BC - 3
Vector Calculus
Name:
Calculator Allowed.
1.
r
r
r
If v  2 i  3 j , find each of the follow.
a.
3v  6iˆ  9 ˆj
d.
a unit vector in the direction of v.
vˆ 
2.
b.
v  22  (3)2  13
c.
4v  4
 13   4
v
2 ˆ 3 ˆ

i
j
v
13
13
Find the position function R(t) for each acceleration vector and initial conditions.
a.
a(t) = -3t i, v(0) = 2j, r(0) = 4i
3 2
3 2
v (t )   3t , 0 dt 
t  C1 , C2 . Since v (0)  0, 2 ,v (t ) 
t ,2
2
2
Thus,
r (t )  
b.
.
3 2
1 3
1 3
t , 2 dt 
t  C1 , 2t  C2 . Since r (0)  4, 0 ,r (t ) 
t  4, 2t
2
2
2
1
2
1
a(t )  (1  t ) i  et j, v(0)  i  j, r(0)  i  j
3
1
3

2

t 
2
v (t )   (1  t ) i  e j dt   (1  t ) 2  C1  i   et  C2  j. .


3

3
2
5
Since v (0)  v(0)  i  j, v (t )   (1  t ) 2   i   et  2  j.
3
3
Thus,
3
5
 2

 4

5
5
r (t )    (1  t ) 2   i   et  2  j  dt   (1  t ) 2  t  C1  i   et  2t  C2  .
3
3

 15

 3
5
4
1
5
1
Since r(0)  i  j, r (t )   (1  t ) 2  t   i  et  2t j
3
3 15 
 15


13
BC - 3
c.
Vector Calculus
Name:
r
r
r
a (t )  1  cos  2t  , sin(3t ) , v  0   1, 2 and r  0   0, 0
1

v (t )   (1  cos(t ))i  sin(3t )t j dt   t  sin(t )  C1  i   cos(3t )  C2  j. .
3

7
 1
Since v (0)  v(0)  i  2 j, v (t )   t  sin(t )  1 i    cos(3t )   j.
3
 3
Thus,

7 
7
 1
1
  1

r (t )    t  sin(t )  1 i    cos(3t )   j. dt   t 2  cos(t )  t  C1  i    sin(3t )  t  C2  .
3 
3
 3
2
  9


7 
1
  1
Since r (0)  0i  0 j, r (t )   t 2  cos(t )  t  1 i    sin(3t )  t 
3 
2
  9
BC - 3
Vector Calculus
Name:
3. (AP 2010 #3) Set up of all derivatives and integrals should be and shown, you may calculate the
answers using calculator.
A particle is moving along a curve so that its position at time t is given by ( x(t ), y (t )), where
x(t )  t 2  4t  8 and y(t) is not explicitly given. Bot x and y are measured in meters and t is measured in
dy
 tet 3  1 .
seconds. It is known that
dt
a. Find the speed of the particle at time t = 3 seconds.
2
2
 dx   dy 
speed =      
 dt   dt 
At t = 3, speed =
 2t  4 
2
  tet 3  1
2
4   3  1  2 2
2
b. Find the total distance traveled by the particle for 0  t  4 seconds.
4
distance =

0
2
2
 dx   dy 
     dt  
 dt   dt 
0
3
 2t  4 
2
  tet 3  1 dt  11.588
2
c. Find the time t , 0  t  4 , when the line tangent to the path of the particle is horizontal. Is
the direction of the particle to the left or right at this time? Give a reason for your answer.
dy
dy dt tet 3  1
slope =
=

 0,  t  2.208
dx dx
2t  4
dt
d. There is a point with x – coordinate 5 through which the particle passes twice. Find each
of the following:
i.
The two values of t when that occurs.
x(t )  t 2  4t  8  5  (t  3)(t  1)  0  t  1 or 3
ii.
The slopes of the two lines tangent to the particles path at these points.
slope =
iii.
dy
dx
t 1 =
e2  1
dy
,slope =
2
dx
t 3
=
3 1
1
2
The y – coordinate of that point given y (2)  3 
3
1
e
y (1)  y (3)  y (2)   (tet 3  1)dt  6.891(from calculator)
2
BC - 3
4.
Vector Calculus
Name:
If f ( x)  x3  2 x  3 , find a unit tangent vector and a unit normal vector to the graph of (x) at the
point where x  2 .
f ( x)  3x 2  2  f (2)  10 . Then the tangent vector must have y-component 10 times the xcomponent: v  1,10  vˆ 
v

v
1
10
,
.
101 101
Since the normal vector n is normal to the tangent vector, nˆ  
5.
10
1
.
,
101 101
Determine each of the following integrals
a.

cos 3t ,  sin 3t dt
 cos 3t,  sin 3t dt 
1
1
sin 3t  C1, cos 3t  C2
3
3
 t i  t j  dt
1
b.
2
3
0
1
1
1
t 2i  t 3 j  dt   t 3i  t 4 j 

0
4
3
0
1
1
1
 i j
3
4
BC - 3
Vector Calculus
Name:
6. At time t, a particle moving in the xy-plane is at position (x(t), y(t)), where x(t) and y(t) are not explicitly given.
For t ≥ 0,
( )
dx
dy
= 4t + 1 and
= sin t 2 . At time t = 0, x(0) = 0 and y(0) = –4.
dt
dt
(a) Find the speed of the particle at time t = 3, and find the acceleration vector of the particle at time t = 3.
2
2
 dx   dy 
speed =      
 dt   dt 
At t = 3, speed =
 4t  1
2
  sin(t 2 ) 
2
169  sin 2 (9)
(b) Find the slope of the line tangent to the path of the particle at time t = 3.
dy
dy dt sin(t 2 )
sin(9)
slope =
=

, so at t  3, slope =
dx dx
4t  1
13
dt
(c) Find the position of the particle at time t = 3.

x(3)  x(0)   (4t  1) dt  0   2t 2  t
0

3
3
  21
0
3
y (3)  y (0)   (sin(t 2 ) dt  4  .774  3.226 (from calculator)
0
 ( x(3), y (3))  (21, 3.226)
(d) Find the total distance traveled by the particle over the time interval 0 ≤ t ≤ 3.
3
distance =

0
2
2
 dx   dy 
     dt  
 dt   dt 
0
3
 4t  1
2
  sin(t 2 )  dt  21.091
2