This document contains solutions to exercises from Rogawski's Calculus textbook. It provides vector parametrizations of curves, determines whether curves intersect or collide, evaluates limits related to derivatives of vector-valued functions, finds solutions to differential equations, and applies the chain rule.

1. 1(ALU) 1IM1 ANGEL DAVID ORTIZ RESENDIZ AD2014-CALVEC-INDUSTRIAL-JLLF
Assignment CURVAS due 09/07/2014 at 11:54pm CDT
1. (1 pt) From Rogawski ET 2e section 13.1, exercise 27.
Use cos(t) and sin(t), with positive coefficients, to parame-trize
the intersection of the surfaces x2+y2 = 4 and z = 5x2.
r(t) = h , , i
Solution:
The points on the cylinder x2 +y2 = 4 and on z = 5x2 can be
written in the form:
x2+y2 = 4
!(2cost
;2sint; z)
z = 5x2 !
x;y;5x2The points (x;y; z) on the intersection curve must satisfy the
following equations:
x = 2cost
y = 2sint
z = 5x2 = 5(2cost)2
We
obtain the vector parametrization:
r(t) =
2cos t;2sin t;5(2cos t)2Answer(s) submitted:
2cost
(correct)
Correct Answers:
2*cos(t); 2*sin(t); 5*[2*cos(t)]ˆ2
2. (1 pt) From Rogawski ET 2e section 13.1, exercise 37.
Find a parametrization, using cos(t) and sin(t), of the follo-wing
curve:
The intersection of the plane y=7 with the sphere x2+y2+z2 =
74
r(t) = h , , i
Solution: Substituting y = 7 in the equation of the sphere
gives:
x2+(7)2+z2 = 74)x2+z2 = 25
This circle in the horizontal plane y = 7 has the parametrization
x =
p
25cost; z =
p
25sint.Therefore, the points on the inter-section
of the plane
y=7 and the sphere x2+y2+z2 =74, can be written in the form
(5cost;7;5sint), yielding the following parametrization:
r(t) = h5cos t;7;5sin ti
Answer(s) submitted:
5sint
(correct)
Correct Answers:
5*cos(t); 7; 5*sin(t)
3. (1 pt) From Rogawski ET 2e section 13.1, exercise 27.
Use cos(t) and sin(t), with positive coefficients, to parame-trize
the intersection of the surfaces x2+y2 = 16 and z = 7x4.
r(t) = h , , i
Solution:
The points on the cylinder x2 +y2 = 16 and on z = 7x4 can be
written in the form:
x2+y2 = 16
!(4cost
;4sint; z)
z = 7x4 !
x;y;7x4The points (x;y; z) on the intersection curve must satisfy the
following equations:
x = 4cost
y = 4sint
z = 7x4 = 7(4cost)4
We
obtain the vector parametrization:
r(t) =
4cos t;4sin t;7(4cos t)4Answer(s) submitted:
4cost
(correct)
Correct Answers:
4*cos(t); 4*sin(t); 7*[4*cos(t)]ˆ4
4. (1 pt) From Rogawski ET 2e section 13.1, exercise 37.
Find a parametrization, using cos(t) and sin(t), of the follo-wing
curve:
The intersection of the plane y=5 with the sphere x2+y2+z2 =
125
r(t) = h , , i
Solution: Substituting y = 5 in the equation of the sphere
gives:
x2+(5)2+z2 = 125)x2+z2 = 100
This circle in the horizontal plane y = 5 has the parametrization
x =
p
100cost; z =
p
100sint.Therefore, the points on the inter-section
of the plane
y = 5 and the sphere x2 +y2 +z2 = 125, can be written in the
form (10cost;5;10sint), yielding the following parametriza-tion:
r(t) = h10cos t;5;10sin ti
Answer(s) submitted:
10cost
(correct)
Correct Answers:
10*cos(t); 5; 10*sin(t)
1

2. 5. (1 pt) From Rogawski ET 2e section 13.1, exercise 14.
The function r (t) traces a circle. Determine the radius, cen-ter,
and plane containing the circle
r (t) = 7i+ (7cos(t)) j+ (7sin(t))k
Plane : x =
Circle’s Center : ( , , )
Radius :
Solution: We have:
x(t) = 7;y(t) = 7cos(t); z(t) = 7sin(t)
Hence,
y(t)2 + z(t)2 = 49cos2(t) + 49sin2(t) = 49(cos2(t) +
sin2(t)) = 49
This is the equation of a circle in the vertical plane x = 7.
pThe circle is centered at the point (7;0;0) and its radius is
49 = 7
Answer(s) submitted:
7
7
0
0
7
(correct)
Correct Answers:
7
7
0
0
7
6. (1 pt) From Rogawski ET 2e section 13.1, exercise 5.
Find a vector parametrization of the line through P =
(7;5;3) in the direction v = h8;7;5i
r(t) =
( + t ) i+
( + t ) j+
( + t ) k
Solution: We use the vector parametrization of the line to
obtain:
r (t) = O˜P + tv = h7;5;3i + t h8;7;5i =
h78t;5+7t;35ti
or in the form:
r(t) = (78t) i+(5+7t) j+(35t)k
Answer(s) submitted:
-7
-8
5
7
-3
-5
(correct)
Correct Answers:
-7
-8
5
7
-3
-5
7. (1 pt) From Rogawski ET 2e section 13.1, exercise 30.
Determine
whether
r1 and
r2 collide or intersect:
r1(t) =
t; t2; t3r2(t) =
(5t+7) ; 16t2;216
The two paths
A. do not collide
B. collide
The two paths
A. do not intersect
B. intersect
Solution:
T
he two
parts collide if there exists a value of t such that:
t; t2; t3=
(5t +7) ;16t2;216
Equating corresponding components we obtain the following
equations:
t = (5t +7)
t2 = 16t2
t3 = 216
The second equation implies that t = 0, but this value does not
satisfy the other equations. Therefore, the equations have no so-lution,
which means that the paths do not collide.
The two paths intersect if there exist values of t and s such
that:
s; s2; s3=
(5t +7) ;16t2;216
Or equivalently:
s = (5t +7)
s2 = 16t2
s3 = 216
The second equation implies that s = 4t, and the third equa-tion
implies s = 6. Hence t = 1;5
We can see that both solutions don’t satisfy the first equation,
hence the two paths do not intersect.
Answer(s) submitted:
A
A
(correct)
Correct Answers:
A
A
8. (1 pt) From Rogawski ET 2e section 13.1, exercise 28.
Given two paths r1 and r2. such that for each t1 and t2
r1(t1)6= r2(t2).
The two paths
A. intersect and collide
B. intersect but do not collide
C. do not intersect
D. collide but do not intersect
2

3. Solution:
Intersection is a geometric property of the curves. Collision de-pends
on the actual parametrization. By definition the two paths
do not intersect.
Answer(s) submitted:
C
(correct)
Correct Answers:
C
9. (1 pt) From Rogawski ET 2e section 13.1, exercise 9.
Which of the space curves above describes the vector-valued function: r(t) = hcos t; sin t;cos t sin 12ti?
Enter number of figure using one of 1 through 6.
Solution:
The curve describing the given vector-valued function is the curve in figure 3.
Answer(s) submitted:
3
(correct)
Correct Answers:
3
10. (1 pt) From Rogawski ET 2e section 13.1, exercise 33.
Find a parametrization of the line through the origin whose projection on the xy-plane is a line of slope 5
and on the yz-plane is a line of slope 8 (i.e., Dz=Dy = 8).
Scale your answer so that the smallest coefficient is 1.
r(t) = h t; t; ti
Solution: We denote by (x;y; z) the points on the line.
The projection of the line on the xy-plane is the line through the origin having slope 5,
that is the line y = 5x in the xy-plane. The projection of the line
on the yz-plane is the line through the origin with slope 8,
that is the line z = 8y. Thus, the points on the desired line satisfy
the following equalities:
y = 5x
)y = 5x; z = 8 5x = 40x
z = 8y
We conclude that the points on the line are all the points in the form (x;5x;40x)
Using x = t as parameter we obtain the following parametrization:
r(t) = ht; 5t; 40ti .
Answer(s) submitted:
8cos(5s)
5
8sin(5s)
(score 0.333333343267441)
Correct Answers:
1
3

4. 13. (1 pt) From Rogawski ET 2e section 13.2, exercise 5.
Evaluate the limit:
l´ımhr(t+h)r(t)
!0
h for r(t) =
t6; sin t;3
r(t) = h , , i
Solution: This limit is the derivative dr
dt . Using componentwise differentiation yields:
l´ımh!0
r(t+h)r(t)
h = dr
dt =
d
dt (t6); d
dt (sint); d
dt (3)
=
D
6
E
t7 ;cost;0
Answer(s) submitted:
-6tˆ(-7)
cost
0
(correct)
Correct Answers:
-6*tˆ(-7)
cos(t)
0
14. (1 pt) From Rogawski ET 2e section 13.2, exercise 50.
Find the solution r(t) of the differential equation with the given initial condition:
r0(t) = hsin 5t; sin 5t; 9ti ;r(0) = h6;9;8i
r(t) = h ; ; i
Solution: We first integrate the vector r0(t) to find the general solution:
r(t) =
Z
hsin 5t; sin 5t; 9ti dt
=
Z
sin5tdt;
Z
sin5tdt;
Z
9tdt
=
1
5
cos5t;
1
5
cos5t;
9
2
t2
+c
Substituting the initial condition we obtain:
r(0) =
1
5
cos0;
1
5
cos0;
9
2
02
+c
= h6;9;8i =
1
5
;
1
5
+c
;0
Hence,
c = h6;9;8i
1
5
;
1
5
=
;0
31
5
;
46
5
;8
Hence the solution to the differential equation with the given initial condition is:
r(t) =
1
5
cos 5t;
1
5
cos 5t;
9
2
t2
+
31
5
;
46
5
;8
=
1
5
(31cos5t) ;
1
5
(46cos5t) ;8+
9
2
t2
Answer(s) submitted:
(-cos(5t))/5+31/5
(-cos(5t))/5+46/5
(9tˆ2)/2+8
(correct)
Correct Answers:
6.2-[cos(5*t)]/5
9.2-[cos(5*t)]/5
8+4.5*tˆ2
4

5. 15. (1 pt) From Rogawski ET 2e section 13.2, exercise 25.
Evaluate d
dt r(g(t)) using the Chain Rule:
r(t) =
et;e2t;6
;g(t) = 8t+4
d
dt r(g(t)) = h , , i
Solution: We first differentiate the two functions:
r0(t) = d
dt
et;e2t;6
=
et;2e2t;0
g0(t) = d
dt (8t +4) = 8
Using the Chain Rule we get:
d
dt r(g(t)) = g0(t)r0(g(t)) = 8
D
e8t+4;2e2(8t+4);0
E
=
D
8e8t+4;16e2(8t+4);0
E
Answer(s) submitted:
8 eˆ(8 t+4)
16 eˆ(16 t+8)
0
(correct)
Correct Answers:
8*eˆ(8*t+4)
16*eˆ[2*(8*t+4)]
0
16. (1 pt) From
Rogawski ET 2e section 13.2, exercise 27.
Let r(t) =
t2;1t; 4t
. Calculate the derivative of r(t) a(t) at t = 9,
assuming that a(9) = h4;3;2i and a0(9) = h5;1;7i
d
dt r(t) a(t)jt=9 =
Solution: By the Product Rule for dot products we have
d
dt r(t) a(t) = r(t) a0(t)+r0(t) a(t)
At t = 9 we have
d
dt r(t) a(t)jt=9 = r(9) a0(9)+r0(9) a(9)
We compute the derivative r0(9) :
r0(t) = d
dt
t2;1t; 4t
= h2t;1;4i)r0(9) = h18;1;4i
Also, r(9) =
92;19;4 9
= h81;8;36i. Substituting the vectors in the equation above, we obtain:
d
dt r(t) a(t)jt=9 = h81;8;36i h5;1;7i+h18;1;4i h4;3;2i
= (4058+252)+(72+3+8) = 588
The derivative of r(t) a(t) at t = 9 is 588 .
Answer(s) submitted:
588
(correct)
Correct Answers:
588
17. (1 pt) From Rogawski ET 2e section 13.2, exercise 17.
Use the
appropriate
Product Rule to evaluate the derivative, where
r1(t) =
10t;7;t6;r2(t) = h4;et;8i
d
(dt r1(t) r2(t)) =
Solution: By the Product Rule for dot products we have:
d
dt r1 r2 = r1 r0
2+r0
1 r2
We compute the derivatives of r1 and r2:
r0
1 = d
dt
10t;7;t6
=
10;0;6t5
r0
2 = d
dt h4;et;8i = h0;et;0i
Then,
d
dt r1(t) r2(t) =
10t;7;t6
h0;et;0i+
10;0;6t5
h4;et;8i
= 7et +48t540
Answer(s) submitted:
48tˆ5+7eˆt-40
5

6. (correct)
Correct Answers:
7*eˆt+48*tˆ5-40
18. (1 pt) From Rogawski ET 2e section 13.2, exercise 50.
Find the solution r(t) of the differential equation with the given initial condition:
r0(t) = hsin 6t; sin 3t; 9ti ;r(0) = h7;4;4i
r(t) = h ; ; i
Solution: We first integrate the vector r0(t) to find the general solution:
r(t) =
Z
hsin 6t; sin 3t; 9ti dt
=
Z
sin6tdt;
Z
sin3tdt;
Z
9tdt
=
1
6
cos6t;
1
3
cos3t;
9
2
t2
+c
Substituting the initial condition we obtain:
r(0) =
1
6
cos0;
1
3
cos0;
9
2
02
+c
= h7;4;4i =
1
6
;
1
3
+c
;0
Hence,
c = h7;4;4i
1
6
;
1
3
=
;0
43
6
;
13
3
;4
Hence the solution to the differential equation with the given initial condition is:
r(t) =
1
6
cos 6t;
1
3
cos 3t;
9
2
t2
+
43
6
;
13
3
;4
=
1
6
(43cos6t) ;
1
3
(13cos3t) ;4+
9
2
t2
Answer(s) submitted:
(-cos(6t))/6+43/6
(-cos(3t))/3+13/3
(9tˆ2)/2+4
(correct)
Correct Answers:
7.16667-[cos(6*t)]/6
4.33333-[cos(3*t)]/3
4+4.5*tˆ2
19. (1 pt) From Rogawski ET 2e section 13.2, exercise 46.
Evaluate the integral:
Z t
0
(7si+21s2j+2k)ds
Answer : i+ j+ k
Solution: We first compute the integral of each component:
Z t
0
7sds =
7
2
s2

10. t
0
=
7
2
t2
Z t
0
21s2 ds =
21
3
s3

14. t
0
= 7t3
6

15. Z t
0
2ds = 2s

19. t
0
= 2t
Hence,
Z t
0
(7si+21s2j+2k)ds =
Z t
0
7sds
i+
Z t
0
21s2 ds
j+
Z t
0
2ds
k
=
7
2
t2
i+
7t3
j+(2t)k
Answer(s) submitted:
(7tˆ2)/2
21tˆ3/(3)
2t
(correct)
Correct Answers:
3.5*tˆ2
7*tˆ3
2*t
20. (1 pt) From Rogawski ET 2e section 13.4, exercise 29.
For a plane curve r(t) = hx(t);y(t)i,
k(t) = jx0(t)y00(t)x00(t)y0(t)j
(x0(t)2+y0(t)2)3=2 :
Use this equation to compute the curvature at the given point.
r(t) = h2t4;4t4i; t = 2:
k(2) =
Solution:
We quickly compute
Function Formula at t = 2
x0(t) 2 4t3 64
x00(t) 2 4 3t2 96
y0(t) 4 4t3 128
y00(t)
4 4 3t2
192
And so
k(2) = j(64)(192)(96)(128)j
((64)2+(128)2)(3=2) :
Answer(s) submitted:
0
(correct)
Correct Answers:
abs((64)*(-192)-(96)*(-128))/(((64))ˆ2+((-128))ˆ2)ˆ(3/2)
21. (1 pt) From Rogawski ET 2e section 13.4, exercise 17.
Find the curvature of the plane curve
y = 4t2
at the point t = 1.
k(1) =
Solution:
By the curvature of a graph of y = f (t) in the plane, we have:
k(t) = j f 00(t)j
(1+( f 0(t))2)3=2 :
7

20. Here f (t) = 4t2, so f 0(t) = 4 2t and f 00(t) = 8. Hence
k(t) =
8
1+((8)t)2
1;5 :
We evaluate this at t = 1 to obtain
k(1) = 8 10
(1+64 12)3=2
0;0152658:
Answer(s) submitted:
8/(65ˆ(3/2))
(correct)
Correct Answers:
0.0152658
22. (1 pt) From Rogawski ET 2e section 13.4, exercise 10.
Calculate k(t) when
r(t) = h3t1;5;6ti
k(t) = .
Solution:
By the formula for curvature we have
k(t) = jjr0(t)r00(t)jj
jjr0(t)jj3 :
We now find
r0(t) = h3t2;0;6i
then
r00(t) = h6t3;0;0i
and so their cross product is
r0(t)r00(t) =

26. i j k
3t2 0 6
6t3 0 0

32. =

36. 0 6
0 0

40. i

44. 3t2 6
6t3 0

48. j+

52. 3t2 0
6t3 0

56. k
= h0;36t3;0i:
We compute the necessary lengths:
jjr0(t)r00(t)jj = 36jtj3
jjr0(t)jj =
p
(3t2)2+02+62 =
p
9t4+36
leaving
k(t) =
36jtj3
(9t4+36)3=2
which can be simplified to
36jtj3
(9+36t4)3=2 :
Answer(s) submitted:
((36t)/(tˆ4))/((sqrt(((-3/(tˆ2))ˆ2)+36))ˆ3)
(incorrect)
Correct Answers:
36*|t|ˆ(-3)/([9*tˆ(-4)+36]ˆ1.5)
8

57. 23. (1 pt) From Rogawski ET 2e section 13.4, exercise 3.
Calculate r0(t) and T(t), where
r(t) = h4t 4;3t;8+5ti:
r0(t) = h ; ; i.
T(t) = h ; ; i.
Solution:
We first find
r0(t) = h4;3;5i
and so
jjr0(t)jj =
q
42+(3)2+52 =
p
50:
The unit tangent vector is therefore:
T(t) = r0(t)
jjr0(t)jj
=
1
p
50
h4;3;5i =
4
p
50
;
3
p
50
;
5
p
50
:
We see that the unit tangent is constant, since the curve is a straight line.
Answer(s) submitted:
4
-3
5
(2sqrt(2))/5
(-3sqrt(2))/10
sqrt(2)/2
(correct)
Correct Answers:
4
-3
5
0.565685
-0.424264
0.707107
24. (1 pt) From Rogawski ET 2e section 13.4, exercise 1.
Calculate r0(t), T(t), and T(4) where
r(t) = h4t2; ti:
r0(t) = h ; i.
T(t) = h ; i.
T(4) = h ; i.
Solution:
We differentiate r(t) to obtain:
r0(t) = h4 2t;1i
and so
jjr0(t)jj =
q
(8t)2+1:
We now find the unit tangent vector:
T(t) = r0(t)
jjr0(t)jj
=
q 1
(8t)2+1
h4 2t;1i:
For t = 4 we obtain the vector:
T(4) =
1
p
1025
h32;1i =
32
p
1025
;
1
p
1025
:
Answer(s) submitted:
8t
1
(8t)/(sqrt((8t)ˆ2+1))
1/(sqrt((8t)ˆ2+1))
9

58. 32/(5sqrt(41))
1/(5sqrt(41))
(correct)
Correct Answers:
4*2*t
1
8*t/[sqrt((8*t)ˆ2+1)]
1/[sqrt((8*t)ˆ2+1)]
0.999512
0.0312348
25. (1 pt) From Rogawski ET 2e section 13.3, exercise 12.
Find the speed: r(t) = hcosh(t) ;cosh(t) ;ti at t = 4.
v(4) =
Solution:
The velocity vector is r0(t) = hsinh(t) ; sinh(t) ;1i.
At t = 4 the velocity p
is r0(4) = h27;2899;27;2899;1i, hence the speed is
v(4) = jjr0(4)jj =
(27;2899)2+(27;2899)2+(1)2 = 38;6067
Answer(s) submitted:
sqrt(sinhˆ2(4)+sinhˆ2(4)+1)
(correct)
Correct Answers:
38.6067
26. (1 pt) From Rogawski ET 2e section 13.3, exercise 6.
Compute the length of the curve r(t) = 5ti+8tj+(t2+2)k over the interval 0 t 4
Hint: use the formula Z p
t2+a2 dt =
1
2
t
p
t2+a2+
1
2
a2 ln
t +
p
t2+a2
+C
L =
Solution: The derivative of r(t) is r0(t) = 5i+8j+2tk.Using the Arc Length Formula we get:
L =
Z 4
0

62. jr0(t)
jdt =
Z 4
0
q
52+82+(2t)2 dt =
Z 4
0
p
4t2+89dt
We substitute u = 2t; du = 2dt and use the given integration formula. This gives:
L =
1
2
Z 8
0
p
u2+89du =
1
4
u
p
u2+89+
1
4
89ln
u+
p
u2+89

66. 8
0
=
1
4
p
82+89+
8
89
4
ln
8+
p
82+89
89
4
ln
p
89
=
4
2
p
153+
89
4
ln
8+
p
153
89
4
ln
p
89 =
4
2
p
153+
89
4
ln
p
153
p
89
8+
41;86
Answer(s) submitted:
41.8647
(correct)
Correct Answers:
41.8647
10

67. 27. (1 pt) From Rogawski ET 2e section 13.3, exercise 1.
Compute the length of the curve r(t) = h9t; 4t1;2t3i over the interval 0 t 8
L =
Solution: We have x(t) = 9t;y(t) = 4t 1; z(t) = 2t 3 hence
x0(t) = 9; y0(t) = 4; z0(t) = 2
We use the Arc Length Formula to obtain:
L =
Z 8
0
k(r0(t)kdt =
Z 8
0
q
x0(t)2+y0(t)2+z0(t)2 dt
=
Z 8
0
q
(9)2+42+(2)2 dt = 8
p
101
Answer(s) submitted:
80.399
(correct)
Correct Answers:
80.399
28. (1 pt) From Rogawski ET 2e section 13.3, exercise 24.
Find an arc length parametrization of r(t) = het sin t;et cos t;1eti
r1(s) = h , , i
Solution:
An arc length parametrization is r1(s) = r(f(s)) where t = f(s) is the inverse of the arc length function. We compute the arc length
function:
s(t) =
Rt0
jjr0(u)du
Differentiating r(t) and computing the norm of r0(t) gives:
r0(t) = het sin t+et cos t;et cos tet sin t;1eti
= et hsint +cost;cost sint;1i
p
jjr0(t)jj = et
(sin t+cos t)2+(cos tsin t)2+12 = et
q
2(sin2 t+cos2 t)+1 =
p
3et
Thus,
s(t) =
Rt0
p
3eudu =
p
3 eujt
0 =
p
3(et 1)
We find the inverse function of s(t) by solving s =
p
3(et 1) for t. We obtain:
ps
3 = et 1
et = 1+ ps
3
t = f(s) = ln
1+ ps
3
D An arc length parametrization for r1(s) = r(f(s)) is:
eln(1+0;57735s) sin(ln(1+0;57735s)) ;eln(1+0;57735s) cos(ln(1+0;57735s)) ;eln(1+0;57735s)
E
= (1+0;57735s) hsin (ln(1+0;57735s)) ;cos (ln(1+0;57735s)) ;1i
Answer(s) submitted:
(incorrect)
Correct Answers:
(1+0.57735*s)*sin(ln(1+0.57735*s))
(1+0.57735*s)*cos(ln(1+0.57735*s))
1+0.57735*s
11

68. 29. (1 pt) From Rogawski ET 2e section 13.3, exercise 10.
Find the
speed at the given
value of t:
r(t) =
et7;3; 7t1; t = 7
v(7) =
Solution: The velocity vector is
et7;0;7t2
and at t = 7
r0(7) =
e77;0;7 72
=
1;0;
1
7
:
The speed is the magnitude of the velocity vector, that is,
v(7) = kr0(7)k =
s
12+02+
1
7
2
=
r
50
49
1;01
Answer(s) submitted:
1.01015
(correct)
Correct Answers:
1.01015
30. (1 pt) From Rogawski ET 2e section 13.3, exercise 23.
Find a path that traces the circle in the plane y = 3 with radius r = 4 and center (2;3;0) with constant speed 12.
r1(s) = h , , i
Solution:
We start with the following parametrization of the circle:
r(t) = h2;3;0i+4hcos t;0; sin ti = h2+4cos t;3;0+4sin ti
We need to reparametrize the curve by making a substitution on t = f(s), so that the new parametrization r1(s) = r(f(s)) satisfies
jjr0
1(s)jj = 12 for all s. We find r0
1 using the Chain Rule:
r0
1(s) = d
ds r(f(s)) = f0(s)r0(f(s))
Next, we differentiate r(t) and then replace t by f(s):
r0(t) = h4sin t;0;4cos ti
r0(f(s)) = h4sin(f(s));0;4cos(f(s))i
Now, we get: r0
1(s) = f0(s) h4sin(f(s));0;4cos(f(s))i = 4f0(s) hsin(f(s));0;cos(f(s))i
Hence,
jjr0
p
(sin(s))2+(cos(s))2 = 4jf0(s)j
1(s)jj = 4jf0(s)j
1(s)jj = 12 for all s, we choose f0(s) = 3.
To satisfy jjr0
We may take the antiderivative of f(s) = 3s and obtain the following parametrization:
r1(s) = r(f(s)) = r(3s) = h2+4cos(3s);3;0+4sin(3s)i
Answer(s) submitted:
4cos(3s)+2
-3
4sin(3s)
(correct)
Correct Answers:
2+4*cos(3*s)
-3
4*sin(3*s)
12

69. 31. (1 pt) From Rogawski ET 2e section 13.3, exercise 3.
Compute the length of the curve r(t) =
2t; ln t; t2
over the interval 1 t 4
L =
Solution: The derivative of r(t) is r0(t) =
2; 1t
; 2t
. We use the Arc Length Formula to obtain:
L =
Z 4
1
kr0(t)kdt =
Z 4
1
s
22+
1
t
2
+(2t)2 dt
=
Z 4
1
s
2t +
1
t
2
dt =
Z 4
1
2t +
1
t
dt = t2+lnt

73. 4
1
= (16+ln4)(1+ln1) = 15+ln4
Answer(s) submitted:
16.3863
(correct)
Correct Answers:
16.3863
32. (1 pt) From Rogawski ET 2e section 13.3, exercise 29.
Evaluate s(t) =
Rt ¥ jjr0(u)jjdu for the Bernoulli spiral r(t) = het cos(3t);et sin(3t)i.
It is convenient to take ¥ as the lower limit since s(¥) = 0. Then use s to obtain an arc length parametrization of r(t).
r1(s) = h , i
Solution:
We differentiate r(t) and compute the norm of the derivative vector. This gives:
r0(t) = het cos(3t)3et sin(3t);et sin(3t)+3et cos(3t)i
= et hcos(3t)p
3sin(3t); sin(3t)+3cos(3t)i
jjr0(t)jj = et
(cos(3t)3sin(3t))2+(sin(3t)+3cos(3t))2
= et
q
cos2(3t)+sin2(3t)+9(cos2(3t)+sin2(3t)) =
p
10et
We now evaluate the improper integral:
s(t) =
Rt ¥ jjr0(u)jjdu
= l´ım
R!¥
RtR
p
10eudu = l´ım
R!¥
p
10eu

75. t
R = l´ım
R!¥
p
10
et eR
=
p
10et
An arc length parametrization of r(t) is r1(s) = r(f(s)) where t = f(s) p
is the inverse function of s(t).
We find t = f(s) by solving s =
10et for t:
t = f(s) = ln ps
10
D An arc length parametrization of r(t) is:
s
3ln
3;16228 cos
s
3;16228
3ln
; s
3;16228 sin
s
3;16228
E
Answer(s) submitted:
(incorrect)
Correct Answers:
s/3.16228*cos(3*ln(s/3.16228))
s/3.16228*sin(3*ln(s/3.16228))
13

76. Generated by
c WeBWorK, http://webwork.maa.org, Mathematical Association of America
14