Math 2511 – Calc III Practice Exam 1
Answers
1.
Definitions: Please state in your own words the meaning of the following terms:
Vector: A vector is a directed line segment.
Unit vector: A unit vector is a vector with length 1, i.e. ||v|| = 1
Tangent vector to a curve: A tangent vector is that vector which is tangent to a curve, i.e. it “just
touches” the curve.
d)
Normal vector to a curve: A normal vector is a vector that is perpendicular to the tangent vector to a
curve.
e)
Smooth Curve: A smooth curve C is the graph of a function r (t )  x(t ), y(t ), z (t )  such that r is
differentiable, r’ is continuous, and r(t) is never zero.
f)
Curvature: The curvature is a number K such that a circle of radius 1/K with center on the normal
T ' (t )
vector “touches” the curve the best. It is defined as K 
r ' (t )
a)
b)
c)
g)
Limit of a function z  f ( x, y ) : A function f has a limit L at a point ( x0 , y 0 ) if given any   0 there
is a   0 such that f ( x, y)  L   whenever 0  ( x, y)  ( x0 , y 0 )  
h)
Continuity of a function z  f ( x, y ) : A function f is continuous at a point ( x0 , y 0 ) if
lim
( x , y )( x0 , y0 )
i)
f ( x, y)  f ( x0 , y 0 )
Arc Length: The arc length of a curve is the actual length of a curve if you stretched it out on a
b
measuring tape. It is mathematically defined as
 r ' (t ) dt
a
2.
True/False questions:
2
a)
Is u  u  u : FALSE (it is true, though, that u  u  u
b)
If r (s)  s 2 , s, s 3  then s can not be the arc length parameter: TRUE (if s was the arc length
parameter, then r ' (s)  1 . But in our case r ' (s)   2s,1,3s 2   4s 2  1  9s 4  1
c)
d)
Is it true that a tanget vector to a curve parameterized by its arc length is always a unit vector?:
TRUE (that is in fact the nice feature of arc length parameter: if s is arc length, then || r’(s) || = 1)
If
lim
f ( x, y )  0 then lim f ( x,0)  0 : TRUE (if (x,y) approach zero, then x and y can not be
e)
zero simultaneously, but one variable can be equal to zero and the other can approach zero)
If lim f (0, y)  0 then
lim
f ( x, y )  0 : FALSE (take, for example, the function
( x , y )( 0,0)
y 0
f ( x, y ) 
f)
x0
( x , y )( 0,0)
y
)
x y
If f is continuous at (0,0), and f(0,0) = 10, then
lim
( x , y )( 0,0)
f ( x, y)  10 TRUE (it’s exactly the
definition of continuity)
3.
Vectors: Suppose u  7,2,3  , v  1,4,5  , and w  2,1,3 
Are u and v orthogonal, parallel, or neither? Let’s compute the dot product between them:
u  v  7,2,3    1,4,5  (7  8  15)  0 so the vectors are orthogonal
b)
Find the angle between v and w: Again the dot product will basically tell the answer.
vw
9
v  w  9, v  42 , w  14 so therefore cos( ) 

 0.3711
v w
42 14
a)
c)
Find u  v (dot product), u  v (cross product), u  (v  w) , and u : u  v  0 ,
u  v  -22, - 38, 26  , u  (v  w)  72 , u  62 :
d)
Find the equation of the plane spanned by v and w that passes through the origin: The plane has
as a normal vector v  w  -17, - 13, 7  , hence –17x – 13y + 7z + D = 0. Since the origin is part of
the plane we get D = 0.
e)
f)
g)
h)
4.
a)
b)
c)
Find the equation of the plane spanned by u and v: The plane has as a normal vector
u  v  -22, - 38, 26  , hence the equation of the plane is –22x – 38y + 26z + D = 0. Since the origin
is part of the plane we get D = 0.
Find the distance between the planes in (d) and (e): The planes are not parallel because their normal
vectors are not parallel. Therefore, they must intersect which means that the distance between them is
zero.
Find the distance of the plane x + y + z = 1 to the point (1, 2, 3): The plane has the normal vector
<1, 1, 1>. Letting x and y be zero we find that z = 1, so that the point (0, 0, 1) is on the plane. Now we
need to find the projection of the vector (1, 2, 3) – (0, 0, 1) = (1, 2, 2) to the normal vector (1, 1, 1).
 1,2,2    1,1,1  1  2  2
5
That is easy: dist 


 1,1,1 
3
3
wu
25
Find the projection of w onto u. proju ( w) 
u
 7,2,3 
2
62
u
Vector valued functions:
Find r ' (t ) if r (t )  6t ,7t 2 , t 3  : r’(t) = <6, -14t, 3t2>
Find r ' (t ) if r (t )  a cos3 (t ), a sin 3 (t ), t sin(t )  : r’(t) = <-3a cos2(t) sin(t), 3a sin2(t) cos(t), sin(t)+t
cos(t)>
d
If r (t )  4t , t 2 , t 3  , find r ' (t ) , r ' ' (t ) ,
r (t ) .
dt
r’(t) = <4, 2t, 3t2>
r’’(t) = <0, 2, 6t>
1 2
d
d
1
|| r (t ) || 
16t 2  t 4  t 6  16t 2  t 4  t 6
(32t  4t 3  6t 5 )
dt
dt
2

d)
If r (t )  e t ,3t 3 ,
2
2
3
 some curve, find
6t
2
 r (t )dt   e dt,  3t
t
1
1

2
3
dt,
1
3
 6t dt  e
2
1
2
 r (t )dt :
1
3
3
 e, (2 4  1), ln( 2)  C
4
6
1
If r (t )  t ,  , find T(t), N(t), at and an
t
T is a unit tangent vector, so we need to find r’(t) = <1, -t-2>. Then T (t ) 
1
1 t
4
 1,t  2 
To find the normal vector, we would have to differentiate T, which is not so easy. But since a normal
vector is perp. to the tangent, we can right away state that a normal vector is given by <t -2, 1>. Then
the principle unit normal vector N (t ) 
1
1 t
4
 t  2 ,1  . That answer could be off by a minus sign,
which we will not consider as incorrect at this stage.
The tangential component is
1
2
3
t~
1
2
The normal component is
e)
t~41
t~
t~41
Repeat (e) for r (t )  e t cos(t ), e t sin(t )  for t 

.
2
Again we first find r ' (t )  et cos(t )  et sin(t ), et sin(t )  et cos(t )  . Since T is a unit vector, we need
to find
r ' (t ) 
 e 2t cos 2 (t )  2e 2t cos(t ) sin(t )  e 2t sin 2 (t )  e 2t sin 2 (t )  2e 2t sin(t ) cos(t )  e 2t cos 2 (t ) 
 2e 2t  2e t
r ' (t )
1
Then T (t ) 

 cos(t )  sin(t ), cos(t )  sin(t ) 
r ' (t )
2
To find the normal vector, we can compute the derivate of T, since the usually difficult term involving
1
the square root has cancelled out. So T ' (t ) 
  sin(t )  cos(t ),  sin(t )  cos(t )  and
2
1
1
T ' (t ) 
sin 2 (t )  2 sin(t ) cos(t )  cos 2 (t )  sin 2 (t )  2 sin(t ) cos(t )  cos 2 (t ) 
2  1.
2
2
Thus T ' (t ) is a unit vector and therefore equal to the principle unit normal vector.
The computation of a N and a T are left as an exercise.
f)
If r (t )  3  3t ,4t  , find the arc length of the curve between 0 and 1. The answer is 5
g)
If r (t )  4t , cos(t ), sin(t )  , find the arc length of the curve between 0 and
 2

16  cos 2 (t )  sin 2 (t ) dt 
0

2

. The answer is
2
17
, but the original problem had a 3 sin(t) instead of a simpler
sin(t) and was much harder.
h)
i)
Find the curvature of the curve r ( s)  1 
1
s,1 
1
s  where s is the arc length parameter.
2
2
This is a straight line, so it’s curvature is zero. Using the formulas confirms that because the curvature
involves r’’, which is <0, 0>.
r ' (t )  r ' ' (t )
t2
Find the curvature of r (t )  t ,3t 2 ,  . We use the formula of curvature K 
.
3
2
r ' (t )
r ' (t )  1,6t , t  and r ' ' (t )  0,6,1  . Then r ' (t )  1  37 t 2 and r 'r ' '  0,1,6  so that – finally
- K
 0,1,6 
1  37 t 2
5.
3

37
1  37t 
2 3/ 2
Surfaces: Find the domain for the following functions
a)
f ( x, y )  4  x 2  y 2 . Domain is the disk x^2 + y^2 <= 4
b)
f ( x, y) 
c)
d)
1
. Domain is all points (x,y) except for x = 0 and y = 0 (the x and y axis)
xy
1
. Domain is all points (x, y) except (0, 0)
f ( x, y )  2
x  y2
1
f ( x, y )  2
. Domain is all points except |x| = |y| (diagonals)
x  y2
6.
Limits and Continuity: Determine the following limits as (x,y) -> (0,0) and state which function is
continuous everywhere and which have a removable point of discontinuity.
xy  1
lim
a)
.
( x , y ) ( 0, 0 ) x 2  y 2  1
Answer: As (x,y) approaches (0,0) the top approaches 1 and so does the bottom. Hence the limit is 1.
b)
c)
d)
e)
lim
xy  1
x2  y2
Answer: As (x,y) approaches (0,0) the top approaches 1 and the bottom approaches zero. Hence the
limit does not exist.
xy
lim
.
2
( x , y ) ( 0, 0 ) x  y 2
Answer: As (x,y) approaches (0,0) the top approaches 0 and so does the bottom. Hence we are stuck.
We try approaches along the axis x = 0, y approaching zero (limit is 0), along the axis y = 0, x
approaching zero (limit is 0), and x = y, approaching zero (limit is ½). Therefore, the limit does not
exist.
( x , y ) ( 0, 0 )
lim
x2 y
.
x2  y2
Answer: As (x,y) approaches (0,0) the top and bottom both approach zero. Even approaches along x =
0, y = 0, and x = y give a zero limit. We are really stuck and need to resort to the definition of
derivatives. Compare example 2 (b), page 833, for details.
( x , y ) ( 0 , 0 )
lim
x2  y2
.
x2  y2
Answer: As (x,y) approaches (0,0) the top and bottom both approach zero. Setting x = 0 and y
approaching 0 gives a limit of –1, while setting y = 0 and x approaching 0 gives a limit of 1. Hence the
original limit does not exist.
( x , y ) ( 0 , 0 )
7.
Picture: Sketch the circle that fits the graph below the best at the points x = 0 and x = 3. At which of the
two points is the curvature smaller?
Answer: At zero there’s a small circle with center on the y-axis, at x = 3 it’s a larger circle with center on the lower
side of the curve. Since the reciprocals of the radii are the curvature, the curvature at x = 5 is smaller because the
radius is larger there.
8.
Picture: Match the following contour plots (level plots) to their corresponding surfaces.
Answer: [1] -> [C], [2] -> [D], [3] -> [A], [4] -> [B]
[1]
[2]
[3]
[4]
[A]
[B]
[C]
[D]
9.
a)
b)
Prove the following facts:
Show that u  v  (v  u)
Answer: Suppose u = <u1, u2, u3> and v = <v1, v2, v3>. Then
u  v  u 2 v3  u3 v2 , u3 v1  u1v3 , u1v2  u 2 v1    u3 v2  u 2 v3 , u1v3  u3 v1 , u 2 v1  u1v2  v  w
Show that u  (v  u )  0
Answer: Suppose u = <u1, u2, u3> and v = <v1, v2, v3>. Then
v  u  u3 v2  u 2 v3 , u1v3  u3 v1 , u 2 v1  u1v2  and
u  v  u   u1 , u 2 , u 3    u 3 v 2  u 2 v3 , u1v3  u 3 v1 , u 2 v1  u1v 2 
 u1u 3 v 2  u1u 2 v3  u 2 u1v3  u 2 u 3 v1  u 3 u 2 v1  u 3 u1v 2  0
c)
Show that if y  f (x) is a function that is twice continuously differentiable, then the curvature of f at a
point x is K 
| f ' ' ( x) |
1   f ' ( x) 
2 32
Answer: We can describe the curve defined by y  f (x) parametrically as r (t )  t, f (t )  . Then
r ' (t )  1, f ' (t )  and r ' ' (t )  0, f ' ' (t )  . To use the formula for curvature we need to find a cross
product, so we need vectors in space. We extend each vector to 3 dimensions by simply adding a 0 as
d)
e)
the z component: r (t )  t , f (t ),0  , r ' (t )  1, f ' (t ),0  , and r ' ' (t )  0, f ' ' (t ),0  . Now the result
follows immediately, using the formula for curvature (left as an exercise).
Prove that the curvature of a line in space is zero.
Answer:
Prove that if f ( x, y)  x then
lim
f ( x, y )  a
( x , y )( a ,b )
Answer: We need to show that given any   0 there is a   0 such that f ( x, y)  a   whenever
0  ( x, y)  (a, b)   . So, first we note that ( x, y)  (a, b) 
xa 
x  a 2

x  a 2   y  b2
a)
and
. Thus, given any number   0 let    . Then if
( x, y)  (a, b)   we have f ( x, y)  a  x  a 
which finishes the proof.
10.
x  a 2   y  b2
x  a 2   y  b2
 ( x, y)  (a, b)    
Story problem (motion)
A baseball is hit 3 feet above ground at 100 feet per second and at an angle of Pi/4 with respect to the
ground. Find the maximum height reached by the baseball. Will it clear a 10-foot high fence located
300 feet from home base?
Answer: We need to find the parametric equation of the curve describing the path of the baseball. We
start with a(t )  r ' ' (t )  0, g  from which it follows by integration that r ' (t )  0, gt    a, b 
1
and r (t )  0, gt 2    at, bt    c, d  . We know that the baseball starts out 3 feet above
2
1
ground, i.e. at the vector <0, 3>. Thus r (t )  0, gt 2    at, bt    0,3  . We also know that the
2


initial velocity is r ' (t )  100  cos( ), sin( )  50 2 ,50 2  so that
4
4
1 2
r (t )  0, gt    50 2t ,50 2t    0,3  . The maximum height is reached when the derivative
2
of the y component of r(t) is zero (or equivalently when the velocity in the y direction is zero). Hence it
50 2
 2.20970869 1 . Substituting that t into the y
g
component of r(t) gives us the maximum height of
1
y max   g 2.20970869 12  50 2 2.20970869 1  3  81.1250000 2 feet. To find out whether it
2
clears a 10 foot high fence 300 feet from home we first find the time when the x component of r(t) =
300
300, i.e. 300  0  50 2t or t 
x  4.24 seconds. After 4.24 seconds the height of the baseball
50 2
1
is y max   g 4.24 2  50 2 4.24   3  15.1716751 feet, so it will clear the fence (another home
2
run).
What is the maximum height and range of a projectile fired at a height of 3 feet above the ground with
an initial velocity of 900 feet/sec and at an angle of 45 degrees above the horizontal?
Answer: Similar to the above problem we compute the position as
1
r (t )  0, gt 2    450 2t ,450 2t    0,3  . Therefore the maximum height is reached at the
2
450 2
time t when t 
 19.8873782 2 . Hence the max. height is
g
1
y max   g 19.8873782 22  450 2 19.8873782 2  3  6331.12500 0 so the maximum height is
2
6331 feet.
is reached when  gt  50 2  0 or when t 
b)