14
VECTOR FUNCTIONS
ET 13
14.1
Vector Functions and Space Curves
ET 13.1
1. The component functions4 w2,h3w, andln(w + 1) are all dened when 4 w2 0 2 w 2 and w + 1 A 0 w A 1, so the domain of r is (1> 2].
3. lim
w0+cos w = cos 0 = 1, limw0+sin w = sin 0 = 0, limw0+w ln w = limw0+
ln w 1@w= limw0+
1@w
1@w2 = limw0+w = 0
[by l’Hospital’s Rule]. Thus lim
w0+hcos w> sin w> w ln wi =
lim
w0+cos w> limw0+sin w> limw0+w ln w = h1> 0> 0i.
5. lim
w0h
3w= h0= 1, lim w0
w2
sin2w = limw0sin12w
w2 = 1 lim w0 sin2w w2 = 1 lim w0 sin w w 2 = 112 = 1 andlim
w0cos 2w = cos 0 = 1. Thus the given limit equals i + j + k.
7. The corresponding parametric equations for this curve are{ = sin w, | = w. We can make a table of values, or we can eliminate the parameter: w = | { = sin |, with | R. By comparing different values of w, we nd the direction in whichw increases as indicated in the graph.
9. The corresponding parametric equations are{ = w, | = cos 2w, } = sin 2w. Note that|2+ }2= cos22w + sin22w = 1, so the curve lies on the circular cylinder|2+ }2= 1. Since { = w, the curve is a helix.
11. The corresponding parametric equations are{ = 1, | = cos w, } = 2 sin w. Eliminating the parameter in| and } gives |2+ (}@2)2= cos2w + sin2w = 1 or|2+ }2@4 = 1. Since { = 1, the curve is an ellipse centered at (1> 0> 0) in the plane{ = 1.
139
13. The parametric equations are{ = w2,| = w4,} = w6. These are positive forw 6= 0 and 0 when w = 0. So the curve lies entirely in the rst quadrant. The projection of the graph onto the{|-plane is | = {2,| A 0, a half parabola. On the{}-plane } = {3,} A 0, a half cubic, and the |}-plane, |3= }2.
15. Takingr0= h0> 0> 0i and r1= h1> 2> 3i, we have from Equation 13.5.4 [ET 12.5.4]
r(w) = (1 w) r0+ w r1= (1 w) h0> 0> 0i + w h1> 2> 3i, 0 w 1 or r(w) = hw> 2w> 3wi, 0 w 1.
Parametric equations are{ = w, | = 2w, } = 3w, 0 w 1. 17. Takingr0= h1> 1> 2i and r1= h4> 1> 7i, we have
r(w) = (1 w) r0+ w r1= (1 w) h1> 1> 2i + w h4> 1> 7i, 0 w 1 or r(w) = h1 + 3w> 1 + 2w> 2 + 5wi, 0 w 1.
Parametric equations are{ = 1 + 3w, | = 1 + 2w, } = 2 + 5w, 0 w 1.
19. { = cos 4w, | = w, } = sin 4w. At any point ({> |> }) on the curve, {2+ }2= cos24w + sin24w = 1. So the curve lies on a circular cylinder with axis the|-axis. Since | = w, this is a helix. So the graph is VI.
21. { = w, | = 1@(1 + w2), } = w2. Note that| and } are positive for all w. The curve passes through (0> 1> 0) when w = 0. Asw , ({> |> }) ( > 0> ), and as w , ({> |> }) ( > 0> ). So the graph is IV.
23. { = cos w, | = sin w, } = sin 5w. {2+ |2 = cos2w + sin2w = 1, so the curve lies on a circular cylinder with axis the }-axis. Each of {, | and } is periodic, and at w = 0 and w = 2 the curve passes through the same point, so the curve repeats itself and the graph is V.
25. If{ = w cos w, | = w sin w, } = w, then {2+ |2= w2cos2w + w2sin2w = w2= }2, so the curve lies on the cone}2= {2+ |2. Since} = w, the curve is a spiral on this cone.
27. Parametric equations for the curve are{ = w, | = 0, } = 2w w2. Substituting into the equation of the paraboloid gives 2w w2= w2 2w = 2w2 w = 0, 1. Since r(0) = 0 and r(1) = i + k, the points of intersection
are(0> 0> 0) and (1> 0> 1).
29. r(w) = hcos w sin 2w> sin w sin 2w> cos 2wi. We include both a regular plot and a plot
showing a tube of radius 0.08 around the curve.
SECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES ET SECTION 13.1 ¤ 141
31.r(w) = hw> w sin w> w cos wi
33. { = (1 + cos 16w) cos w, | = (1 + cos 16w) sin w, } = 1 + cos 16w. At any point on the graph,
{2+ |2= (1 + cos 16w)2cos2w + (1 + cos 16w)2sin2w
= (1 + cos 16w)2= }2, so the graph lies on the cone{2+ |2= }2.
From the graph at left, we see that this curve looks like the projection of a leaved two-dimensional curve onto a cone.
35.Ifw = 1, then { = 1, | = 4, } = 0, so the curve passes through the point (1> 4> 0). If w = 3, then { = 9, | = 8, } = 28, so the curve passes through the point(9> 8> 28). For the point (4> 7> 6) to be on the curve, we require | = 1 3w = 7 w = 2= But then } = 1 + (2)3= 7 6= 6, so (4> 7> 6) is not on the curve.
37.Both equations are solved for}, so we can substitute to eliminate }: s{2+ |2= 1 + | {2+ |2= 1 + 2| + |2 {2 = 1 + 2| | =1
2({2 1). We can form parametric equations for the curve F of intersection by choosing a
parameter{ = w, then | =12(w2 1) and } = 1 + | = 1 +21(w2 1) =12(w2+ 1). Thus a vector function representing F isr(w) = w i +12(w2 1) j +12(w2+ 1) k.
39. The projection of the curveF of intersection onto the
{|-plane is the circle {2+ |2= 4> } = 0. Then we can write
{ = 2 cos w, | = 2 sin w, 0 w 2 . Since F also lies on the surface} = {2, we have} = {2= (2 cos w)2= 4 cos2w. Then parametric equations forF are { = 2 cos w, | = 2 sin w, } = 4 cos2w, 0 w 2 .
41.For the particles to collide, we requirer1(w) = r2(w) w2> 7w 12> w2= 4w 3> w2> 5w 6. Equating components givesw2= 4w 3, 7w 12 = w2, andw2= 5w 6. From the rst equation, w2 4w + 3 = 0 (w 3)(w 1) = 0 so w = 1 orw = 3. w = 1 does not satisfy the other two equations, but w = 3 does. The particles collide when w = 3, at the
point(9> 9> 9). 43. (a) lim wdu(w) + limwdv(w) = G lim wdx1(w)> limwdx2(w)> limwdx3(w) H
+Gwdlimy1(w)> limwdy2(w)> limwdy3(w)
H
and the limits of these
component functions must each exist since the vector functions both possess limits asw d. Then adding the two vectors
TX.10
and using the addition property of limits for real-valued functions, we have that
lim
wdu(w) + limwdv(w) =
G lim
wdx1(w) + limwdy1(w)> limwdx2(w) + limwdy2(w)> limwdx3(w) + limwdy3(w)
H =Glim
wd[x1(w) + y1(w)] > limwd[x2(w) + y2(w)] > limwd[x3(w) + y3(w)]
H
= limwdhx1(w) + y1(w)> x2(w) + y2(w)> x3(w) + y3(w)i [using (1) backward]
= lim
wd[u(w) + v(w)]
(b)lim
wdfu(w) = limwdhfx1(w)> fx2(w)> fx3(w)i =
G lim
wdfx1(w)> limwdfx2(w)> limwdfx3(w)
H =Gf limwdx1(w)> f limwdx2(w)> f limwdx3(w)
H
= fGwdlimx1(w)> limwdx2(w)> limwdx3(w)
H = f lim
wdhx1(w)> x2(w)> x3(w)i = f limwdu(w)
(c)lim wdu(w) · limwdv(w) = G lim wdx1(w)> limwdx2(w)> limwdx3(w) H
·Gwdlimy1(w)> limwdy2(w)> limwdy3(w)
H =kwdlimx1(w) l k lim wdy1(w) l +kwdlimx2(w) l k lim wdy2(w) l +kwdlimx3(w) l k lim wdy3(w) l = lim
wdx1(w)y1(w) + limwdx2(w)y2(w) + limwdx3(w)y3(w)
= limwd[x1(w)y1(w) + x2(w)y2(w) + x3(w)y3(w)] = limwd[u(w) · v(w)]
(d)lim wdu(w) × limwdv(w) = G lim wdx1(w)> limwdx2(w)> limwdx3(w) H ×Gwdlimy1(w)> lim wdy2(w)> limwdy3(w) H =Gkwdlimx2(w) l k lim wdy3(w) l kwdlimx3(w) l k lim wdy2(w) l > k lim wdx3(w) l k lim wdy1(w) l klim wdx1(w) l k lim wdy3(w) l > k lim wdx1(w) l k lim wdy2(w) l kwdlimx2(w) l k lim wdy1(w) lH =Gwdlim[x2(w)y3(w) x3(w)y2(w)] > limwd[x3(w)y1(w) x1(w)y3(w)] >
lim
wd[x1(w)y2(w) x2(w)y1(w)]
H = lim
wdhx2(w)y3(w) x3(w)y2(w)> x3(w) y1(w) x1(w)y3(w)> x1(w)y2(w) x2(w)y1(w)i
= lim
wd[u(w) × v(w)]
45. Letr(w) = hi (w) > j (w) > k (w)i and b = he1> e2> e3i. If lim
wdr(w) = b, then limwdr(w) exists, so by (1),
b = limwdr(w) =Gwdlimi(w)> limwdj(w)> limwdk(w)H. By the denition of equal vectors we havelim
wdi(w) = e1,wdlimj(w) = e2
andlim
wdk(w) = e3. But these are limits of real-valued functions, so by the denition of limits, for every% A 0 there exists
1A 0, 2A 0, 3 A 0 so that if 0 ? |w d| ? 1then|i(w) e1| ? %@3, if 0 ? |w d| ? 2then|j(w) e2| ? %@3, and
if0 ? |w d| ? 3then|k(w) e3| ? %@3. Letting = minimum of {1> 2> 3}, then if 0 ? |w d| ? we have
|i(w) e1| + |j(w) e2| + |k(w) e3| ? %@3 + %@3 + %@3 = %. But
|r(w) b| = |hi(w) e1> j(w) e2> k(w) e3i| =s(i(w) e1)2+ (j(w) e2)2+ (k(w) e3)2
SECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ET SECTION 13.2 ¤ 143
Thus for every% A 0 there exists A 0 such that if 0 ? |w d| ? then
|r(w) b| |i(w) e1| + |j(w) e2| + |k(w) e3| ? %. Conversely, suppose for every % A 0, there exists A 0 such
that if0 ? |w d| ? then |r(w) b| ? % |hi(w) e1> j(w) e2> k(w) e3i| ? %
s
[i(w) e1]2+ [j(w) e2]2+ [k(w) e3]2? % [i(w) e1]2+ [j(w) e2]2+ [k(w) e3]2? %2. But each term on the left side of the last inequality is positive, so if0 ? |w d| ? , then [i(w) e1]2? %2,[j(w) e2]2? %2and [k(w) e3]2? %2or, taking the square root of both sides in each of the above,|i(w) e1| ? %, |j(w) e2| ? % and
|k(w) e3| ? %. And by denition of limits of real-valued functions we have lim
wdi(w) = e1,wdlimj(w) = e2and
lim
wdk(w) = e3. But by (1),wdlim r(w) =
G lim
wdi(w)> limwdj(w)> limwdk(w)
H , solim
wdr(w) = he1> e2> e3i = b.
14.2
Derivatives and Integrals of Vector Functions
ET 13.2
1. (a)
(b) r(4=5) r(4)
0=5 = 2[r(4=5) r(4)], so we draw a vector in the same direction but with twice the length of the vectorr(4=5) r(4). r(4=2) r(4)
0=2 = 5[r(4=2) r(4)], so we draw a vector in the same direction but with5 times the length of the vector r(4=2) r(4). (c) By Denition 1,r0(4) = lim k0 r(4 + k) r(4) k . T(4) = r 0(4) |r0(4)|.
(d)T(4) is a unit vector in the same direction as r0(4), that is, parallel to the tangent line to the curve at r(4) with length 1.
TX 10
RIVATIVE RIVATIVE
3. Since({ + 2)2= w2= | 1 | = ({ + 2)2 1, the curve is a parabola. (a), (c) (b)r0(w) = h1> 2wi, r0(1) = h1> 2i 5. { = sin w, | = 2 cos w so {2+ (|@2)2= 1 and the curve is
an ellipse.
(a), (c) (b)r0(w) = cos w i 2 sin w j,
r0 4 = 2 2 i 2 j 7. Since| = h3w= (hw)3= {3, the curve is part of a cubic cuve. Note that here,{ A 0. (a), (c) (b)r0(w) = hwi + 3h3wj, r0(0) = i + 3 j 9. r0(w) = g gw[w sin w] > ggw w2> g
gw[w cos 2w] = hw cos w + sin w> 2w> w( sin 2w) · 2 + cos 2wi = hw cos w + sin w> 2w> cos 2w 2w sin 2wi
11. r(w) = i j + h4wk r0(w) = 0 i + 0 j + 4h4wk = 4h4wk 13. r(w) = hw2i j + ln(1 + 3w) k r0(w) = 2whw2i + 3
1 + 3wk 15. r0(w) = 0 + b + 2w c = b + 2w c by Formulas 1 and 3 of Theorem 3.
17. r0(w) = whw+ hw> 2@(1 + w2)> 2hw r0(0) = h1> 2> 2i. So |r0(0)| =12+ 22+ 22=9 = 3 and T(0) = r|r00(0)(0)|= 1 3h1> 2> 2i = 1 3>23>23 .
19. r0(w) = sin w i + 3 j + 4 cos 2w k r0(0) = 3 j + 4 k. Thus T(0) = r|r00(0)(0)|= 1 02+ 32+ 42(3 j + 4 k) = 1 5(3 j + 4 k) = 35j +45k. 21. r(w) = w> w2> w3 r0(w) = 1> 2w> 3w2. Thenr0(1) = h1> 2> 3i and |r0(1)| =12+ 22+ 32=14, so T(1) = r|r00(1)(1)|= 1 14h1> 2> 3i = G 1 14>214>314 H . r00(w) = h0> 2> 6wi, so r0(w) × r00(w) = i j k 1 2w 3w2 0 2 6w = 2w 3w2 2 6w i 1 3w2 0 6w j + 1 2w 0 2 k = (12w2 6w2) i (6w 0) j + (2 0) k = 6w2> 6w> 2
SECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ET SECTION 13.2 ¤ 145
23.The vector equation for the curve isr(w) = 1 + 2w> w3 w> w3+ w, sor0(w) = 1@w> 3w2 1> 3w2+ 1. The point (3> 0> 2) corresponds to w = 1, so the tangent vector there is r0(1) = h1> 2> 4i. Thus, the tangent line goes through the point
(3> 0> 2) and is parallel to the vector h1> 2> 4i. Parametric equations are { = 3 + w, | = 2w, } = 2 + 4w. 25.The vector equation for the curve isr(w) = hwcos w> hwsin w> hw, so
r0(w) = hw( sin w) + (cos w)(hw), hwcos w + (sin w)(hw), (hw)
= hw(cos w + sin w)> hw(cos w sin w)> hw
The point(1> 0> 1) corresponds to w = 0, so the tangent vector there is
r0(0) = h0(cos 0 + sin 0)> h0(cos 0 sin 0)> h0= h1> 1> 1i. Thus, the tangent line is parallel to the vector
h1> 1> 1i and parametric equations are { = 1 + (1)w = 1 w, | = 0 + 1 · w = w, } = 1 + (1)w = 1 w. 27.r(w) = w> hw> 2w w2 r0(w) = 1> hw> 2 2w. At(0> 1> 0),
w = 0 and r0(0) = h1> 1> 2i. Thus, parametric equations of the tangent
line are{ = w, | = 1 w, } = 2w.
29.r(w) = hw cos w> w> w sin wi r0(w) = hcos w w sin w> 1> w cos w + sin wi. At( > > 0), w = and r0( ) = h1> 1> i. Thus, parametric equations of the tangent line are{ = w, | = + w, } = w.
31.The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of
intersection. Sincer01(w) = 1> 2w> 3w2andw = 0 at (0> 0> 0), r01(0) = h1> 0> 0i is a tangent vector to r1at(0> 0> 0). Similarly, r0
2(w) = hcos w> 2 cos 2w> 1i and since r2(0) = h0> 0> 0i, r02(0) = h1> 2> 1i is a tangent vector to r2at(0> 0> 0). If is the angle between these two tangent vectors, thencos = 1
16h1> 0> 0i · h1> 2> 1i =16 and = cos1 1 6 66. 33.U01(16w3i 9w2j + 25w4k) gw =U01 16w3gw i U01 9w2gwj +U1 0 25w4gw k =4w41 0 i 3w31 0 j + 5w51 0 k = 4 i 3 j + 5 k
35.U0 @2(3 sin2w cos w i + 3 sin w cos2w j + 2 sin w cos w k) gw =U0 @23 sin2w cos w gwi +U @2 0 3 sin w cos2w gw j +U0 @22 sin w cos w gwk =sin3w @2 0 i + cos3w @2 0 j+ sin2w @2 0 k = (1 0) i + (0 + 1) j + (1 0) k = i + j + k
TX 10
RIVATIVE RIVATIVE37. U hwi + 2w j + ln w kgw =Uhwgwi +U2w gwj +Uln w gwk
= hwi + w2j + (w ln w w) k + C, where C is a vector constant of integration.
39. r0(w) = 2w i + 3w2j +w k r(w) = w2i + w3j +23w3@2k + C, where C is a constant vector. Buti + j = r (1) = i + j +23k + C. Thus C = 23k and r(w) = w2i + w3j +23w3@223k.
For Exercises 41–44, letu(w) = hx1(w)> x2(w)> x3(w)iandv(w) = hy1(w)> y2(w)> y3(w)i.In each of these exercises, the procedure is to apply Theorem 2 so that the corresponding properties of derivatives of real-valued functions can be used.
41. g gw[u(w) + v(w)] = ggwhx1(w) + y1(w)> x2(w) + y2(w)> x3(w) + y3(w)i = g gw[x1(w) + y1(w)] > ggw[x2(w) + y2(w)] > ggw[x3(w) + y3(w)] = hx0 1(w) + y10(w)> x02(w) + y02(w)> x03(w) + y03(w)i = hx0 1(w)> x02(w) > x03(w)i + hy01(w)> y02(w)> y30(w)i = u0(w) + v0(w) 43. g
gw[u(w) × v(w)] = ggwhx2(w)y3(w) x3(w)y2(w)> x3(w)y1(w) x1(w)y3(w)> x1(w)y2(w) x2(w)y1(w)i = hx0
2y3(w) + x2(w)y03(w) x03(w)y2(w) x3(w)y02(w)>
x0
3(w)y1(w) + x3(w)y10(w) x01(w)y3(w) x1(w)y30(w)>
x0
1(w)y2(w) + x1(w)y20(w) x02(w)y1(w) x2(w)y10(w)i
= hx0
2(w)y3(w) x03(w)y2(w) > x03(w)y1(w) x01(w)y3(w)> x01(w)y2(w) x02(w)y1(w)i
+ hx2(w)y30(w) x3(w)y02(w)> x3(w)y10(w) x1(w)y30(w)> x1(w)y02(w) x2(w)y10(w)i
= u0(w) × v(w) + u(w) × v0(w)
Alternate solution: Let r(w) = u(w) × v(w). Then
r(w + k) r(w) = [u(w + k) × v(w + k)] [u(w) × v(w)]
= [u(w + k) × v(w + k)] [u(w) × v(w)] + [u(w + k) × v(w)] [u(w + k) × v(w)] = u(w + k) × [v(w + k) v(w)] + [u(w + k) u(w)] × v(w)
(Be careful of the order of the cross product.) Dividing through byk and taking the limit as k 0 we have r0(w) = lim k0 u(w + k) × [v(w + k) v(w)] k + limk0 [u(w + k) u(w)] × v(w) k = u(w) × v0(w) + u0(w) × v(w) by Exercise 14.1.43(a) [ET 13.1.43(a)] and Denition 1.
45. g
gw[u(w) · v(w)] = u0(w) · v(w) + u(w) · v0(w) [by Formula 4 of Theorem 3]
= hcos w> sin w> 1i · hw> cos w> sin wi + hsin w> cos w> wi · h1> sin w> cos wi = w cos w cos w sin w + sin w + sin w cos w sin w + w cos w
SECTION 14.3 ARC LENGTH AND CURVATURE ET SECTION 13.3 ¤ 147
47. g
gw[r(w) × r0(w)] = r0(w) × r0(w) + r(w) × r00(w) by Formula 5 of Theorem 3. But r0(w) × r0(w) = 0 (by Example 2 in Section 13.4 [ET 12.4]). Thus, g
gw[r(w) × r0(w)] = r(w) × r00(w). 49. g gw|r(w)| = ggw[r(w) · r(w)]1@2= 12[r(w) · r(w)]1@2[2r(w) · r0(w)] = 1|r(w)|r(w) · r0(w) 51.Sinceu(w) = r(w) · [r0(w) × r00(w)], u0(w) = r0(w) · [r0(w) × r00(w)] + r(w) · g gw[r0(w) × r00(w)] = 0 + r(w) · [r00(w) × r00(w) + r0(w) × r000(w)] [sincer0(w) r0(w) × r00(w)] = r(w) · [r0(w) × r000(w)] [sincer00(w) × r00(w) = 0]
14.3
Arc Length and Curvature
ET 13.3
1.r(w) = h2 sin w> 5w> 2 cos wi r0(w) = h2 cos w> 5> 2 sin wi |r0(w)| =s(2 cos w)2+ 52+ (2 sin w)2=29. Then using Formula 3, we haveO =U1010 |r0(w)| gw =U1010 29 gw = 29 w1010= 2029.
3.r(w) =2 w i + hwj + hwk r0(w) =2 i + hwj hwk |r0(w)| =t22+ (hw)2+ (hw)2=2 + h2w+ h2w=s(hw+ hw)2= hw+ hw [sincehw+ hwA 0]. ThenO =U01 |r0(w)| gw =U01(hw+ hw) gw =hw hw10= h h1. 5.r(w) = i + w2j + w3k r0(w) = 2w j + 3w2k |r0(w)| =4w2+ 9w4= w4 + 9w2 [sincew 0]. ThenO =U01|r0(w)| gw =U01w4 + 9w2gw = 181 ·23(4 + 9w2)3@2l1 0= 1 27(133@2 43@2) = 271(133@2 8). 7.r(w) = w> w> w2 r0(w) = 1 2w> 1> 2w |r0(w)| = u 1 2w 2 + 12+ (2w)2=t1 4w+ 1 + 4w2, so O =U14 |r0(w)| gw =U4 1 t 1 4w+ 1 + 4w2gw 15=3841.
9.r(w) = hsin w> cos w> tan wi r0(w) = cos w> sin w> sec2w |r0(w)| =scos2w + ( sin w)2+ (sec2w)2=1 + sec4w and O =U @4
0 |r0(w)| gw =
U @4 0
1 + sec4w gw 1=2780.
11.The projection of the curveF onto the {|-plane is the curve {2 = 2| or | =12{2,} = 0. Then we can choose the parameter { = w | = 1
2w2. SinceF also lies on the surface 3} = {|, we have } =13{| = 13(w)(12w2) = 16w3. Then parametric
equations forF are { = w, | = 12w2,} =16w3and the corresponding vector equation isr(w) = w>12w2>16w3. The origin corresponds tow = 0 and the point (6> 18> 36) corresponds to w = 6, so
O =U06 |r0(w)| gw =U6 0 1>w>12w2 gw = U06 t 12+ w2+1 2w2 2 gw =U06t1 + w2+1 4w4gw =U6 0 t (1 +1 2w2)2gw = U6 0(1 +12w2) gw = w +1 6w3 6 0= 6 + 36 = 42 SEC SEC
13. r(w) = 2w i + (1 3w) j + (5 + 4w) k r0(w) = 2 i 3 j + 4 k andgvgw = |r0(w)| =4 + 9 + 16 =29. Then v = v(w) =U0w|r0(x)| gx =Uw
0
29 gx =29 w. Therefore, w = 1
29v, and substituting for w in the original equation, we
haver(w(v)) = 2 29v i + 1 3 29v j +5 +4 29v k.
15. Herer(w) = h3 sin w> 4w> 3 cos wi, so r0(w) = h3 cos w> 4> 3 sin wi and |r0(w)| =s9 cos2w + 16 + 9 sin2w =25 = 5. The point(0> 0> 3) corresponds to w = 0, so the arc length function beginning at (0> 0> 3) and measuring in the positive direction is given byv(w) =U0w|r0(x)| gx =U0w5 gx = 5w. v(w) = 5 5w = 5 w = 1, thus your location after moving 5 units along the curve is(3 sin 1> 4> 3 cos 1).
17. (a)r(w) = h2 sin w> 5w> 2 cos wi r0(w) = h2 cos w> 5> 2 sin wi |r0(w)| =s4 cos2w + 25 + 4 sin2w =29. ThenT(w) = r 0(w) |r0(w)| = 129h2 cos w> 5> 2 sin wi or G 2 29cos w>529> 229sin w H . T0(w) = 1 29h2 sin w> 0> 2 cos wi |T0(w)| =129 s 4 sin2w + 0 + 4 cos2w = 2 29. Thus N(w) = T|T00(w)(w)|= 1@ 29
2@29h2 sin w> 0> 2 cos wi = h sin w> 0> cos wi. (b)(w) = |T 0(w)| |r0(w)| = 2@ 29 29 = 229 19. (a)r(w) = 2 w> hw> hw r0(w) = 2> hw> hw |r0(w)| =2 + h2w+ h2w=s(hw+ hw)2= hw+ hw. Then T(w) = r|r00(w)(w)| =hw+ h1 w 2> hw> hw= 1 h2w+ 1 2hw> h2w> 1 after multiplying byhw hw and T0(w) = 1 h2w+ 1 2hw> 2h2w> 0 2h2w (h2w+ 1)2 2hw> h2w> 1 = (h2w1+ 1)2(h2w+ 1) 2hw> 2h2w> 0 2h2w 2hw> h2w> 1= 1 (h2w+ 1)2 2hw1 h2w> 2h2w> 2h2w Then |T0(w)| = 1 (h2w+ 1)2 s 2h2w(1 2h2w+ h4w) + 4h4w+ 4h4w= 1 (h2w+ 1)2 s 2h2w(1 + 2h2w+ h4w) = (h2w1+ 1)2t2h2w(1 + h2w)2=2hw(1 + h2w) (h2w+ 1)2 = 2 hw h2w+ 1 Therefore N(w) = T0(w) |T0(w)|= h 2w+ 1 2 hw 1 (h2w+ 1)2 2 hw(1 h2w)> 2h2w> 2h2w = 1 2 hw(h2w+ 1) 2 hw(1 h2w)> 2h2w> 2h2w= 1 h2w+ 1 1 h2w>2 hw>2 hw (b)(w) = |T 0(w)| |r0(w)| = 2 hw h2w+ 1·hw+ h1 w = 2 hw h3w+ 2hw+ hw = 2 h2w h4w+ 2h2w+ 1= 2 h2w (h2w+ 1)2
SECTION 14.3 ARC LENGTH AND CURVATURE ET SECTION 13.3 ¤ 149 21.r(w) = w2i + w k r0(w) = 2w i + k, r00(w) = 2 i, |r0(w)| =s(2w)2+ 02+ 12=4w2+ 1, r0(w) × r00(w) = 2 j, |r0(w) × r00(w)| = 2. Then (w) = |r0(w) × r00(w)| |r0(w)|3 = 2 4w2+ 13 = 2 (4w2+ 1)3@2.
23.r(w) = 3w i + 4 sin w j + 4 cos w k r0(w) = 3 i + 4 cos w j 4 sin w k, r00(w) = 4 sin w j 4 cos w k, |r0(w)| =s9 + 16 cos2w + 16 sin2w =9 + 16 = 5, r0(w) × r00(w) = 16 i + 12 cos w j 12 sin w k,
|r0(w) × r00(w)| =s256 + 144 cos2w + 144 sin2w =400 = 20. Then (w) = |r0(w) × r00(w)|
|r0(w)|3 = 2053 = 425.
25.r(w) = w> w2> w3 r0(w) = 1> 2w> 3w2. The point(1> 1> 1) corresponds to w = 1, and r0(1) = h1> 2> 3i |r0(1)| =1 + 4 + 9 =14. r00(w) = h0> 2> 6wi r00(1) = h0> 2> 6i. r0(1) × r00(1) = h6> 6> 2i, so
|r0(1) × r00(1)| =36 + 36 + 4 =76. Then (1) = |r0(1) × r00(1)| |r0(1)|3 = 76 143 = 17 u 19 14. 27.i({) = 2{ {2, i0({) = 2 2{, i00({) = 2, ({) = [1 + (i|i000({))({)|2]3@2 =[1 + (2 2{)|2| 2]3@2 =(4{2 8{ + 5)2 3@2 29.i({) = 4{5@2, i0({) = 10{3@2, i00({) = 15{1@2, ({) = [1 + (i|i000({))({)|2]3@2 = 15{1@2 [1 + (10{3@2)2]3@2 = 15{ (1 + 100{3)3@2
31.Since|0= |00= h{, the curvature is({) = ||
00({)|
[1 + (|0({))2]3@2 =
h{
(1 + h2{)3@2 = h{(1 + h2{)3@2.
To nd the maximum curvature, we rst nd the critical numbers of({): 0({) = h{(1 + h2{)3@2+ h{3 2 (1 + h2{)5@2(2h2{) = h{1 + h2{ 3h2{ (1 + h2{)5@2 = h{ 1 2h2{ (1 + h2{)5@2. 0({) = 0 when 1 2h2{= 0, so h2{=1
2 or{ = 12ln 2. And since 1 2h2{A 0 for { ? 12ln 2 and
1 2h2{? 0 for { A 1
2ln 2, the maximum curvature is attained at the point
1 2ln 2> h( ln 2)@2 =1 2ln 2>12 . Since lim {h {(1 + h2{)3@2= 0> ({) approaches 0 as { .
33. (a)F appears to be changing direction more quickly at S than T, so we would expect the curvature to be greater at S . (b) First we sketch approximate osculating circles atS and T. Using the
axes scale as a guide, we measure the radius of the osculating circle
atS to be approximately 0=8 units, thus = 1
= 1 10=8 1=3. Similarly, we estimate the radius of the osculating circle atT to be 1=4 units, so = 1
11=4 0=7.
35. | = {2 |0= 2{3, |00= 6{4, and
({) = [1 + (|||000)|2]3@2 = [1 + (2{6{43)2]3@2 ={4(1 + 4{6 6)3@2.
The appearance of the two humps in this graph is perhaps a little surprising, but it is
explained by the fact that| = {2increases asymptotically at the origin from both directions, and so its graph has very little bend there. (Note that(0) is undened.)
37. Notice that the curvee has two inection points at which the graph appears almost straight. We would expect the curvature to be0 or nearly 0 at these values, but the curve d isn’t near 0 there. Thus, d must be the graph of | = i({) rather than the graph of curvature, ande is the graph of | = ({).
39. Using a CAS, we nd (after simplifying)
(w) = 6
4 cos2w 12 cos w + 13
(17 12 cos w)3@2 . (To compute cross
products in Maple, use the VectorCalculus package and the CrossProduct(a,b) command; in Mathematica, use Cross[a,b].) Curvature is largest at integer multiples of2 .
41. { = hwcos w { = hw(cos w sin w) ¨{ = hw( sin w cos w) + hw(cos w sin w) = 2hwsin w, | = hwsin w | = hw(cos w + sin w) ¨| = hw( sin w + cos w) + hw(cos w + sin w) = 2hwcos w. Then
(w) = | {¨| |¨{|[ {2+ |2]3@2 =hw(cos w sin w)(2hwcos w) hw(cos w + sin w)(2hwsin w) ([hw(cos w sin w)]2+ [hw(cos w + sin w)]2)3@2
= 2h2w(cos2w sin w cos w + sin w cos w + sin2w)
h2w(cos2w 2 cos w sin w + sin2w + cos2w + 2 cos w sin w + sin2w)3@2 =
2h2w(1) [h2w(1 + 1)]3@2 = 2h2w h3w(2)3@2 = 2 h1 w 43. 1>23> 1corresponds tow = 1. T(w) = r 0(w) |r0(w)|= 2w> 2w2> 1 4w2+ 4w4+ 1= 2w> 2w2> 1 2w2+ 1 , soT(1) = 2 3>23>13 .
T0(w) = 4w(2w2+ 1)2 2w> 2w2> 1+ (2w2+ 1)1h2> 4w> 0i (by Formula 3 of Theorem 14.2 [ET 13.2])
= (2w2+ 1)2 8w2+ 4w2+ 2> 8w3+ 8w3+ 4w> 4w= 2(2w2+ 1)2 1 2w2> 2w> 2w N(w) = T0(w) |T0(w)|= 2(2w2+ 1)2 1 2w2> 2w> 2w 2(2w2+ 1)2s(1 2w2)2+ (2w)2+ (2w)2 = 1 2w2> 2w> 2w 1 4w2+ 4w4+ 8w2 = 1 2w2> 2w> 2w 1 + 2w2 N(1) = 1 3>23> 23 andB(1) = T(1) × N(1) = 4929> 49 +19>49 +29= 32>13>23. 45. (0> > 2) corresponds to w = . r(w) = h2 sin 3w> w> 2 cos 3wi
T(w) = r|r00(w)(w)|= s h6 cos 3w> 1> 6 sin 3wi 36 cos23w + 1 + 36 sin23w =
1
37h6 cos 3w> 1> 6 sin 3wi.
T( ) =1
SECTION 14.3 ARC LENGTH AND CURVATURE ET SECTION 13.3 ¤ 151
plane is6 ({ 0) + 1(| ) + 0(} + 2) = 0 or | 6{ = . T0(w) =1
37h18 sin 3w> 0> 18 cos 3wi |T0(w)| =
s
182sin23w + 182cos23w
37 = 1837
N(w) = T|T00(w)(w)|= h sin 3w> 0> cos 3wi. So N( ) = h0> 0> 1i and B( ) = 1
37h6> 1> 0i × h0> 0> 1i = 137h1> 6> 0i.
SinceB( ) is a normal to the osculating plane, so is h1> 6> 0i.
An equation for the plane is1({ 0) + 6(| ) + 0(} + 2) = 0 or { + 6| = 6 .
47.The ellipse9{2+ 4|2= 36 is given by the parametric equations { = 2 cos w, | = 3 sin w, so using the result from Exercise 40,
(w) = | {¨| ¨{ ||[ {2+ |2]3@2 = |(2 sin w)(3 sin w) (3 cos w)(2 cos w)|
(4 sin2w + 9 cos2w)3@2 = (4 sin2w + 9 cos6 2w)3@2.
At(2> 0), w = 0. Now (0) = 276 = 29, so the radius of the osculating circle is 1@(0) = 9
2 and its center is
5
2> 0
. Its equation is therefore{ +522+ |2=814. At(0> 3), w = 2, and 2=68 = 34. So the radius of the osculating circle is43and its center is0>53. Hence its equation is{2+| 532=169.
49.The tangent vector is normal to the normal plane, and the vectorh6> 6> 8i is normal to the given plane. ButT(w) k r0(w) and h6> 6> 8i k h3> 3> 4i, so we need to nd w such that r0(w) k h3> 3> 4i.
r(w) = w3> 3w> w4 r0(w) = 3w2> 3> 4w3k h3> 3> 4i when w = 1. So the planes are parallel at the point (1> 3> 1).
51. =gT gv =gT@gwgv@gw = |gT@gw|gv@gw andN = gT@gw |gT@gw|, soN = gTgwgTgw
gTgwgvgw = gT@gwgv@gw = gTgv by the Chain Rule. 53. (a)|B| = 1 B · B = 1 g gv(B · B) = 0 2 gBgv · B = 0 gBgv B (b)B = T × N gB gv = ggv(T × N) = ggw(T × N) 1gv@gw= ggw(T × N) 1|r0(w)|= [(T0× N) + (T × N0)] 1|r0(w)| = T0× T0 |T0| + (T × N0) 1 |r0(w)| = T × N 0 |r0(w)| gBgv T
(c)B = T × N T N, B T and B N. So B, T and N form an orthogonal set of vectors in the three-dimensional spaceR3. From parts (a) and (b),gB@gv is perpendicular to both B and T, so gB@gv is parallel to N. Therefore,gB@gv = (v)N, where (v) is a scalar.
(d) SinceB = T × N, T N and both T and N are unit vectors, B is a unit vector mutually perpendicular to both T and N. For a plane curve, T and N always lie in the plane of the curve, so that B is a constant unit vector always
perpendicular to the plane. ThusgB@gv = 0, but gB@gv = (v)N and N 6= 0, so (v) = 0.
55. (a)r0= v0T r00= v00T + v0T0= v00T + v0gT
gvv0= v00T + (v0)2N by the rst Serret-Frenet formula. (b) Using part (a), we have
r0× r00= (v0T) × [v00T + (v0)2N]
= [(v0T) × (v00T)] +(v0T) × ((v0)2N) (by Property 3 of Theorem 13.4.8 [ ET 12.4.8])
= (v0v00)(T × T) + (v0)3(T × N) = 0 + (v0)3B = (v0)3B
(c) Using part (a), we have
r000= [v00T + (v0)2N]0= v000T + v00T0+ 0(v0)2N + 2v0v00N + (v0)2N0
= v000T + v00gT
gv v0+ 0(v0)2N + 2v0v00N + (v0)2 gNgv v0
= v000T + v00v0 N + 0(v0)2N + 2v0v00N + (v0)3( T + B) [by the second formula]
= [v000 2(v0)3] T + [3v0v00+ 0(v0)2] N + (v0)3B
(d) Using parts (b) and (c) and the facts thatB · T = 0, B · N = 0, and B · B = 1, we get (r0× r00) · r000 |r0× r00|2 = (v0)3B · [v000 2(v0)3] T + [3v0v00+ 0(v0)2] N + (v0)3B |(v0)3B|2 = (v 0)3 (v0)3 [(v0)3]2 = . 57. r = w>12w2>13w3 r0 = 1> w> w2, r00= h0> 1> 2wi, r000= h0> 0> 2i r0× r00= w2> 2w> 1 = (r0× r00) · r000 |r0× r00|2 = w2> 2w> 1· h0> 0> 2i w4+ 4w2+ 1 = w4+ 4w22+ 1
59. For one helix, the vector equation isr(w) = h10 cos w> 10 sin w> 34w@(2 )i (measuring in angstroms), because the radius of each helix is10 angstroms, and } increases by 34 angstroms for each increase of 2 in w. Using the arc length formula, letting w go from0 to 2=9 × 108× 2 , we nd the approximate length of each helix to be
O =U02=9×108×2 |r0(w)| gw =U2=9×108×2 0 t (10 sin w)2+ (10 cos w)2+34 2 2gw =t100 +34 2 22=9×10 8×2 = 2=9 × 108× 2 t100 +34 2
2 2=07 × 1010Å — more than two meters!
14.4
Motion in Space: Velocity and Acceleration
ET 13.4
1. (a) Ifr(w) = {(w) i + | (w) j + }(w) k is the position vector of the particle at time t, then the average velocity over the time interval[0> 1] is
vave= r(1) r(0)
1 0 = (4=5 i + 6=0 j + 3=0 k) (2=7 i + 9=8 j + 3=7 k)1 = 1=8 i 3=8 j 0=7 k. Similarly, over the other intervals we have
[0=5> 1] : vave= r(1) r(0=5)1 0=5 = (4=5 i + 6=0 j + 3=0 k) (3=5 i + 7=2 j + 3=3 k)0=5 = 2=0 i 2=4 j 0=6 k
[1> 2] : vave= r(2) r(1)2 1 = (7=3 i + 7=8 j + 2=7 k) (4=5 i + 6=0 j + 3=0 k)1 = 2=8 i + 1=8 j 0=3 k
SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ET SECTION 13.4 ¤ 153
(b) We can estimate the velocity atw = 1 by averaging the average velocities over the time intervals [0=5> 1] and [1> 1=5]: v(1) 1
2[(2 i 2=4 j 0=6 k) + (2=8 i + 0=8 j 0=4 k)] = 2=4 i 0=8 j 0=5 k. Then the speed is
|v(1)| s(2=4)2+ (0=8)2+ (0=5)2 2=58. 3.r(w) = 12w2> w Atw = 2: v(w) = r0(w) = hw> 1i v(2) = h2> 1i a(w) = r00(w) = h1> 0i a(2) = h1> 0i |v(w)| =w2+ 1 5.(w) = 3 cos w i + 2 sin w j Atw = @3: v(w) = 3 sin w i + 2 cos w j v 3 = 33 2 i + j
a(w) = 3 cos w i 2 sin w j a 3 = 3 2i 3 j |v(w)| =s9 sin2w + 4 cos2w =s4 + 5 sin2w
Notice that{2@9 + |2@4 = sin2w + cos2w = 1, so the path is an ellipse. 7.r(w) = w i + w2j + 2 k Atw = 1:
v(w) = i + 2w j v(1) = i + 2 j
a(w) = 2 j a(1) = 2 j
|v(w)| =1 + 4w2
Here{ = w, | = w2 | = {2and} = 2, so the path of the particle is a parabola in the plane} = 2.
9.r(w) = w2+ 1> w3> w2 1 v(w) = r0(w) = 2w> 3w2> 2w, a(w) = v0(w) = h2> 6w> 2i, |v(w)| =s(2w)2+ (3w2)2+ (2w)2=9w4+ 8w2= |w|9w2+ 8.
11.r(w) =2 w i + hwj + hwk v(w) = r0(w) =2 i + hwj hwk, a(w) = v0(w) = hwj + hwk, |v(w)| =2 + h2w+ h2w=s(hw+ hw)2= hw+ hw.
13.r(w) = hwhcos w> sin w> wi
v(w) = r0(w) = hwhcos w> sin w> wi + hwh sin w> cos w> 1i = hwhcos w sin w> sin w + cos w> w + 1i
a(w) = v0(w) = hwhcos w sin w sin w cos w> sin w + cos w + cos w sin w> w + 1 + 1i
= hwh2 sin w> 2 cos w> w + 2i
|v(w)| = hwscos2w + sin2w 2 cos w sin w + sin2w + cos2w + 2 sin w cos w + w2+ 2w + 1
= hww2+ 2w + 3
15. a(w) = i + 2 j v(w) =Ua(w) gw =U(i + 2 j) gw = w i + 2w j + C and k = v (0) = C,
soC = k and v(w) = w i + 2w j + k. r(w) =Uv(w) gw =U(w i + 2w j + k) gw = 12w2i + w2j + w k + D. Buti = r (0) = D, so D = i and r(w) =12w2+ 1i + w2j + w k.
17. (a)a(w) = 2w i + sin w j + cos 2w k
v(w) =U(2w i + sin w j + cos 2w k) gw = w2i cos w j +1
2sin 2w k + C
andi = v (0) = j + C, so C = i + j
andv(w) =w2+ 1i + (1 cos w) j +12sin 2w k. r(w) =U[w2+ 1i + (1 cos w) j +1 2sin 2w k]gw =1 3w3+ w i + (w sin w) j 1 4cos 2w k + D
Butj = r (0) = 14k + D, so D = j +14k and r(w) =13w3+ wi + (w sin w + 1) j +1414cos 2wk. (b)
19. r(w) = w2> 5w> w2 16w v(w) = h2w> 5> 2w 16i, |v(w)| =4w2+ 25 + 4w2 64w + 256 =8w2 64w + 281 and g
gw|v(w)| =12(8w2 64w + 281)1@2(16w 64). This is zero if and only if the numerator is zero, that is,
16w 64 = 0 or w = 4. Since ggw|v(w)| ? 0 for w ? 4 and ggw|v(w)| A 0 for w A 4, the minimum speed of153 is attained atw = 4 units of time.
21. |F(w)| = 20 N in the direction of the positive }-axis, so F(w) = 20 k. Also p = 4 kg, r(0) = 0 and v(0) = i j. Since20k = F(w) = 4 a(w), a(w) = 5 k. Then v(w) = 5w k + c1wherec1= i j so v(w) = i j + 5w k and the
speed is|v(w)| =1 + 1 + 25w2 =25w2+ 2. Also r(w) = w i w j +52w2k + c2and0 = r(0), so c2 = 0
andr(w) = w i w j + 52w2k.
23. |v(0)| = 500 m@s and since the angle of elevation is 30, the direction of the velocity is123 i + j. Thus
v(0) = 2503 i + jand if we set up the axes so the projectile starts at the origin, thenr(0) = 0. Ignoring air resistance, the only force is that due to gravity, soF(w) = pj j where j 9=8 m@s2. Thusa(w) = j j and v(w) = jw j + c1. But 2503 i + j= v(0) = c1, sov(w) = 2503 i + (250 jw) j and r(w) = 2503 w i +250w 12jw2j + c2where 0 = r(0) = c2. Thusr(w) = 2503 w i +250w 12jw2j.
(a) Setting250w 12jw2= 0 gives w = 0 or w = 500j 51=0 s. So the range is 2503 ·500j 22 km. (b)0 = g gw 250w 1 2jw2
= 250 jw implies that the maximum height is attained when w = 250@j 25=5 s. Thus, the maximum height is(250)(250@j) j(250@j)2 12 = (250)2@(2j) 3=2 km.
(c) From part (a), impact occurs atw = 500@j 51=0. Thus, the velocity at impact is
SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ET SECTION 13.4 ¤ 155
25.As in Example 5,r(w) = (y0cos 45)w i +(y0sin 45)w 12jw2j =12y02 w i +y02 w jw2j. Then the ball lands atw = y0
2
j s. Now since it lands90 m away, 90 = 12y0
2 y0
2
j ory20= 90j and the initial velocity
isy0 =90j 30 m@s.
27.Let be the angle of elevation. Then y0= 150 m@s and from Example 5, the horizontal distance traveled by the projectile is g = y20sin 2
j . Thus150
2sin 2
j = 800 sin 2 = 800j1502 0=3484 2 20=4or180 20=4 = 159=6.
Two angles of elevation then are 10=2and 79=8.
29.Place the catapult at the origin and assume the catapult is 100 meters from the city, so the city lies between(100> 0) and(600> 0). The initial speed is y0= 80 m@s and let be the angle the catapult is set at. As in Example 5, the trajectory of the catapulted rock is given byr (w) = (80 cos )w i +(80 sin )w 4=9w2j. The top of the near city wall is at (100> 15), which the rock will hit when(80 cos ) w = 100 w = 5
4 cos and(80 sin )w 4=9w2= 15 80 sin · 4 cos 5 4=9
5 4 cos
2
= 15 100 tan 7=65625 sec2 = 15. Replacing sec2 with tan2 + 1 gives
7=65625 tan2 100 tan + 22=62625 = 0. Using the quadratic formula, we have tan 0=230324, 12=8309
13=0,85=5. So for13=0? ? 85=5, the rock will land beyond the near city wall. The base of the far wall is
located at(600> 0) which the rock hits if (80 cos )w = 600 w = 15
2 cos and(80 sin )w 4=9w2= 0 80 sin · 152 cos 4=9 15 2 cos 2 = 0 600 tan 275=625 sec2 = 0
275=625 tan2 600 tan + 275=625 = 0. Solutions are tan 0=658678, 1=51819 33=4,56=6. Thus the
rock lands beyond the enclosed city ground for33=4? ? 56=6, and the angles that allow the rock to land on city ground are13=0? ? 33=4,56=6? ? 85=5. If you consider that the rock can hit the far wall and bounce back into the city, we calculate the angles that cause the rock to hit the top of the wall at(600> 15): (80 cos )w = 600 w = 15
2 cos and (80 sin )w 4=9w2= 15 600 tan 275=625 sec2 = 15 275=625 tan2 600 tan + 290=625 = 0.
Solutions aretan 0=727506, 1=44936 36=0,55=4, so the catapult should be set with angle where 13=0? ? 36=0,55=4? ? 85=5.
31. (a) Afterw seconds, the boat will be 5w meters west of point D. The velocity of the water at that location is4003 (5w)(40 5w) j. The velocity of the boat in still water is5 i> so the resultant velocity of the boat is v(w) = 5 i + 3 400(5w)(40 5w) j = 5i + 3 2w 163w2 j. Integrating, we obtain r(w) = 5w i +3 4w2161w3
j + C. If we place the origin at D (and consider j
to coincide with the northern direction) thenr(0) = 0 C = 0 and we have r(w) = 5w i +34w2161w3j. The boat
reaches the east bank after8 s, and it is located at r(8) = 5(8)i +34(8)2161(8)3j = 40 i + 16 j. Thus the boat is 16 m downstream.
(b) Let be the angle north of east that the boat heads. Then the velocity of the boat in still water is given by 5(cos ) i + 5(sin ) j. At w seconds, the boat is 5(cos )w meters from the west bank, at which point the velocity of the water is4003 [5(cos )w][40 5(cos )w] j. The resultant velocity of the boat is given by
v(w) = 5(cos ) i +5 sin + 3 400(5w cos )(40 5w cos ) j = (5 cos ) i +5 sin +3 2w cos 163w2cos2 j. Integrating,r(w) = (5w cos ) i +5w sin +34w2cos 161w3cos2j (where we have again placed
the origin atD). The boat will reach the east bank when 5w cos = 40 w = 40
5 cos = 8cos . In order to land at pointE(40> 0) we need 5w sin + 34w2cos 161w3cos2 = 0
5 8 cos sin +3 4 8 cos 2 cos 1 16 8 cos 3 cos2 = 0 1 cos (40 sin + 48 32) = 0 40 sin + 16 = 0 sin = 2 5. Thus = sin1 2 5
23=6, so the boat should head23=6south of
east (upstream). The path does seem realistic. The boat initially heads upstream to counteract the effect of the current. Near the center of the river, the current is stronger and the boat is pushed downstream. When the boat nears the eastern bank, the current is slower and the boat is able to progress upstream to arrive at pointE.
33. r(w) = (3w w3) i + 3w2j r0(w) = (3 3w2) i + 6w j,
|r0(w)| =s(3 3w2)2+ (6w)2 =9 + 18w2+ 9w4 =s(3 3w2)2 = 3 + 3w2,
r00(w) = 6w i + 6 j, r0(w) × r00(w) = (18 + 18w2) k. Then Equation 9 gives
dW = r 0(w) · r00(w) |r0(w)| = (3 3w 2)(6w) + (6w)(6) 3 + 3w2 = 18w + 18w 3 3 + 3w2 = 18w(1 + w 2) 3(1 + w2) = 6w k or by Equation 8, dW= y0= g gw
3 + 3w2= 6w and Equation 10 givesd Q = |r 0(w) × r00(w)| |r0(w)| = 18 + 18w 2 3 + 3w2 = 18(1 + w 2) 3(1 + w2) = 6.
35. r(w) = cos w i + sin w j + w k r0(w) = sin w i + cos w j + k, |r0(w)| =ssin2w + cos2w + 1 =2, r00(w) = cos w i sin w j, r0(w) × r00(w) = sin w i cos w j + k.
ThendW = r
0(w) · r00(w)
|r0(w)| = sin w cos w sin w cos w2 = 0 and dQ= |r
0(w) × r00(w)| |r0(w)| = s sin2w + cos2w + 1 2 = 2 2= 1. 37. r(w) = hwi +2 w j + hwk r0(w) = hwi +2 j hwk, |r(w)| =h2w+ 2 + h2w=s(hw+ hw)2= hw+ hw, r00(w) = hwi + hwk. Then d W = h 2w h2w hw+ hw = (h w+ hw)(hw hw) hw+ hw = hw hw= 2 sinh w anddQ=2h wi 2 j 2hwk hw+ hw = s 2(h2w+ 2 + h2w) hw+ hw = 2 hhww+ h+ hww =2.
CHAPTER 14 REVIEW ET CHAPTER 13 ¤ 157
39.The tangential component ofa is the length of the projection of a onto T, so we sketch the scalar projection ofa in the tangential direction to the curve and estimate its length to be4=5 (using the fact that a has length 10 as a guide). Similarly, the normal component of a is the length of the projection of a onto N, so we sketch the scalar projection of a in the normal direction to the curve and estimate its length to be9=0. Thus dW 4=5 cm@s2and dQ 9=0 cm@s2.
41.If the engines are turned off at timew, then the spacecraft will continue to travel in the direction of v(w), so we need a w such that for some scalarv A 0, r(w) + v v(w) = h6> 4> 9i. v(w) = r0(w) = i + 1
w j +(w28w+ 1)2 k r(w) + v v(w) = 3 + w + v> 2 + ln w + vw> 7 w24+ 1+(w28vw+ 1)2 3 + w + v = 6 v = 3 w, so7 4 w2+ 1+ 8(3 w)w(w2+ 1)2 = 9 24w 12w 2 4 (w2+ 1)2 = 2 w4+ 8w2 12w + 3 = 0.
It is easily seen thatw = 1 is a root of this polynomial. Also 2 + ln 1 + 3 1
1 = 4, so w = 1 is the desired solution.
14
Review
ET 13
1.A vector function is a function whose domain is a set of real numbers and whose range is a set of vectors. To nd the derivative or integral, we can differentiate or integrate each component of the vector function.
2.The tip of the moving vectorr(w) of a continuous vector function traces out a space curve.
3.The tangent vector to a smooth curve at a pointS with position vector r(w) is the vector r0(w). The tangent line at S is the line throughS parallel to the tangent vector r0(w). The unit tangent vector is T(w) = r
0(w)
|r0(w)|.
4.(a) – (f ) See Theorem 14.2.3 [ET 13.2.3].
5.Use Formula 14.3.2 [ ET 13.3.2], or equivalently, 14.3.3 [ ET 13.3.3].
6. (a) The curvature of a curve is =gT gv
where T is the unit tangent vector. (b)(w) =T 0(w) r0(w) (c)(w) = |r 0(w) × r00(w)| |r0(w)|3 (d)({) = |i00({)| [1 + (i0({))2]3@2
7. (a) The unit normal vector: N(w) = T
0(w)
|T0(w)|. The binormal vector: B(w) = T(w) × N(w).
(b) See the discussion preceding Example 7 in Section 14.3 [ ET 13.3].
8. (a) Ifr(w) is the position vector of the particle on the space curve, the velocity v(w) = r0(w), the speed is given by |v(w)|, and the accelerationa(w) = v0(w) = r00(w).
(b)a = dWT + dQN where dW = y0anddQ= y2.
9. See the statement of Kepler’s Laws on page 880 [ET page 844].
1. True. If we reparametrize the curve by replacingx = w3, we haver(x) = x i + 2x j + 3x k, which is a line through the origin with direction vectori + 2 j + 3 k.
3. False. By Formula 5 of Theorem 14.2.3[ ET 13.2.3], g
gw[u(w) × v(w)] = u0(w) × v(w) + u(w) × v0(w).
5. False. is the magnitude of the rate of change of the unit tangent vector T with respect to arc length v, not with respect to w. 7. True. At an inection point wherei is twice continuously differentiable we must have i00({) = 0, and by Equation 14.3.11
[ET 13.3.11], the curvature is0 there.
9. False. Ifr(w) is the position of a moving particle at time w and |r(w)| = 1 then the particle lies on the unit circle or the unit sphere, but this does not mean that the speed|r0(w)| must be constant. As a counterexample, let r(w) = w>1 w2, then r0(w) = 1> w@1 w2and|r(w)| =w2+ 1 w2 = 1 but |r0(w)| =s1 + w2@(1 w2) = 1@1 w2which is not
constant.
11. True. See the discussion preceding Example 7 in Section 14.3 [ ET 13.3].
1. (a) The corresponding parametric equations for the curve are{ = w, | = cos w, } = sin w. Since |2+ }2= 1, the curve is contained in a
circular cylinder with axis the{-axis. Since { = w, the curve is a helix. (b)r(w) = w i + cos w j + sin w k
r0(w) = i sin w j + cos w k
r00(w) = 2cos w j 2sin w k
3. The projection of the curveF of intersection onto the {|-plane is the circle {2+ |2= 16> } = 0. So we can write { = 4 cos w, | = 4 sin w, 0 w 2 . From the equation of the plane, we have } = 5 { = 5 4 cos w, so parametric equations forF are { = 4 cos w, | = 4 sin w, } = 5 4 cos w, 0 w 2 , and the corresponding vector function is r(w) = 4 cos w i + 4 sin w j + (5 4 cos w) k, 0 w 2 .
CHAPTER 14 REVIEW ET CHAPTER 13 ¤ 159
5.U01(w2i + w cos w j + sin w k) gw =U01w2gwi +U01w cos w gwj +U01sin w gwk =1 3w3 1 0 i + w sin w 1 0 U1 0 1 sin w gw j +1 cos w 1 0 k =1 3i + 1 2cos w 1 0 j + 2k = 13i 22j +2 k
where we integrated by parts in the|-component.
7.r(w) = w2> w3> w4 r0(w) = 2w> 3w2> 4w3 |r0(w)| =4w2+ 9w4+ 16w6and O =U03|r0(w)| gw =U3
0
4w2+ 9w4+ 16w6gw. Using Simpson’s Rule with i(w) =4w2+ 9w4+ 16w6andq = 6 we
havew =306 =12 and O w 3 i(0) + 4i1 2 + 2i(1) + 4i3 2 + 2i(2) + 4i5 2 + i(3) =1 6 0 + 0 + 0 + 4 ·t41 2 2 + 91 2 4 + 161 2 6 + 2 ·s4(1)2+ 9(1)4+ 16(1)6 + 4 ·t43 2 2 + 93 2 4 + 163 2 6 + 2 ·s4(2)2+ 9(2)4+ 16(2)6 + 4 ·t45 2 2 + 95 2 4 + 165 2 6 +s4(3)2+ 9(3)4+ 16(3)6 86=631
9.The angle of intersection of the two curves,, is the angle between their respective tangents at the point of intersection. For both curves the point(1> 0> 0) occurs when w = 0.
r0
1(w) = sin w i + cos w j + k r01(0) = j + k and r02(w) = i + 2w j + 3w2k r02(0) = i.
r0
1(0) · r02(0) = (j + k) · i = 0. Therefore, the curves intersect in a right angle, that is, = 2.
11. (a)T(w) = r 0(w) |r0(w)|= w2> w> 1 |hw2> w> 1i| = w2> w> 1 w4+ w2+ 1 (b)T0(w) = 12(w4+ w2+ 1)3@2(4w3+ 2w) w2> w> 1+ (w4+ w2+ 1)1@2h2w> 1> 0i =(w4+ w2w23+ 1) w3@2 w2> w> 1+ 1 (w4+ w2+ 1)1@2h2w> 1> 0i = 2w5 w3> 2w4 w2> 2w3 w+ 2w5+ 2w3+ 2w> w4+ w2+ 1> 0 (w4+ w2+ 1)3@2 = 2w> w4+ 1> 2w3 w (w4+ w2+ 1)3@2 |T0(w)| =4w2+ w8 2w4+ 1 + 4w6+ 4w4+ w2 (w4+ w2+ 1)3@2 = w8+ 4w6+ 2w4+ 5w2 (w4+ w2+ 1)3@2 and N(w) = 2w> 1 w4> 2w3 w w8+ 4w6+ 2w4+ 5w2. (c)(w) = |T 0(w)| |r0(w)| = w8+ 4w6+ 2w4+ 5w2 (w4+ w2+ 1)2 13.|0= 4{3,|00= 12{2and({) = || 00| [1 + (|0)2]3@2 = 12{2 (1 + 16{6)3@2, so(1) = 12173@2.
15.r(w) = hsin 2w> w> cos 2wi r0(w) = h2 cos 2w> 1> 2 sin 2wi T(w) = 1
5h2 cos 2w> 1> 2 sin 2wi
T0(w) =1
5h4 sin 2w> 0> 4 cos 2wi N(w) = h sin 2w> 0> cos 2wi. So N = N( ) = h0> 0> 1i and
TX.10
B = T × N =1
5h1> 2> 0i. So a normal to the osculating plane is h1> 2> 0i and an equation is
1({ 0) + 2(| ) + 0(} 1) = 0 or { 2| + 2 = 0.
17. r(w) = w ln w i + w j + hwk, v(w) = r0(w) = (1 + ln w) i + j hwk,
|v (w)| =s(1 + ln w)2+ 12+ (hw)2 =s2 + 2 ln w + (ln w)2+ h2w, a(w) = v0(w) = 1
wi + hwk
19. We set up the axes so that the shot leaves the athlete’s hand7 ft above the origin. Then we are given r(0) = 7j, |v(0)| = 43 ft@s, and v(0) has direction given by a 45angle of elevation. Then a unit vector in the direction ofv(0) is
1
2(i + j) v(0) =432(i + j). Assuming air resistance is negligible, the only external force is due to gravity, so as
in Example 14.4.5 [ET 13.4.5] we havea = j j where here j 32 ft@s2. Sincev0(w) = a(w), we integrate, giving v(w) = jw j + C where C = v(0) = 43 2(i + j) v (w) = 432i + 43 2 jw j. Since r0(w) = v(w) we integrate again, sor(w) =43 2w i + 43 2w 12jw2 j + D. But D = r(0) = 7 j r(w) = 43 2w i + 43 2w 12jw2+ 7 j. (a) At2 seconds, the shot is at r(2) =43
2(2) i + 43 2(2) 12j(2)2+ 7
j 60=8 i + 3=8 j, so the shot is about 3=8 ft above the ground, at a horizontal distance of60=8 ft from the athlete.
(b) The shot reaches its maximum height when the vertical component of velocity is0: 43
2 jw = 0
w = 43
2 j 0=95 s. Then r(0=95) 28=9 i + 21=4 j, so the maximum height is approximately 21=4 ft. (c) The shot hits the ground when the vertical component ofr(w) is 0, so43
2w 12jw2+ 7 = 0
16w2+43
2w + 7 = 0 w 2=11 s. r(2=11) 64=2 i 0=08 j, thus the shot lands approximately 64=2 ft from the
athlete.
21. (a) Instead of proceeding directly, we use Formula 3 of Theorem 14.2.3 [ ET 13.2.3]: r(w) = w R(w) v = r0(w) = R(w) + w R0(w) = cos $w i + sin $w j + w v
g.
(b) Using the same method as in part (a) and starting withv = R(w) + w R0(w), we have a = v0= R0(w) + R0(w) + w R00(w) = 2 R0(w) + w R00(w) = 2 v
g+ w ag.
(c) Here we haver(w) = hwcos $w i + hwsin $w j = hwR(w). So, as in parts (a) and (b), v = r0(w) = hwR0(w) hwR(w) = hw[R0(w) R(w)]
a = v0= hw[R00(w) R0(w)] hw[R0(w) R(w)] = hw[R00(w) 2 R0(w) + R(w)]
= hwa
g 2hwvg+ hwR
PROBLEMS PLUS
1. (a)r(w) = U cos $w i + U sin $w j v = r0(w) = $U sin $w i + $U cos $w j, so r = U(cos $w i + sin $w j) and v = $U( sin $w i + cos $w j). v · r = $U2( cos $w sin $w + sin $w cos $w) = 0, so v r. Since r points along a
radius of the circle, andv r, v is tangent to the circle. Because it is a velocity vector, v points in the direction of motion. (b) In (a), we wrotev in the form $U u, where u is the unit vector sin $w i + cos $w j. Clearly |v| = $U |u| = $U. At
speed$U, the particle completes one revolution, a distance 2 U, in time W = 2 U $U = 2 $. (c)a = gv
gw = $2U cos $w i $2U sin $w j = $2U(cos $w i + sin $w j), so a = $2r. This shows that a is proportional tor and points in the opposite direction (toward the origin). Also, |a| = $2|r| = $2U.
(d) By Newton’s Second Law (see Section 14.4 [ET 13.4]),F = pa, so |F| = p |a| = pU$2= p ($U)
2
U = p |v|
2
U . 3. (a) The projectile reaches maximum height when0 = g|
gw = ggw[(y0sin )w 12jw2] = y0sin jw; that is, when
w = y0sin j and| = (y0sin ) y0sin j 1 2j y0sin j 2 = y20sin2
2j . This is the maximum height attained when the projectile is red with an angle of elevation. This maximum height is largest when = 2. In that case,sin = 1 and the maximum height isy
2 0
2j.
(b) LetU = y20j. We are asked to consider the parabola {2+ 2U| U2= 0 which can be rewritten as | = 1
2U{2+ U2. The points on or inside this parabola are those for whichU { U and 0 | 1
2U{2+ U2. When the projectile is red at angle of elevation, the points ({> |) along its path satisfy the relations { = (y0cos ) w and
| = (y0sin )w 12jw2, where0 w (2y0sin )@j (as in Example 14.4.5 [ET 13.4.5]). Thus
|{| y0cos 2y0sin j =y02 j sin 2 y02 j
= |U|. This shows that U { U. Forw in the specied range, we also have | = wy0sin 12jw= 12jw
2y0sin j w 0 and | = (y0sin )y { 0cos j2 { y0cos 2 = (tan ) { 2y2 j 0cos2{ 2= 1
2U cos2{2+ (tan ) {. Thus
| 1 2U{2+ U2 = 2U cos12{2+ 1 2U{2+ (tan ) { U2 = {2 2U 1 1 cos2 + (tan ) { U 2 = { 2(1 sec2) + 2U (tan ) { U2 2U = (tan2) {2+ 2U (tan ) { U2U 2 = [(tan ) { U]2U 2 0 We have shown that every target that can be hit by the projectile lies on or inside the parabola| = 1
2U{2+ U2. 161
Now let(d> e) be any point on or inside the parabola | = 1
2U{2+ U2. ThenU d U and 0 e 12Ud2+ U2. We seek an angle such that (d> e) lies in the path of the projectile; that is, we wish to nd an angle such that
e = 1
2U cos2d2+ (tan ) d or equivalently e = 12U(tan2 + 1)d2+ (tan ) d. Rearranging this equation we get
d2 2Utan2 d tan + d2 2U+ e
= 0 or d2(tan )2 2dU(tan ) + (d2+ 2eU) = 0 ( ) . This quadratic equation
fortan has real solutions exactly when the discriminant is nonnegative. Now E2 4DF 0 (2dU)2 4d2(d2+ 2eU) 0 4d2(U2 d2 2eU) 0 d2 2eU + U2 0
e 12U(U2 d2) e 1
2Ud2+ U2. This condition is satised since(d> e) is on or inside the parabola | = 12U{2+ U
2. It follows that(d> e) lies in the path of the projectile when tan satises ( ), that is, when tan = 2dU ± s 4d2(U2 d2 2eU) 2d2 = U ± U2 2eU d2 d .
(c) If the gun is pointed at a target with heightk at a distance G downrange, then tan = k@G. When the projectile reaches a distance G downrange (remember we are assuming that it doesn’t hit the ground rst), we haveG = { = (y0cos )w, sow = G
y0cos and| = (y0sin )w 1 2jw2= G tan jG 2 2y2 0cos2 .
Meanwhile, the target, whose{-coordinate is also G, has fallen from height k to height k 1
2jw2= G tan jG 2
2y2 0cos2
. Thus the projectile hits the target.
5. (a)a = j j v = v0 jw j = 2 i jw j s = s0+ 2w i 21jw2j = 3=5 j + 2w i 12jw2j
s = 2w i +3=5 1 2jw2
j. Therefore | = 0 when w =s7@j seconds. At that instant, the ball is 2s7@j 0=94 ft to the right of the table top. Its coordinates (relative to an origin on the oor directly under the table’s edge) are(0=94> 0). At impact, the velocity isv = 2 i 7j j, so the speed is |v| =4 + 7j 15 ft@s.
(b) The slope of the curve whenw = u 7 j isg|g{ = g|@gwg{@gw = jw2 =j s 7@j 2 = 7j 2 . Thuscot = 7j 2 and 7=6.
(c) From (a),|v| =4 + 7j. So the ball rebounds with speed 0=84 + 7j 12=08 ft@s at angle of inclination 90 82=3886. By Example 14.4.5 [ET 13.4.5], the horizontal distance traveled between bounces is
g = y02sin 2
j , wherey0 12=08 ft@s and 82=3886. Therefore,g 1=197 ft. So the ball strikes the oor at about2s7@j + 1=197 2=13 ft to the right of the table’s edge.
PROBLEMS PLUS ¤ 163
7. The trajectory of the projectile is given byr(w) = (y cos )w i +(y sin )w 12jw2j, so v(w) = r0(w) = y cos i + (y sin jw) j and
|v(w)| =t(y cos )2+ (y sin jw)2=sy2 (2yj sin ) w + j2w2=
v j2 w2 2y j (sin ) w + y 2 j2 = j v w yjsin 2 + yj22 yj22sin2 = j v w yjsin 2 + yj22cos2
The projectile hits the ground when(y sin )w 12jw2= 0 w = 2yj sin , so the distance traveled by the projectile is O() =](2y@j) sin
0 |v(w)| gw = ] (2y@j) sin 0 j v w yjsin 2 + yj22 cos2 gw = j 5 7w (y@j) sin 2 v w yjsin 2 + y jcos 2 + [(y@j) cos ]2 2ln 3 Cw y jsin + v w yjsin 2 + y jcos 24 D 6 8 (2y@j) sin 0
[using Formula 21 in the Table of Integrals]
= j2 5 7y jsin v y jsin 2 + y jcos 2 + y jcos 2 ln 3 Cy jsin + v y jsin 2 + y jcos 24 D + y jsin v y jsin 2 +y jcos 2 y jcos 2 ln 3 Cy jsin + v y jsin 2 +y jcos 24 D 6 8 = j2 y jsin · yj+ y 2 j2 cos2 ln y jsin + yj + yjsin · yj yj22 cos2 lny jsin + yj
= yj2sin + y2j2 cos2 ln (y@j) sin + y@j
(y@j) sin + y@j
= yj2 sin + y2j2 cos2 ln1 + sin
1 sin
We want to maximizeO() for 0 @2. O0() = y2 j cos + y 2 2j cos2 · 1 sin
1 + sin ·(1 sin )2 cos 2 2 cos sin ln
1 + sin 1 sin = yj2 cos + y2j2 cos2 · 2
cos 2 cos sin ln 1 + sin 1 sin = yj2 cos + yj2 cos 1 sin ln 1 + sin 1 sin = yj2cos 2 sin ln 1 + sin 1 sin
O() has critical points for 0 ? ? @2 when O0() = 0 2 sin ln1 + sin 1 sin
= 0 (since cos 6= 0). Solving by graphing (or using a CAS) gives 0=9855. Compare values at the critical point and the endpoints: O(0) = 0, O( @2) = y2@j, and O(0=9855) 1=20y2@j. Thus the distance traveled by the projectile is maximized
