Math 6C – Chapter 11 Quiz
* Solutions*
DEJA VU: No, you’re not crazy… The first 3 problems also occur on the Chapter 10 Quiz.
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1. Find a unit vector in the direction of ( 3, 1 ).
The y–component equals what?
Just divide this vector by it’s length, and you have a unit vector in the same direction:
(3,1)
(3,1)  3 1 


,

10  10 10 
32 12
The y-component of this vector is thus
1
1
 10 2 .
10
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2. A = (–1, 2) and B = (2, 3). The vector projection of A onto B has x–component equal to what?
The vector projection of A onto B is given by:

A






B B
B B
  A B
B   1
 
2
B   2 
 2
 
 3
 2 
  2
2
 3  2  3
The x-component of this projection vector is thus

 2  6
 2
 
 3
 13
8
4  2  
    13 
13  3 12 
 
13 
8
13
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3. A = (1, 2, 1) and B = (1, 1, –1). Find a unit vector in the direction of B x A.
Its z–component equals what?
 1  1 
     1
 1    2  
 1 1   1
   
 3
2
1 1 1 1 
 
,
,
 1  (2), (1 (1)),2 1  2
1 1 1 1 2 
 1
 
Dividing this vector by its length yields a unit vector in this direction:
(3, 2,1)
(3, 2,1)  3 2 1 

 
,
,

14
14
14
14
32  (2)2 12


The z-component of this vector is thus
1
1
 14 2 .
14
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4. a(t) = (2t, 4t3) ,
v(0) = (1,0),
r(0) = (0,1).
Then r(1) has y–component equal to what?
t2  C 
 v (t )  a (t )dt   4 1 
t  C2 
Use the given initial condition to solve for the two constants of integration:
 02  C   
C   
v (0)   4 1   1   1   1
C2  0
0
 
0  C2   
 2t 
a (t )   3 
 4t 



And now we integrate again to find the position as a function of time:
t3




t

C
2
t 1
3
3



v (t )  4
 r (t )  v (t )dt   5
.

 t 
t



 C4 
 5


Use the given initial condition to solve for the two constants of integration:
 03

 0C   
3
0  C3   0
r (0)   3 5


 .
 1 


1
C
0
 4   

 C4   
 5

And so we have
 t3

r (t )   35
t
 5
Hence, the y-component of r(1) is
6
5

13

  r (1)   3

 5

1
1

 5
t

4
 
  3 .
 6
1  5 
  
1
.
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5. Find the unit tangent vector T for the plane curve r(t) = (cos t, et).
component of T is what?
cos t 
r (t )   t 
 e 


Tˆ (t )  v (t ) 
v (t )
  sin t 

 et 


  sin t 
1


sin 2 t  e2t  e t 

v (t )  
The absolute value of the x–
And the absolute value of the x-component of T is thus
sin t
sin 2 t  e2t
.
6. Find the principal unit normal vector N for the plane curve r(t) = (cos t, et).
of the x–component of N is what?
The absolute value
In problem #5 we found that
Tˆ (t ) 
  sin t 
1

.
sin 2 t  e2t  e t 
The principal unit normal vector for this curve is given by
dTˆ
Nˆ  dt
dTˆ
dt
.
Although we could (somewhat tediously) compute the principal normal via this formula, it’s easier to
proceed as follows.
First note that this normal vector is simply a unit vector perpendicular to T for each t. There are two
such perpendicular unit vectors – one is the “principal” one and the other its negative. (Since this
problem asks for the absolute value of the principal normal vector’s x-component, we needn’t worry
about which of these two vectors we find.)
 B 
Secondly notice that 
 is perpendicular to
 A 
 A
  , since their dot products is zero.
 B 
It follows that
Nˆ 
 et 
1

,
2
2
t


sin
t
sin t  e 

and so the absolute value of the x–component of N is
et
sin 2 t  e2t
.
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7. Find the principal unit binormal vector B for the plane curve r(t) = (cos t, et).
value of the x–component of B is what?
Bˆ  Tˆ  Nˆ , and we will use the results of problems # 5 and 6 to compute this.
The absolute
  sin t 


1
1
 et  
Bˆ  Tˆ  Nˆ 
sin 2 t  e2t  0  sin 2 t  e2t


  sin t   e t 

 

 2 1 2t  e t   sin t 
sin t  e  0   0 

 


 


 sin t et 
 2 1 2t  0, 0,
sin t  e 
et
sin t 

 sin 2 t  e2t 

  0, 0,
  0, 0, 1
sin 2 t  e2t 




 et 


sin t 


 0 



Hence the absolute value of the x–component of B is 0.
Of course it’s easier to simply realize that since this is a plane curve were T and N are always in the
xy-plane, B must be perpendicular to the xy-plane!

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8. Find the curvature,  for r(t) = ( cos t, et ) at t =.

v a
v
cos t 
r (t )   t 
 e 

 (t ) 

  sin t    cos t 

t  
t 
 e  e

 0   0 


 
sin 2 t  e 2t
And so
  sin t 
 
 et 


v (t )  


3

et sin t  et cos t



3
3
2
2
t

sin t  e



 0,



  cos t 

 e t 
a (t )  
 sin t  cos t 
0,
et
et 



3
2
2
t

sin t  e


et cos t  sin t



3
2
2
t

sin t  e


 ( ) 
e cos  sin 

3
e 1 0
e  1

3
e 3 e 2

 sin 2   e 2 
 0  e 2








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9. Find the radius of curvature (the radius of the osculating circle) of
at the point (0, 1).
y  1 x 2
Since the given parabolic curve is symmetric about the y-axis, and the given point is on the y-axis, it
follows that the best-fit circle has its center on the y-axis. Suppose the center is (0,c). Then the
equation of the osculating circle is
x2  y  c 2  r 2 .


This circle passes through the point (0, 1), so we have that
02  1 c   r 2 ,
2
1 c  r. Let’s assume this is positive for now (visually this is reasonable, looking at even a
crude graph!), so 1 c  r.
hence
Thus the osculating circle is given by
x2   y  c 2  1 c 
2
.
The slope of the circle at (0, 1) is 0 (since this is the vertex of the parabola). Implicitly differentiating
the above equation with respect to x, we get:
2 x  2  y  c  dy  0
dx
dy  x
dx c  y
The first derivatives of the parabola’s equation and this circle are both 0 at x=0. The second
derivative of the parabola’s equation is –2, which must also be the second derivative of the osculating
circle at (0, 1).
dx  x d c  y c  y  x dy
c

y




d y
dx dx
dx

2
2
2
dx
c  y 
c  y 
2
At the point (0, 1), this becomes
and since this must equal –2, we have
d2y  1 ,
dx2 c 1
1  2  c  1 .
2
c 1
Thus, the radius of curvature is
r  1  c  1 1  1
2
2
.
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 2cos t 


10. Find length of the parametrized curve, r (t )   2sin t  , from


 5t 


0  t  .
 2sin t 


v(t )   2cos t  , so




5


L



0
 2sin t    2cos t    5  dt
2
2
2



  4sin 2 t  4cos2 t  5 dt   4  5 dt  3
0
0
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 2t 1


11. Find cosine of the angle between the velocity and acceleration vectors of r (t )   2 t  , at


 t3 


t=1.
 2 
0


 
v(t )   2   a(t )   0  ,


 
3t 2 
6t 


hence


v(1)  



2

2

3 
&
0
 
a(1)  0 .
 
6
The cosine of the angle between two vectors is given by
cos  v
a
va





2  0

2  0
3  6
 18  3
4  2  9 0  0  36 15  6
15
