BC Calc III
Sample Vector Quiz
Name:
Calculator allowed.
You must show enough work so that I can recreate your results.
#1.
Let v  5i  3 j . Find
a. the magnitude of v
v  52  32  34
b. a unit vector in the direction of v
v
5 ˆ
3 ˆ
vˆ  
i
j
v
34
34
#2.
32
 j , v (0)  15 i  20 j , r (0)  15 i , where
t 1
a (t ), v (t ), and r (t ) represent the acceleration, velocity, and position vectors
respectively.
Suppose that a (t ) 
a. Find v (t ) and r (t ) .
v (t )   0,
32
, dt  C1 , 64 t  1  C2 .
t 1
.
Since v (0)  15, 20 ,v (t )  15, 64 t  1  84
Thus,
r (t )   15, 64 t  1  84 dt  15t  C1 , 
Since r (0)  15, 0 ,r (t )  15t  15, 
128
3
128
3
 t  1
 t  1
3
3
 84t  C2 .
 84t 
128
.
3
b. Find the speed at time t = 0.
speed = v  0   15, 20 = 152  202 = 25
c. Set up an integral that gives the total distance traveled for 0  t  1.
1

0
BC CALC III
1


2
15, 64 t  1  84 dt   152  64 t  1  84 dt
0
#3. A diver leaps from the edge of a diving platform into a pool below. The figure below
shows the initial position of the diver and her position at a later time. At time t seconds after
she leaps, the horizontal distance from the front edge of the platform to the diver’s
shoulders is given by x(t ) , and the vertical distance from the water to her shoulders is given
by y (t ) where x(t ) and y (t ) are measured in meters. Suppose that the diver’s shoulders are
11.4 meters above the water when she makes her leap and that
dx
dy
 0.8 and
 3.6  9.8t for 0  t  A , where A is the time when the diver’s shoulders
dt
dt
enter the water. Use your calculator on this problem, but show set-up clearly.
a. Find the maximum vertical distance
from the water surface to the diver’s
shoulders.
dy
 0  t  .367
At maximum height,
dt
.367
y (.367)  11.4 
  3.6  9.8t dt
0
x(t)
 12.061
y(t)
b. Find A, the time when the diver’s shoulders enter the water.
t
y (t )  11.4    3.6  9.8t dt  11.4  3.6t  4.9t 2  0  t  1.936
0
c. Find the total distance travelled by the diver’s shoulders for 0  t  A .
1.936
s

1.936
v(t ) dt 
0

.82   3.6  9.8t  dt  12.946 meters
2
0
d. Find the angle  , 0   

2
, between the path of the diver and the water at the time
t  A.
At time t = 1.936 ,
x(1.936), y(1.936)  .8,3.6  9.8 1.936  .8, 15.372 
 15.372 
  87.021 or 1.519 radians
 .8 
  tan 1 
BC CALC III