Review guide for the final exam in Math 233
1
Basic material.
This review includes the remainder of the material for math 233. The final exam will
be a cumulative exam with a majority of the problems coming from the material covered
beginning with chapter 15.4 of the book.
We first recall polar coordinates formulas given in chapter 10.3. The coordinates of a
point (x, y) ∈ R3 can be described by the equations:
x = r cos(θ)
y = r sin(θ),
(1)
p
where r = x2 + y 2 is the distance from the origin and ( xr , yr ) is (cos(θ), sin(θ)) on the
unit circle. Note that r ≥ 0 and θ can be taken to lie in the interval [0, 2π).
To find r and θ when x and y are known, we use the equations:
r2 = x2 + y 2
tan(θ) =
y
.
x
(2)
Example 1 Convert the point (2, π3 ) from polar to Cartesian coordinates.
Solution : Since r = 2 and θ = π3 , Equations 1 give
π
1
=2· =1
3
2
√
π
3 √
y = r sin(θ) = 2 sin = 2 ·
= 3.
3
2
√
Therefore, the point is (1, 3) in Cartesian coordinates.
x = r cos(θ) = 2 cos
Example 2 Represent the point with Cartesian coordinates (1, −1) in terms of polar
coordinates.
Solution : If we choose r to be positive, then Equations 2 give
p
p
√
r = x2 + y 2 = 12 + (−1)2 = 2
y
= −1.
x
Since the point (1, −1) lies in the fourth quadrant, we can choose θ = − π4 or θ = 7π
4 .
√
√ 7π
π
Thus, one possible answer is ( 2, − 4 ); another is (r, θ) = ( 2, 4 ).
The next theorem describes how to calculate the integral of a function f (x, y) over a
polar rectangle. Note that dA = r dr dθ.
tan(θ) =
1
BASIC MATERIAL.
2
Theorem 3 (Change to Polar Coordinates in a Double Integral) If f is continuous on a polar rectangle R given by 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β, where 0 ≤ β − α ≤ 2π,
then
Z Z
Z βZ b
f (x, y)dA =
f (r cos(θ), r sin(θ))r dr dθ.
R
α
a
RR
2
Example 4 Evaluate
R (3x + 4y )dA, where R is the region in the upper half-plane
2
2
bounded by the circles x = y = 1 and x2 + y 2 = 4.
Solution : The region R can be described as
R = {(x, y) | y ≥ 0, 1 ≤ x2 + y 2 ≤ 4}.
It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by 1 ≤ r ≤ 2,
0 ≤ θ ≤ π. Therefore, by Theorem 3,
Z Z
Z πZ 2
(3r cos(θ) + 4r2 sin2 (θ))r dr dθ
(3x + 4y 2 )dA =
0
R
πZ
Z
=
Z
=
(3r2 cos(θ) + 4r3 sin2 (θ)) dr dθ
1
0
π
2
3
4
Z
2
[r cos(θ) + r sin
0
1
Z
(θ)]r=2
r=1
π
dθ =
(7 cos(θ) + 15 sin2 (θ)) dθ
0
π
15
=
[7 cos(θ) + (1 − cos(2θ))] dθ
2
0
¸π
15π
15θ 15
−
sin(2θ) =
.
= 7 sin(θ) +
2
4
2
0
Example 5 Find the volume of the solid bounded by the plane z = 0 and the paraboloid
z = 1 − x2 − y 2 .
Solution : If we put z = 0 in the equation of the paraboloid, we get x2 + y 2 = 1. This
means that the plane intersects the paraboloid in the circle x2 + y 2 = 1, so the solid
lies under the paraboloid and above the circular disk D given by x2 + y 2 ≤ 1. In polar
coordinates D is given by 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π. Since 1 − x2 − y 2 = 1 − r2 , the volume is
Z Z
Z 2π Z 1
2
2
V =
(1 − x − y )dA =
(1 − r2 )r dr dθ
D
Z
2π
0
Z
=
0
0
1
0
·
(r − r3 ) dr dθ = 2π
r2 r4
−
2
4
¸1
=
0
π
.
2
1
BASIC MATERIAL.
3
If we had used rectangular coordinates instead of polar coordinates, then we would have
obtained
Z Z
Z 1 Z √1−x2
2
2
V =
(1 − x − y )dA =
(1 − x2 − y 2 ) dy dx
√
D
−1
− 1−x2
which is not easy to evaluate because it involves finding the following integrals:
Z p
Z
Z
p
3
2
2
2
1 − x dx
x 1 − x dx
(1 − x2 ) 2 dx.
The next theorem extends our previous application of Fubini’s theorem from section
15.3 for type II regions.
Theorem 6 If f continuous on a polar region of the form
D = {(r, θ) | α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)}
then
Z Z
Z
β
Z
h2 (θ)
f (x, y)dA =
D
f (r cos(θ), r sin(θ))r dr dθ.
α
h1 (θ)
The next definition describes the notion of a vector field. We have already seen an
example of a vector field associated to a function f (x, y) defined on a domain D ⊂ R2 ,
namely the gradient vector field ∇f (x, y) = hfx (x, y), fy (x, y)i. In nature and in physics,
we have the familiar examples of the velocity vector field in weather and force vector fields
that arise in gravitational fields, electric and magnetic fields.
Definition 7 Let D be a set in R2 (a plane region). A vector field on R2 is a function
F that assigns to each point (x, y) in D a two-dimensional vector F(x, y).
Definition 8 Let E be a subset of R3 . A vector field on R3 is a function F that assigns
to each point (x, y, z) in E a three-dimensional vector F(x, y, z).
Note that a vector field F on R3 can be expressed by its component functions. So if
F = (P, Q, R), then:
F(x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k.
We now describe our first kind of ”line integral”. These type of integrals arise form
integrating a function along a curve C in the plane or in R3 . The type of line integral
described in the next definition is called a line integral with respect to arc length.
1
BASIC MATERIAL.
4
Definition 9 If f is defined on a smooth curve C in R2 , then the line integral of f
along C is
Z
n
X
f (x, y) ds = lim
f (x∗i , yi∗ )∆si
n→∞
C
j=1
if this limit exists.
The following formula can be used to evaluate a line integral.
Theorem 10 Suppose f (x, y) is a continuous function on a differentiable curve C(t),
C : [a, b] → R2 . Then
sµ ¶
µ ¶2
Z
Z b
dx 2
dy
f (x, y) ds =
f (x(t), y(t))
+
dt.
dt
dt
a
C
Note that in the above formula,
sµ
dx
dt
¶2
µ
+
dy
dt
¶2
,
is the speed of C(t) at time t.
R
Example 11 Evaluate C 2x ds, where C consists of the arc C of the parabola y = x2
from (0, 0) to (1, 1).
Solution : We can choose x as the parameter and the equations for C become
y = x2
x=x
Therefore
sµ
0 ≤ x ≤ 1.
¶
dy 2
2x ds =
2x
+
dx
dx
C
0
√
¸1
Z 1 p
3
2
5−1
1
5
2
.
=
2x 1 + 4x2 dx = · (1 + 4x ) 2 =
4
3
6
0
0
Z
Z
1
dx
dx
¶2
µ
Actually for what we will studying next, another type of line integral will be important.
These line integrals are called line integrals of f along C with respect to x and y.
They are defined respectively for x and y by the following limits:
Z
f (x, y) dx = lim
C
n→∞
n
X
i=1
f (x∗i , yi∗ )∆xi
1
BASIC MATERIAL.
5
Z
f (x, y) dy = lim
n→∞
C
n
X
f (x∗i , yi∗ )∆yi .
i=1
The following formulas show here to calculate these new type line integrals . Note
that these integrals depend on the orientation of the curve C i.e., the initial and terminal
points.
Theorem 12
Z
Z
b
f (x, y) dx =
Z
C
Z
a
b
f (x, y) dy =
C
Example 13 Evaluate
R
C
f (x(t), y(t))x0 (t) dt
f (x(t), y(t))y 0 (t) dt.
a
y 2 dx + x dy, where
1. C = C1 is the line segment from (−5, −3) to (0, 2) and
2. C = C2 is the arc of the parabola x = 4 − y 2 from (−5, −3) to (0, 2).
Solution :
1. A parametric representation for the line segment is
x = 5t − 5,
y = 5t − 3,
0 ≤ t ≤ 1.
Then dx = 5 dt, dy = 5 dt, and Theorem 12 gives
Z
Z 1
2
y dx + x dy =
(5t − 3)2 (5dt) + (5t − 5)(5 dt)
C1
0
Z
=5
·
1
(25t2 − 25t + 4) dt
0
25t3 25t2
−
+ 4t
=5
3
2
¸1
0
5
=− .
6
2. Since the parabola is given as a function of y, let’s take y as the parameter and write
C2 as
x = 4 − y 2 y = y,
−3 ≤ y ≤ 2.
Then dx = −2y dy and by Theorem 12 we have
Z
Z 2
y 2 dx + x dy =
y 2 (−2y) dy + (4 − y 2 ) dy
C2
−3
1
BASIC MATERIAL.
6
Z
2
=
(−2y 3 − y 2 + 4) dy
−3
· 4
¸2
y
y3
5
= − −
+ 4y
= 40 .
2
3
6
−3
One can also define in a similar manner the line integral of a function f along a curve
C in R3 .
Theorem 14
Z
Z
b
f (x, y, z) ds =
C
a
sµ ¶
µ ¶2 µ ¶2
dx 2
dy
dz
f (x(t), y(t), z(t))
+
+
dt.
dt
dt
dt
R
Example 15 Evaluate c y dx + z dy + x dz,, where C consists of the line segment C1
from (2, 0, 0) to (3, 4, 5) followed by the vertical line segment C2 from (3, 4, 5) to (3, 4, 0).
Solution : We write C1 as
r(t) = (1 − t)h2, 0, 0i + th3, 4, 5i = h2 + t, 4t, 5ti
or, in parametric form, as
x=2+t
Thus
y = 4t
Z
Z
z = 5t
1
y dx + z dy + x dz =
C1
0 ≤ t ≤ 1.
(4t) dt + (5t)4 dt + (2 + t)5 dt
0
Z
=
0
1
t2
(10 + 29t) dt = 10t + 29
2
¸1
= 24.5.
0
Likewise, C2 can be written in the form
r(t) = (1 − t)h3, 4, 5i + th3, 4, 0i = h3, 4, 5 − 5ti
or
x = 3 y = 4 z = 5 − 5t 0 ≤ t ≤ 1.
Then dx = 0 = dy, so
Z
Z
y dx + z dy + x dz =
C2
1
2(−5) dt = −15.
0
1
BASIC MATERIAL.
7
Adding the values of these integrals, we obtain
Z
y dx + z dy + x dz = 24.5 − 15 = 9.5.
C
We now get to our final type of line integral which can be considered to be a line
integral of a vector field. This type of integral is used to calculate the work W done by a
force field F in moving a particle along a smooth curve C.
Theorem 16 If C is given by the vector equation r(t) = x(t)i + y(t)j + z(t)k on the
interval [a, b], then the work W can be calculated by
Z
W =
b
F(r(t)) · r0 (t) dt,
a
where · is the dot product.
In general, we make the following definition which is related to the formula in the
above theorem.
Definition 17 Let F be a continuous vector field defined on a smooth curve C given by
a vector function r(t), a ≤ t ≤ b. Then the line integral of F along C is
Z
Z
F · dr =
C
Z
b
0
F(r(t)) · r (t) dt =
F · T ds.
a
C
Example 18 Find the work done by the force field F(x, y) = x2 i − xyj in moving a
particle along the quarter-circle r(t) = cos t i + sin t j, 0 ≤ t ≤ π2 .
Solution : Since x = cos t and y = sin t, we have
F(r(t)) = cos2 ti − cos t sin tj
and
r0 (t) = − sin ti + cos tj.
Therefore, the work done is
Z
Z
F · dr =
C
π
2
Z
0
F(r(t)) · r (t)dt =
0
π
2
0
cos3 t
=2
3
¸ π2
0
2
=− .
3
(−2 cos2 t sin t)dt
1
BASIC MATERIAL.
Example 19 Evaluate
cubic given by
8
R
C
F · dr, where F(x, y, z) = xyi + yzj + zxk and C is the twisted
y = t2
x=t
z = t3
0 ≤ t ≤ 1.
Solution : We have
r(t) = ti + t2 j + t3 k
r0 (t) = i + 2tj + 3t2 k
F(r(t)) = t3 i + t5 j + t4 k.
Thus,
Z
Z
F · dr =
C
Z
1
=
0
1
F(r(t)) · r0 (t)dt
0
t4 5t7
(t + 5t )dt =
+
4
7
3
¸1
6
=
0
27
.
28
Theorem 20 If C in R3 is parametrized by r(t) and F = P i + Qj + Rk, then
Z
Z
F · dr =
P dx + Qdy + Rdz.
C
C
We now apply the material covered so far on line integrals to obtain several versions
of the fundamental theorem of calculus in the multivariable setting. Recall that the
fundamental theorem calculus can be written as
Z b
F0 (x)dx = F (b) − F (a),
a
when F0 (x) is continuous on [a, b].
Theorem 21 Let C be a smooth curve given by the vector function r(t), a ≤ t ≤ b.
Let f be a differentiable function of two or three variables whose gradient vector ∇f is
continuous on C. Then
Z
∇f · dr = f (r(b)) − f (r(a)).
C
For further discussion, we make the following definitions.
Definition 22 A curve r : [a, b] → R3 (or R3 ) closed if r(a) = r(b).
Definition 23 A domain D ⊂ R3 (or R2 ) is open if for any point p in D, a small ball
(or disk) centered at p in R3 (in R2 ) is contained in D.
1
BASIC MATERIAL.
9
Definition 24 A domain D ⊂ R3 (or R2 ) is connected if any two points in D can be
joined by a path contained inside D.
Definition 25 A curve r : [a, b] → R3 (or R2 ) is a simple curve if it doesn’t intersect
itself anywhere between its end points (r(t1 ) 6= r(t2 ) when a < t1 < t2 < b).
Definition 26 An open, connected region D ⊂ R2 is a simply-connected region if any
simple closed curve in D encloses only points that are in D.
Definition 27 A vector field F is called a conservative vector field if it is the gradient
of some scalar function f (x, y); the function f (x, y) is called a potential function for F.
For example, for f (x, y) = xy + y 2 , ∇f = hy, x + 2yi and so, F(x, y) = yi + (x + 2y)j is a
conservative vector field.
Definition
with domain
D, we say that the line
R 28 If F is a continuous vector field
R
R
integral C F · dr is independent of path if C1 F · dr = C2 F · dr for any two paths C1
and C2 in D with the same initial and the same terminal points.
We now states several theorems that you should know for the final exam.
R
R
Theorem 29 C F · dr is independent of path in D if and only if C F · dr = 0 for every
closed path C in D.
Theorem
30 Suppose F is a vector field that is continuous on an open connected region
R
D. If C F · dr is independent of path in D, then F is a conservative vector field on D;
that is, there exists a function f such that ∇f = F.
Theorem 31 If F(x, y) = P (x, y)i + Q(x, y)j is a conservative vector field, where P and
Q have continuous first-order partial derivatives on a domain D, then throughout D we
have
∂P
∂Q
=
.
∂y
∂x
Theorem 32 Let F = P i + Qj be a vector field on an open simply-connected region D.
Suppose that P and Q have continuous first-order derivatives and
∂P
∂Q
=
∂y
∂x
throughout D.
Then F is conservative.
Example 33 Determine whether or not the vector field F(x, y) = (x − y)i + (x − 2)j is
conservative.
1
BASIC MATERIAL.
10
Solution : Let P (x, y) = x − y and Q(x, y) = x − 2. Then
∂P
= −1
∂y
Since
∂P
∂y
6=
∂Q
∂x ,
∂Q
= 1.
∂x
F is not conservative by Theorem 31.
Example 34 Determine whether or not the vector field F(x, y) = (3 + 2xy)i + (x2 − 3y 2 )j
is conservative.
Solution : Let P (x, y) = 3 + 2xy and Q(x, y) = x2 − 3y 2 . Then
∂P
∂Q
= 2x =
.
∂y
∂x
Also, the domain of F is the entire plane (D = R2 ), which is open and simply-connected.
Therefore, we can apply Theorem 32 and conclude that F is conservative.
Attention! You will likely have a problem on the final exam which is similar to the
one described in the next example.
Example 35 (a) If F(x, y) = (3+2xy)i+(x2 −3y 2 )j, find a function f such that F = ∇f .
R
(b) Evaluate the line integral C F · dr, where C is the curve given by
r(t) = et sin t i + et cos t j, 0 ≤ t ≤ π.
Solution :
(a) From Example 34 we know that F is conservative and so there exists a function f
with ∇f = F, that is,
fx (x, y) = 3 + 2xy
(3)
fy (x, y) = x2 − 3y 2 .
(4)
Integrating (3) with respect to x, we obtain
f (x, y) = 3x + x2 y + g(y).
(5)
Notice that the constant of integration is a constant with respect to x, that is, a
function of y, which we have called g(y). Next we differentiate both sides of (5) with
respect to y:
fy (x, y) = x2 + g 0 (y).
(6)
Comparing (4) and (6), we see that
g 0 (y) = −3y 2 .
1
BASIC MATERIAL.
11
Integrating with respect to y, we have
g(y) = −y 3 + K
where K is a constant. Putting this in (5), we have
f (x, y) = 3x + x2 y − y 3 + K
as the desired potential function.
(b) To use Theorem 21 all we have to know are the initial and terminal points of C,
namely, r(0) = (0, 1) and r(π) = (0, −eπ ). In the expression for f (x, y) in part (a),
any value of the constant K will do, so let’s choose K = 0. Then we have
Z
Z
F · dr =
∇f · dr = f (0, −eπ ) − f (0, 1)
C
C
= e3π − (−1) = e3π + 1.
This method is much shorter than the straightforward method for evaluating line
integrals that we learned in Section 16.2.
Definition 36 A simple closed parametrized curve C in R2 always bounds a bounded
simply-connected domain D. We say that C is positively oriented if for the parametrization r(t) of C, the region D is always on the left at r(t) traverses C. Note that this
parametrization is the counterclockwise one one the boundary of unit disk D = {(x, y) |
x2 + y 2 ≤ 1}.
The next theorem is a version of the fundamental theorem of calculus since it allows one
to carry out a two dimensional integral on a domain D by calculating a related integral on
the boundary of D. There will be at least one final exam problem related to the following
theorem.
Theorem 37 (Green’s Theorem) Let C be a positively oriented, piecewise-smooth,
simple closed curve in the plane and let D be the region bounded by C. If P and Q have
continuous partial derivatives on an open region that contains D, then
¶
Z
Z Z µ
∂Q ∂P
−
dA.
P dx + Qdy =
∂x
∂y
C
D
An immediate consequence of Green’s Theorem are the area formulas described in the
next Theorem.
2
PRACTICE PROBLEMS FOR MIDTERM 2.
12
Theorem 38 Let D be a simply-connected domain in the plane with simple closed oriented boundary curve C. Let A be the area of D. Then:
I
I
I
1
A=
x dy = −
y dx =
x dy − y dx.
(7)
2 C
C
C
Example 39 Find the area enclosed by the ellipse
x2
a2
+
y2
b2
= 1.
Solution : The ellipse has parametric equations x = a cos t and y = b sin t, where 0 ≤
t ≤ 2π. Using the third formula in in Equation 7, we have
Z
1
A =
x dy − y dx
2 C
Z
1 2π
=
(a cos t)(b cos t) dt − (b sin t)(−a sin t) dt
2 0
Z
ab 2π
=
dt = πab.
2 0
p
H
Example 40 Evaluate C (3y − esin x ) dx + (7x + y 4 + 1) dy, where C is the circle
x2 + y 2 = 9.
Solution : The region D bounded by C is the disk x2 + y 2 ≤ 9, so let’s change to
polar coordinates after applying Green’s Theorem:
I
p
(3y − esin x ) dx + (7x + y 4 + 1) dy
C
¸
Z Z ·
p
∂
∂
=
(7x + y 4 + 1) −
(3y − esin x ) dA
∂x
∂y
D
Z 2π Z 3
=
(7 − 3) r dr dθ
0
0
Z 2π Z 3
= 4
dθ
r dr = 36π.
0
2
0
Practice problems for midterm 2.
These problems are essentially take from the final exam from math 233 in the fall 2006
class, which is also listed on the main course web site for math 233.
1. Find parametric equations for the line in which the planes x − 2y + z = 1 and
2x + y + z = 1 intersect.
3
MORE PRACTICE PROBLEMS AND AN ANSWER GUIDE.
13
2. Consider the surface x2 + y 2 − 2z 2 = 0 and the point P (1, 1, 1) which lies on the
surface.
(i) Find the equation of the tangent plane to the surface at P .
(ii) Find the equation of the normal line to the surface at P .
3. Find the maximum and minimum values of the function
f (x, y) = x2 + y 2 − 2x
on the disc x2 + y 2 ≤ 4.
4. Evaluate the iterated integral
Z 1Z
0
√
1−x2
p
x2 + y 2 dy dx.
0
5. Find the volume of the solid under the surface z = 4 − x2 − y 2 and above the region
in the xy plane between the circles x2 + y 2 = 1 and x2 + y 2 = 4.
6. Determine whether the following vector fields are conservative or not. Find a potential function for those which are indeed conservative.
(a) F(x, y) = (x2 + xy)i + (xy − y 2 )j.
(b) F(x, y) = (3x2 y + y 2 )i + (x3 + 2xy + 3y 2 )j.
7. (15 points) Evaluate the line integral
Z
(x2 + y)dx + (xy + 1)dy
C
where C is the curve starting at (0, 0), traveling along a line segment to (1, 2) and
then traveling along a second line segment to (0, 3).
8. (15 points) Use Green’s Theorem to evaluate the line integral
Z
F · dr
C
where F =< y 3 + sin 2x, 2xy 2 + cos y > and C is the unit circle x2 + y 2 = 1 which
is oriented counterclockwise.
3
More practice problems and an answer guide.
Here are the practice problems that are also listed on the main course web site.
3
MORE PRACTICE PROBLEMS AND AN ANSWER GUIDE.
3.1
14
1st set of problems.
1. (a) Express the double integral
Z Z
x2 y − x dA
R
as an integrated integral and evaluate it, where R is the first quadrant region
enclosed by the curves y = 0, y = x2 and y = 2 − x.
(b) Find an equivalent iterated integral expression for the double integral in 1a,
where the order of integration is reversed from the order used in part 1a. (Do
not evaluate this integral.)
2. Calculate the line integral
Z
F · dr,
C
where F(x, y) = y 2 xi + xyj, and C is the path starting at (1, 2), moving along a line
segment to (3, 0) and then moving along a second line segment to (0, 1).
3. Evaluate the integral
Z Z
y
p
x2 + y 2 dA
R
with R the region {(x, y) : 1 ≤ x2 + y 2 ≤ 2, 0 ≤ y ≤ x}.
4. (a) Show that the vector field
F(x, y) =
¿
1
x
+ 2x, − 2 + 1
y
y
À
is conservative by finding a potential function f (x, y).
(b) Let C be the path described by the parametric curve r(t) = h1 + 2t, 1 + t2 i for
0 ≤ t ≤ 1. Use your answer from 4a to determine the value of the line integral
Z
F · dr.
C
5. (a) Find the equation of the tangent plane at the point P = (1, 1, −1) in the level
surface f (x, y, z) = 3x2 + xyz + z 3 = 1.
(b) Find the directional derivative of the function f (x, y, z) at P = (1, 1, −1) in the
direction of the tangent vector to the space curve r(t) = h2t2 − t, t−2 , t2 − 2t3 i
at t = 1.
6. Find the absolute maxima and minima of the function
f (x, y) = x2 − 2xy + 2y 2 − 2y
in the region bounded by the lines x = 0, y = 0 and x + y = 7.
3
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3.2
15
2nd set of problems
1. Consider the function f (x, y) = xexy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in the direction of the
point (3, 2).
(b) Give a direction in terms of a unit vector (there are two possibilities) for which
the rate of change of f at P in that direction is zero.
2. (a) Find the work done by the vector field F(x, y) = hx, −y, xi over the circle
r(t) = hcos t, sin th,, 0 ≤ t ≤ 2π.
R
(b) Use Green’s Theorem to calculate the line integral C (−y 2 ) dx + xy dy, over the
positively (counterclockwise) oriented closed curve defined by x = 1, y = 1 and
the coordinate axis.
3. (a) Show that the vector field F(x, y) = hx2 y, 13 x3 i is conservative and find a function f such that F = ∇f .
R
(b) Using the result in the part 3a calculate the line integral C F · dr, along the
curve C which is the arc of y = x4 from (0, 0) to (2, 16).
√
4. Consider the surface x2 + y 2 − 14 z 2 = 0 and the point P (1, 2, −2 5) which lies on
the surface.
(a) Find the equation of the tangent plane to the surface at the point P .
(b) Find the equation of the normal line to the surface at the point P .
5. A flat circular plate has the shape of the region x2 + y 2 ≤ 1. The plate (including
the boundary x2 + y 2 = 1). is heated o that the temperature at any point (x, y) on
the plate is given by
T (x, y) = x2 + 2y 2 − x
Find the temperatures at the hottest and the coldest points on the plate, including
the boundary x2 + y 2 = 1.
6. The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any, are the velocity
and the acceleration of the particle orthogonal?
7. Find parametric equations for the line in which the planes 3x − 6y − 2z = 15 and
2x + y − 2z = 5 intersect.
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3.3
16
3rd set of problems.
1. (a) Find the equation of the plane containing the points P (1, 3, 0), Q(2, −1, 2) and
R(0, 0, 1).
(b) Find all points of intersection of the parametric curve r(t) = h2t2 −2, t, 1−t−t2 i
and the plane x + y + z = 3.
of the function f (x, y) = x2 + 2y 2 − 2y
2. Find the absolute maximum and minimum
√
on the closed disc x2 + y 2 ≤ 5 of radius 5.
3. Evaluate
Z Z
xy dA
R
where R is the region in the first quadrant bounded by the line y = 2x and the
parabola y = x2 .
4. Consider the vector field F(x, y) = h2xy + sin y, x2 + x cos y + 1i.
(a) Show that F(x, y) is conservative by finding a potential function f (x, y) for
F(x, y).
(b) Use your answer to 4a to evaluate the line integral
Z
F · dr
C
where C is the arc of the parabola y = x2 going from (0, 0) to (2, 4).
5. Evaluate the line integral
Z
F · dr
C
where F = hy 2 + sin x, xyi and C is the unit circle oriented counterclockwise.
6. Evaluate the line integral
Z
F · dr
C
where F = hy 2 , 2xy + xi and C is the curve starting at (0, 0), traveling along a line
segment to (2, 1) and then traveling along a second line segment to (0, 3).
Here the answers to the above practice problems.
1st set.
3
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3.
Z Z
y
p
Z
x2 + y 2 dA =
R
π
4
0
Z
√
1
2
µ 4√ ¶
³
√
π´
√
r 2
3
r sin θ r2 r dr dθ = − cos θ 04
= (2 − 2).
1
4
8
4 (a). f (x, y) = xy + x2 + y
(b). f (3, 2) − f (1, 1) = 19
2 .
5 (a).
5(x − 1) − (y − 1) + 4(z + 1) = 0.
(b).
1
1
D~u f (1, 1, −1) = ∇f (1, 1, −1) · ~u = h5, −1, 4i · √ h3, −2, −4i = √ .
29
29
6. max = 84 at (7, 0) and min = −1 at (1, 1).
2nd set.
1. (a).
1
4
D~u f (1, 0) = h1, 1i · √ h2, 2i = √ .
8
8
(b). ~u = h± √12 , ± √12 i.
2 (a). 2π.
(b). 32
3 (a). f (x, y) = 13 x3 y.
(b). f (2, 16) − f (0, 0) =
17
128
3 .
p
√
4 (a). 2(x − 1) + 4(y√− 2) + (z +√2 5) = 0.
(b). r(t0 = h1, 2, −2 5i + t · h2, 4, 5i.
6. t = 0.
7. The line is given by
r(t) = h3, −1, 0i + th14, 2, 15i
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.
3rd set.
1 (a). 2(x − 1) − 3(y − 3) − 7z = 0.
(b). (6, 2, −5) when t = 2 and (6, −2, −1) when t = −2.
√
2. max = 10 + 2 5 and min = − 12 .
3.
x4
2
−
x6 2
12 0
' 2.67.
4 (a). f (x, y) = x2 y + x sin y + y.
(b). f (2, 4) − f (0, 0) = 20 + 2 sin 4.
5. 0.
6. 3.
18