Answers and Solutions to Section 10.3 Homework
Problems 1, 3, 7-27 (odd), 35, 37, and 39
S. F. Ellermeyer
1.
The curve rt  2 sint, 5t, 2 cost , 10  t  10 (which is a piece of a helix)
has length
10
10
 10 |r  t| dt   10

2 cost 2  5 2  2 sint 2 dt
10
 10
29 dt
 20 29 .
3.
2 t, e t , e t , 0  t  1 has length
The curve rt 
1
1
 0 |r  t| dt   0
2
1
2
 e t  2  e t  2 dt

0

0

 0 e t  e t  dt
e t  2  2e t e t  e t  2 dt
1
e t  e t  2 dt
1
 e t  e t | t1
t0
 e 1  e 1   e 0  e 0 
 e  e 1   1  1
 e  e 1 .
7.
For rt  e t sinti  e t costj, we have
r  t  e t cost  sinti  e t  sint  costj
and
|r  t| 

e t cost  sint 2  e t  sint  cost 2
2e 2t
 2 et.
Thus, the arc length function starting from t  0 and measured in the
direction of increasing t is
s

t
 0 |r  u| du
t
0
2 e u du
 2 e t  1.
Solving the equation 2 e t  1  s for t, we obtain
1
t  ln 1  1 s .
2
Thus, using arc length (s) as our parameter, we can write the equation of this
curve as
1  1 s sin ln 1  1 s
2
2
i
 1  1 s cos ln 1  1 s
2
2
j.
qs 
9.
For rt  3 sinti  4tj 3 costk, we have
r  t  3 costi  4j  3 sintk
and
|r  t| 
3 cost 2  4 2  3 sint 2  5
Thus, the arc length function starting from t  0 and measured in the
direction of increasing t is
s

t
 0 |r  u| du
t
 0 5 du
 5t.
Thus, using arc length (s) as our parameter, we can write the equation of this
curve as
qs  3 sin 1 s i  4 sj 3 cos 1 s k
5
5
5
11. For rt  2 sint, 5t, 2 cost , the tangent vector is
r  t  2 cost, 5, 2 sint 
and the length of this tangent vector is
|r  t| 
2 cost 2  5 2  2 sint 2 
The unit tangent vector is
Tt   1 r  t 
|r t|
29 .
1 2 cost, 5, 2 sint .
29
Note that
T  t 
1 2 sint, 0, 2 cost 
29
and the length of this vector is
|T  t|  1
2 sint 2  0 2  2 cost 2 
29
Thus, the unit normal vector is
2 .
29
2
Nt 

1 T  t
|T t|

29
2
1 2 sint, 0, 2 cost 
29
  sint, 0,  cost .
In order to find the curvature using the formula
  dT ,
ds
we must first find the relationship between the parameter t and the arc length
s. This is similar to the work done in problem 1 of this exercise set. We obtain
s  29 t.
Now note (by the Chain Rule) that
dT  ds dT
dt ds
dt
which means that
dT 
ds
dT
dt
ds
dt
1
1 2 sint, 0, 2 cost 
29
29
 1 2 sint, 0, 2 cost 
29
  2 sint, 0, cost .
29

We now obtain
dT  2 .
29
ds
Since this curve is a helix, it makes sense that it has constant curvature.
Also, since we did all of the work of computing |T  t| and |r  t|, it is much
easier to compute the curvature using the formula

|T  t|

 
|r t|
2
29
 2
29
29
since that way we don’t have to deal with s at all.
13. For rt  13 t 3 , t 2 , 2t , the tangent vector is
r  t  t 2 , 2t, 2 
and the length of this tangent vector is
3
2
t 2   2t 2  2 2
|r  t| 

t 4  4t 2  4

2
t 2  2
 t 2  2.
The unit tangent vector is
1 r  t
|r  t|
 2 1 t 2 , 2t, 2 
t 2
t 2 , 2t , 2

.
2
t  2 t2  2 t2  2
Tt 
Note that
4  2t 2 , 4t
4t
,
2
2
2
t 2  2 t 2  2 t 2  2
and the length of this vector is
T  t 
4t
2
t 2  2

|T t| 
1
2
t  2
1

2
t 2  2

2
2

4  2t 2
2
t 2  2
2

4t
2
t 2  2
2
16t 2  16  16t 2  4t 4   16t 2
4t 4  16t 2  16
2
2
t 2  2
2
t  2
 22 .
t 2
Thus, the unit normal vector is
1 T  t
Nt 

|T t|

2
2
 t 2
2
2
4t
, 4  2t 2 , 4t 2
2
2
t  2 t  2 t 2  2
2
2t , 2  t 2 , 2t .
t  2 t2  2 t2  2
The curvature using the formula

2
2
|T  t|
2
2
 2t 2 
 
.
2
2
|r t|
t 2
t  2
15. For rt  t 2 i  tk we have
4
r  t  2ti  k
2t 2  1 2 
|r  t| 
4t 2  1
r  t  2i
r  t  r  t  2ti  k  2i
 4ti  i  2k  i
 2j


|r t  r t|  2.
Thus, the curvature is
|r  t  r  t|
2


.
3/2
|r  t| 3
4t 2  1
17. For rt  sinti  costj  sintk we have
r  t  costi  sintj  costk
|r  t| 

cost 2   sint 2  cost 2
1  cos 2 t
r  t   sinti  costj  sintk
r  t  r  t 
i
j
k
cost
 sint
cost
 sint  cost  sint
 i  k.
|r  t  r  t| 
1 2  1 2 
2.
Thus, the curvature is

19. For rt 
|r  t  r  t|
2

.
3
3/2

|r t|
1  cos 2 t
2 ti  e t j e t k we have
5
r  t 
2 i  e t j  e t k
|r  t| 
2
2
 e t  2  e t  2

e t  2  2e t e t  e t  2

e t  e t  2
 e t  e t
r  t  e t j  e t k
i
r  t  r  t 
j
k
2 e t e t
0
et
e t
 2i  2 e t j  2 e t k.
2 2   2 e t
|r  t  r  t| 
2

2 et

4  2e t  2  2e t  2

2 e t  2  2e t e t  e t  2

2 e t  e t  2

2 e t  e t .
2
Thus, the curvature is
2 e t  e t 
|r  t  r  t|
2



.
3
t
t
t 3

e  e t  2
e  e 
|r t|
Since the point 0, 1, 1 on this curve corresponds to the parameter value
t  0, we see that the curvature at this point is
2
2
0 

.
0
0 2
4
e  e 
21. For the curve fx  x 3 , we have
f  x  3x 2
f  x  6x
so the curvature is
x 
|f  x|

1  f x
2
3/2

6|x|
.
3/2
1  9x 4 
23 For the curve fx  4x 5/2 , we have
f  x  10x 3/2
f  x  15x 1/2
6
so the curvature is
x 
|f  x|
3/2
1  f  x 2
15x 1/2
3/2
1  100x 3  .

25. For the curve fx  e x , we have
f  x  e x
f  x  e x
so the curvature is
x 
|f  x|
1  f  x 2
3/2

ex
3/2
1  e 2x  .
We would like to find at what value of x the curvature is maximum: Since
3/2
1/2
1  e 2x  e x  e x 32 1  e 2x  2e 2x 

 x 
3
1  e 2x 



3/2
1/2
1  e 2x  e x  e x 31  e 2x  e 2x 
1  e 2x 
3
e x 1  e 2x  1/2 1  2e 2x 
1  e 2x 
e x 1  2e 2x 
3
5/2
1  e 2x 
we see that   x  0 when 1  2e 2x  0. Solving this equation for x gives us
e 2x  1/2 or e x  2  1/2 or e x  2 /2 or
2
.
x  ln
2
Since this x gives us the only critical point of  and since it is clear that the
curvature must be maximum at some point for the graph of fx  e x , we
conclude that the curvature is maximum at the point
2
2
ln
,
 0. 34637, 0. 70711.
2
2
7
27. The curvature of the curve (drawn on page 724 of the textbook) seems to be
greater at point P.
35. For rt  t 2 , 23 t 3 , t , we have
r  t  2t, 2t 2 , 1 
|r  t| 
4t 2  4t 4  1
 2 t4  t2  1
4
t2  1
2
 2 t2  1
2
2
 2t  1
Tt   1 r  t
|r t|
2

2
2t , 2t 2 ,
1
2t  1 2t 2  1 2t 2  1
2
4t
4t
2  4t 2 ,
,
2
2
2
2
2
2
2t  1 2t  1 2t  1
The point 1, 23 , 1 corresponds to the parameter value t  1 so
T  t 
8
2, 2, 1
3 3 3
1 T  1
N1 

|T 1|
T1 

1
2
9
2
2
4
9

4
9

 2 , 4 , 4
9 9 9
2
  1 , 2 , 2
3 3 3
and
B1  T1  N1

i
j
k
2
3
 13
2
3
2
3
1
3
 23
  2 i  1 j  2 k.
3
3
3
37. For rt  2 sin3t, t, 2 cos3t , we have
r  t  6 cos3t, 1, 6 sin3t 
|r  t| 
37
Tt  1 6 cos3t, 1, 6 sin3t 
37
T  t  1 18 sin3t, 0, 18 cos3t 
37
The point 0, , 2 corresponds to the parameter value t   so
T  1 6, 1, 0 
37
1 T  
N 

|T |
 0, 0, 1 
and
B  T  N
i


6
37
0
j
k
1
37
0
0
1
1 i  6 j.
37
37
Since the binormal vector is orthogonal to the osculating plane, the
osculating plane has equation

9
x  6y    0
and since the unit tangent vector is orthogonal to the normal plane, the
normal plane has equation
 6x  y    0.
39. The ellipse 9x 2  4y 2  36 can be written as
x 2  y 2  1.
4
9
Parametric equations of this ellipse are
x  2 cost
y  3 sint.
For this parametric curve (r) we have
r  t  2 sint, 3 cost 
|r  t| 
4 sin 2 t  9 cos 2 t
r  t  2 cost, 3 sint 
i
r  t  r  t 
2 sint
j
k
3 cost 0
 6k
2 cost 3 sint 0
|r  t  r  t|  6
|r  t  r  t|


|r  t| 3
6
4 sin t  9 cos 2 t
2
3/2
.
The point 2, 0 on the ellipse corresponds to the parameter value t  0 so at
this point the curvature is
6

 2
3/2
2
9
2
4 sin 0  9 cos 0
The osculating circle at this point thus has radius 9/2 and is centered at the
point  52 , 0 . An equation for the osculating circle is
2
 y 2  81 .
x 5
4
2
Parametric equations of this circle (used to graph the circle) are
x   5  9 cost
2
2
9
y
sint.
2
The point 0, 3 on the ellipse corresponds to the parameter value t  /2 so
at this point the curvature is
6
 3

3/2
2 
4
2 
4 sin 2  9 cos 2
The osculating circle at this point thus has radius 4/3 and is centered at the
10
point 0,
5
3
. An equation for the osculating circle is
2
 16 .
x2  y  5
3
9
Parametric equations of this circle (used to graph the circle) are
x  4 cost
3
y  5  4 sint.
3
3
11