
T
v
The unit normal vector was dened in the preceding section as N T and Equation 5 becomes
7
T
T , so (6) gives
TN
vN
a
vT
v2 N
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SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
911
aT
T a N
Writing a T and a N for the tangential and normal components of acceleration, we have a where
8
aTT
aN N
aT
v
and
aN
v2
aN
FIGURE 7
This resolution is illustrated in Figure 7. Lets look at what Formula 7 says. The rst 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 cara 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, its 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 vv vT v2 N v3 T
T
N
(since T T 1 and T N 0)
Therefore
9
aT
v
va
v
rt
rt rt
Using the formula for curvature given by Theorem 14.3.10, we have
10
aN
v2
rt rt
rt
3
rt
2
rt rt
rt
EXAMPLE 7 A particle moves with position function r t
t 2, t 2, t 3 . Find the tangential
and normal components of acceleration.
SOLUTION
rt rt rt rt
t2 i 2t i 2i s8t 2
t2 j 2t j 2j 9t 4
t3 k 3t 2 k 6t k
Therefore, Equation 9 gives the tangential component as aT rt rt rt 8 t 18 t 3 s8 t 2 9 t 4
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CHAPTER 14 VECTOR FUNCTIONS
Since
rt
rt
ijk 2 t 2 t 3t 2 2 2 6t
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 Keplers 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 (15711630) formulated the following three laws.
Keplers 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 Keplers 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 rst show that the planet moves in one plane. By equating the expressions for F in Newtons two laws, we nd that a GM r r3
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SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
913
and so a is parallel to r. It follows that r to write d r dt v r v Therefore r v v
a
0. We use Formula 5 in Theorem 14.2.3
r r v h
v a 0 0 0
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 Keplers First Law we rewrite the vector h as follows: h r ru ru Then a h GM u r2 r2u u u uu GM u u u
2
v
r ru u
r ru
ru r2 u
ru u rr u u
GM u u u But u u that u u u 2 1 and, since u t 0. Therefore a and so
z
(by Theorem 13.4.8, Property 6)
1, it follows from Example 5 in Section 14.2
h h
GM u v h a h GM u
v
Integrating both sides of this equation, we get
11
h c r
x y
v
h
GM u
c
v u
FIGURE 8
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 xy-plane. Since both v h and u are perpendicular to h, Equation 11 shows that c lies in the xy-plane. This means that we can choose the x- and y-axes 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 c r GM r u c cos rc GMr rc cos
GMr u u
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CHAPTER 14 VECTOR FUNCTIONS
where c
c . Then r rvh GM c cos 1r GM 1 vh e cos
where e
c GM . But r v h r v h hh h
2
h2
where h
h . So r h 2 GM 1 e cos eh 2 c 1 e cos
Writing d
12
h 2 c, we obtain the equation r ed e cos
1
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 Keplers 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
(d) Draw an approximation to the vector v(2) and estimate the speed of the particle at t 2.
y
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.
t 0 0.5 1.0 1.5 2.0 x 2.7 3.5 4.5 5.9 7.3 y 9.8 7.2 6.0 6.4 7.8 z 3.7 3.3 3.0 2.8 2.7
r(2.4)
2 1 0
r(2) r(1.5)
1 2 x
2. The gure 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).
38 |||| 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 specied value of t.
3. r t 4. r t 5. r t
t2 2 ei
t
1, t ,
t
1 1 0
t, 4 st , t e j,
t
t
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S ECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
915
6. r t 7. r t
sin t i sin t i ti
2 cos t j, tj t2 j
t t 1
6 0
27. A gun has muzzle speed 150 m s. Find two angles of elevation
cos t k, t 3 k, t
that can be used to hit a target 800 m away.
28. A batter hits a baseball 3 ft above the ground toward the center
; 8. r t
914 |||| Find the velocity, acceleration, and speed of a particle with the given position function.
eld 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. r t 10. r t 11. r t 12. r t 13. r t 14. r t
t2
1, t 3, t 2 et j ln t j
1 e tk tk tk t2 k
; 29. Water traveling along a straight portion of a river normally
ows fastest in the middle, and the speed slows to almost zero at the banks. Consider a long stretch of river owing 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 ow 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, nd the angle at which the boat should head. Then graph the actual path the boat follows. Does the path seem realistic?
30. Another reasonable model for the water speed of the river in
2 cos t, 3 t, 2 sin t s2 t i ti
t 2
e cos t i t sin t i
sin t j t cos t j
1516 |||| Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.
15. a t 16. a t
k,
v0 10 k,
i v0
j, i
r0 j
0 k, r 0
2i
3j
1718
||||
(a) Find the position vector of a particle that has the given acceleration and the specied initial velocity and position. ; (b) Use a computer to graph the path of the particle.
17. a t 18. a t
i ti
2j tj
2 t k,
v0
0, v0
r0 i
i k,
k r0
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.
3136 |||| Find the tangential and normal components of the acceleration vector.
2
cos 2 t k,
j
19. The position function of a particle is given by
31. r t 32. r t 33. r t 34. r t 35. r t 36. r t
3t 1
t3 i ti
3t 2 j t2 2t j tk
rt
t 2, 5 t, t 2
16 t . When is the speed a minimum? t3 i t2 j t 3 k?
20. What force is required so that a particle of mass m has the
cos t i ti ei ti
sin t j
2
position function r t
tj s2 t j cos 2t j
3t k e tk sin 2t k
21. A force with magnitude 20 N acts directly upward from the
t
xy-plane 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.
23. A projectile is red with an initial speed of 500 m s and angle
37. The magnitude of the acceleration vector a is 10 cm s2. Use the
gure to estimate the tangential and normal components of a.
y
of elevation 30 . Find (a) the range of the projectile, (b) the maximum height reached, and (c) the speed at impact.
24. Rework Exercise 23 if the projectile is red 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 red with angle of elevation 30 . What is the
0 x
muzzle speed if the maximum height of the shell is 500 m?
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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 dened 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.)
39. The position function of a spaceship is
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 Newtons Second Law of Motion that m dv dt ln dm ve dt
rt
3
ti
2
ln t j
7
4 t2 1
k
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?
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? (a) Show that v t v0
APPLIED PROJECT
Keplers 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.
Keplers 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 tted the astronomical data. He wasnt 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 Keplers three laws from two of Newtons own laws, the Second Law of Motion and the Law of Universal Gravitation. In Section 14.4 we proved Keplers First Law using the calculus of vector functions. In this project we guide you through the proofs of Keplers Second and Third Laws and explore some of their consequences.
1. Use the following steps to prove Keplers 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 (b) Deduce that r 2
y
r2 d dt
d k. dt h. r t in the time interval t0 , t
(c) If A A t is the area swept out by the radius vector r as in the gure, show that
r (t)
r (t) A(t)
dA dt (d) Deduce that
x
1 2
r2
d dt
0
dA dt
1 2
h
constant
This says that the rate at which A is swept out is constant and proves Keplers Second Law.
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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 (b) Show that h GM
2
2 a b h.
ed
b . a 423 a. GM
2
2
(c) Use parts (a) and (b) to show that T 2
This proves Keplers Third Law. [Notice that the proportionality constant 4 independent of the planet.]
GM is
3. The period of the Earths orbit is approximately 365.25 days. Use this fact and Keplers
Third Law to nd the length of the major axis of the Earths 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. Its possible to place a satellite into orbit about the Earth so that it remains xed above a
given location on the equator. Compute the altitude that is needed for such a satellite. The Earths 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 rst proposed the idea in 1948. The rst such satellite, Syncom II, was launched in July 1963.)
||||
14 Review
CONCEPT CHECK
1. What is a vector function? How do you nd its derivative and
6. (a) What is the denition of curvature?
its integral?
2. What is the connection between vector functions and space
curves?
3. (a) What is a smooth curve?
(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 .
7. (a) Write formulas for the unit normal and binormal vectors of
(b) How do you nd the tangent vector to a smooth curve at a point? How do you nd the tangent line? The unit tangent vector?
4. If u and v are differentiable vector functions, c is a scalar, and
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 nd the velocity, speed, and acceleration of a
f is a real-valued 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
5. How do you nd the length of a space curve given by a
particle that moves along a space curve? (b) Write the acceleration in terms of its tangential and normal components.
9. State Keplers Laws.
vector function r t ?
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CHAPTER 14 VECTOR FUNCTIONS
TRUE-FALSE QUIZ
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. 1. The curve with vector equation r t
6. If r t is a differentiable vector function, then
t3 i
2t 3 j
3t 3 k curvature is
d rt dt d T dt .
rt
is a line.
2. The curve with vector equation r t 3. The curve with vector equation r t
t, t 3, t 5 is smooth. cos t, t 2, t 4 is smooth.
7. If T t is the unit tangent vector of a smooth curve, then the 8. The binormal vector is B t
Nt
Tt.
4. The derivative of a vector function is obtained by differen-
tiating each component function.
5. If u t and v t are differentiable vector functions, then
9. The osculating circle of a curve C at a point has the same
tangent vector, normal vector, and curvature as C at that point.
10. Different parametrizations of the same curve result in identical
d ut dt
vt
ut
vt
tangent vectors at a given point on the curve.
EXERCISES
1. (a) Sketch the curve with vector function
11. For the curve given by r t
rt
ti
cos t j
sin t k
t
0
t , t , t , nd (a) the unit tangent vector, (b) the unit normal vector, and (c) the curvature. 3 cos t, y 4 sin t at the
1312 3 2
(b) Find r t and r t .
2. Let r t
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 y2
t
1.
points 3, 0 and 0, 4 .
13. Find the curvature of the curve y
x 4 at the point 1, 1 .
; 14. Find an equation of the osculating circle of the curve
z 5. 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
3. Find a vector function that represents the curve of intersection
16 and the plane x
; 4. Find parametric equations for the tangent line to the curve
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.
5. If r t
x
sin 2 t, y
t, z
cos 2 t at the point 0, , 1 .
t2 i
t cos t j
sin t k, evaluate x01 r t dt.
16. The gure shows the curve C traced by a particle with position
6. Let C be the curve with equations x
2 t 3, y 2 t 1, z ln t. Find (a) the point where C intersects the xz-plane, (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 . 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 . 2 t 3 2, cos 2 t, sin 2 t , 0 t 1.
7. Use Simpsons Rule with n
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
8. Find the length of the curve r t 9. The helix r1 t
C
1
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 .
0
r(3) r(3.2)
10. Reparametrize the curve r t
1
x
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C HAPTER 14 REVIEW
919
17. A particle moves with position function
rt t ln t i t j e t k. Find the velocity, speed, and acceleration of the particle.
18. A particle starts at the origin with initial velocity i
(c) Determine the Coriolis acceleration of a particle that moves on a rotating disk according to the equation rt e t cos t i e t sin t j
Its acceleration is a t function.
6t i
12 t 2 j
j 3 k. 6 t k. Find its position
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
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?
20. Find the tangential and normal components of the acceleration
vector of a particle with position function rt ti 2t j t2 k
road tracks, its 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 gure. It looks reasonable at rst glance. Show that the function Fx 1 s1 s2 x2 x if x if 0 if x 0 x 1 s2 1 s2
21. A disk of radius 1 is rotating in the counterclockwise direction
at a constant angular speed . A particle starts at the center of the disk and moves toward the edge along a xed radius so that its position at time t, t 0, is given by r t t R t , where Rt 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
is continuous and has continuous slope, but does not have continuous curvature. Therefore, f is not an appropriate transfer curve. (b) Find a fth-degree 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 fourth-degree 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 gure.
y 1 y
y=F(x)
y=x
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 merry-go-round.
y=0
0
1 2
transfer curve
1 x
x
0
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PROBLEMS PLUS
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 T 2R v 2
y
v r
vt
x
(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
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 lm 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 gure.) 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 Newtons Second Law and part (d) of Problem 1, F sin
2 mvR R
mg
FIGURE FOR PROBLEM 2
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 red 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 red 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 gure. Show that the projectile can hit any target inside or on the boundary
y y
_R
0
Rx
0
D
x
920
5E-14(pp 920-921)
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of the region bounded by the parabola and the x-axis, and that it cant 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 red. 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 red 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 red up an inclined plane whose angle of inclination is . Show that, in order to maximize the (uphill) range, the projectile should be red 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 ring a projectile to hit a target a distance R up an inclined plane. Show that the angle at which the projectile should be red 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 re the projectile is proportional to the square of the initial speed, so minimizing the energy is equivalent to minimizing the initial speed.) ity, assume that air resistance provides a force that is proportional to the velocity and that opposes the motion. Then, by Newtons Second Law, the total force acting on the projectile satises the equation
1
5. A projectile of mass m is red from the origin at an angle of elevation . In addition to grav-
m
d 2R dt 2
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 dR dt where v0 v0 k R m 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 left-hand side of the resulting equation is the derivative of the product e k m t R t . Then integrate to nd an expression for the position vector R t .
6. Find the curvature of the curve with parametric equations
x
y
t
0
sin ( 1 2
2
)d
y
y
t
0
cos ( 1 2
2
)d
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 oor and nd 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 gure.) (c) Suppose the ball rebounds from the oor at the same angle with which it hits the oor, but loses 20% of its speed due to energy absorbed by the ball on impact. Where does the ball strike the oor 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?
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