Download HOMEWORK 1 – SOLUTION

United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 1 – SOLUTION
Section 10.1 Vectors in Plane
Section 10.2 Vectors in Space
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Monday, September 15, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
1ST HOMEWORK – SOLUTION
Fall, 2008
1. Compute −3a + 2b for a = 〈 3, 2 〉 and b = 〈 3, −13 〉.
Answer.
−3a + 2b = −3 〈 3, 2 〉 + 2 〈 3, −13 〉 = 〈 −9, −6 〉 + 〈 6, −26 〉
ä
= 〈 −9 + 6, −6 − 26 〉 = 〈 −3, −32 〉 .
2. Determine whether the vectors a = 〈 1, −2 〉 and b = 〈 2, 1 〉 are parallel.
Answer. We recall that a and b are parallel if and only if there is a scalar s such that
a = sb. If a = sb, then
〈 1, −2 〉 = s 〈 2, 1 〉 = 〈 2s, s 〉 ,
1 = 2s
and
− 2 = s.
There is no such a scalar s satisfying both 2s = 1 and s = −2. Therefore, given two vectors
cannot be parallel.
ä
3. Find the vector with initial point A = (2, 3) and terminal point B = (5, 4).
−−→
Answer. AB = 〈 5 − 2, 4 − 3 〉 = 〈 3, 1 〉 .
ä
4. Find a vector with the magnitude 4 in the same direction as the vector v = 2i − j .
Answer. Let u be such a vector, i.e., the one with the magnitude 4 in the same direction as
the vector v = 2i − j . Since u is parallel to v , there should be a positive scalar s (negative
scalar gives the opposite direction) such that u = sv . Since u has the magnitude 4, we
have
i.e.,
p
4 = kuk = k sv k = | s|kv k = | s|k2i − j k = | s| 5,
p
p
4
4 5
4 5
| s| = p =
,
i.e., s =
.
5
5
5
p
4 5
Therefore, we deduce such a vector u =
(2i − j ) .
5
ä
5. The thrust of an airplane’s engines produces a speed of 600 mph in still air. The wind
velocity is given by 〈 −30, 60 〉. In what direction should the airplane head to fly due west?
Answer. Let v = 〈 x, y 〉 and w = 〈 −30, 60 〉 be the velocities of the airplane and the wind,
respectively. Since we want the airplane to move due west, the sum of two vectors v and
w should satisfy v + w = 〈 c, 0 〉, where the vector 〈 c, 0 〉 points due west with the negative
constant c. The equation implies
〈 c, 0 〉 = v + w = 〈 x, y 〉 + 〈 −30, 60 〉 = 〈 x − 30, y + 60 〉 ,
i.e.,
c = x − 30,
and
y = −60.
Page 1 of 5
Calculus II for Engineering
1ST HOMEWORK – SOLUTION
Fall, 2008
The airplane produces a speed of 600 mph and so we get kv k = 600, which implies
q
6002 = x2 + y2 = x2 + (−60)2 ,
600 = kv k = 〈 x, y 〉 = x2 + y2 ,
p
p
x = ± 6002 − 602 = ±180 11.
p
Because of the condition that x − 30 = c < 0, we choose x = −180 11, hence,
p
p
v = 〈 x, y 〉 = 〈 −180 11, −60 〉 = −60 〈 3 11, 1 〉 .
µ
This points down and left at an angle of tan
−1
¶
1
≈ 5.73917◦ south of west.
p
3 11
ä
6. If vector a has magnitude kak = 3 and vector b has magnitude kbk = 4, what is the largest
possible magnitude for the vector a + b? What is the smallest possible magnitude for the
vector a + b? What will be the magnitude of a + b if a and b are perpendicular?
Answer. The largest magnitude of a + b is 7 (if the vectors point in the same direction).
The smallest magnitude is 1 (if the vectors point in the opposite directions). If the vectors
are perpendicular, then a + b p
can be viewed as the hypotenuse of a right triangle with
sides a and b, so it has length 32 + 42 = 5.
ä
7. For the vector 4i − 2j + 4k, (1) find two unit vectors parallel to the given vector and (2)
write the given vector as the product of its magnitude and a unit vector.
Answer. Let v = 4i − 2j + 4k = 2 (2i − j + 2k). Then it has the magnitude
kv k = k2 (2i − j + 2k) k = 2k (2i − j + 2k) k = 2
p
p
22 + (−1)2 + 22 = 2 9 = 6.
(1) The unit vector in the same direction as v can be found by
v
2 (2i − j + 2k) 1
=
= (2i − j + 2k) .
kv k
6
3
The unit vector in the opposite direction as v is
−
1
v
= − (2i − j + 2k) .
kv k
3
1
Thus, two unit vectors parallel to v are ± (2i − j + 2k).
3
(2) Using the results above, we have
µ
¶
v
2
1
2
v = kv k
= 6 i− j + k .
kv k
3
3
3
ä
8. Identify the plane y = 4 as parallel to the x y–plane, xz–plane or yz–plane and sketch a
graph.
Page 2 of 5
Calculus II for Engineering
1ST HOMEWORK – SOLUTION
Fall, 2008
Answer. The plane y = 4 is parallel to the xz–plane and passes through (0, 4, 0).
−−→
ä
−−→
9. Find the displacement vectors PQ and QR and determine whether the points P = (2, 3, 1),
Q = (0, 4, 2) and R = (4, 1, 4) are colinear (on the same line).
Answer. The displacement vectors are
−−→
PQ = 〈 0 − 2, 4 − 3, 2 − 1 〉 = 〈 −2, 1, 1 〉 ,
−−→
QR = 〈 4 − 0, 1 − 4, 4 − 2 〉 = 〈 4, −3, 2 〉 .
−−→
−−→
There does not exist any scalar s such that PQ = sQR, because no scalar s satisfies simultaneously
−2 = 4s,
1 = −3s,
1 = 2s.
−−→
−−→
It implies that two vectors PQ and QR are not parallel. That is, the points P, Q and R
are not colinear.
ä
10. Use vectors to determine whether the points (2, 1, 0), (5, −1, 2), (0, 3, 3) and (3, 1, 5) form a
square.
Answer.
We recall that if a square has the side length s, then it has the diagonal of length
p
s 2 by the Pythagorean Theorem.
Let P = (2, 1, 0), Q = (5, −1, 2), R = (0, 3, 3) and S = (3, 1, 5). There are six pairs of vectors
−−→ −−→ −−→ −−→ −−→ −−→
−−→ −−→
(PQ, PR, PS, QR, QS, RS) and four of them will correspond to sides. Further PQ, PR
−−→
and PS should form two sides and one diagonal of the square, because they have the same
initial point. When we compute the lengths, we get
p
−−→
kPQ k = k 〈 3, −2, 2 〉 k = 17,
p
−−→
kPS k = k 〈 1, 0, 5 〉 k = 26.
p
−−→
kPR k = k 〈 −2, 2, 3 〉 k = 17,
p
p p
−−→
−−→
−−→ p
It implies PS should be the diagonal. However, since 26 6= 17 2, i.e., kPS k 6= kPQ k 2,
−−→
so PS cannot be the diagonal of a square. That is, those given points cannot form a
square.
ä
11. In the accompanying figure, two ropes are attached to a 300–pound crate. Rope A exerts
a force of 〈 10, −130, 200 〉 pounds on the crate, and rope B exerts a force of 〈 −20, 180, 160 〉
pounds on the crate.
(11.1) If no further ropes are added, find the net force on the crate and the direction it will
move.
Answer. Let the force due to rope A be a = 〈 10, −130, 200 〉, the force due to rope B
be b = 〈 −20, 180, 160 〉, and write the force due to gravity as w = 〈 0, 0, −300 〉. Then
the net force is
a + b + w = 〈 10, −130, 200 〉 + 〈 −20, 180, 160 〉 + 〈 0, 0, −300 〉 = 〈 −10, 50, 60 〉 .
ä
(11.2) If a third rope C is added to balance the crate, what force must this rope exert on the
crate?
Page 3 of 5
Calculus II for Engineering
1ST HOMEWORK – SOLUTION
Fall, 2008
Answer. In order to compensate, rope C must exert a force of 〈 10, −50, −60 〉 or 78.74
(= k 〈 10, −50, −60 〉 k) pounds in direction 〈 1, −5, −6 〉.
ä
(11.3) We want to move the crate up and to the right with a constant force of 〈 0, 30, 20 〉
pounds. If a third rope C is added to accomplish this, what force must the rope exert
on the crate?
Answer. Let the force due to rope C be c. We want the net force to be
a + b + c + w = 〈 0, 30, 20 〉 ,
i.e.,
c + 〈 −10, 50, 60 〉 = 〈 0, 30, 20 〉 ,
where a, b and w defined in (1) above are used. That is,
c = 〈 0, 30, 20 〉 − 〈 −10, 50, 60 〉 = 〈 10, −20, −40 〉 .
So rope C must exert a force of 〈 10, −20, −40 〉 or 45.8 (= k 〈 10, −20, −40 〉 k) pounds
in direction 〈 1, −2, −4 〉.
ä
12. If v ∈ V2 lies in the first quadrant of the x y–plane and makes the angle θ = π/3 with the
positive x–axis and the magnitude 4, then find v in the component form.
Answer. The unit vector making the angle θ = π/3 with the positive x–axis is given by
〈 cos(π/3), sin(π/3) 〉. (Why?) Hence, the desired vector v ∈ V2 is obtained by
p
p
1 3
〉 = 〈 2, 2 3 〉 .
v = 4 〈 cos(π/3), sin(π/3) 〉 = 4 〈 ,
2 2
ä
13. (Think!) Draw the vectors a = 〈 3, 2 〉, b = 〈 2, −1 〉 and c = 〈 7, 1 〉. By using the sketch
involved with those vectors, show that there exist scalars s and t such that c = sa + tb.
Can you prove the existence of such s and t algebraically? Justify your answer.
Answer. The equation c = sa + tb implies 〈 7, 1 〉 = s 〈 3, 2 〉 + t 〈 2, −1 〉, i.e., 7 = 3s + 2t and
9
11
1 = 2s − t. Solving the equations for s and t, we get s = and t = .
ä
7
7
Page 4 of 5
Calculus II for Engineering
1ST HOMEWORK – SOLUTION
Fall, 2008
14. Find the correct figure of the sum a + b with a = −2i − j and b = −3i + 2j , where i and j
are standard basis vectors of V2 .
Answer. A simple computation shows
a + b = −2i − j − 3i + 2j = −5i + j = 〈 −5, 1 〉 .
That is, a + b should point in the direction 〈 −5, 1 〉. It is easy to see that the red–colored
vector in (2) is the vector 〈 −5, 1 〉. Hence, the answer is (2).
ä
H1L
5
-5
-5
-4
-4
-3
-3
-2
-2
H2 L
5
4
4
3
3
2
2
1
1
1
-1
2
3
4
5
-5
-4
-3
-2
1
-1
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
H3 L
5
H4L
5
4
4
3
3
2
2
1
1
1
-1
2
3
4
5
-5
-4
-3
-2
1
-1
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
Page 5 of 5
2
2
3
3
4
4
5
5
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 2 – SOLUTION
Section 10.3 The Dot Product
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Monday, September 22, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
2ND HOMEWORK – SOLUTION
Fall, 2008
1. Compute a · b.
(1.1) a = 〈 3, 2, 0 〉 and b = 〈 −2, 4, 3 〉.
Answer.
a · b = 〈 3, 2, 0 〉 · 〈 −2, 4, 3 〉 = 3(−2) + 2(4) + 0(3) = 2.
ä
(1.2) a = 2i − k and b = 4j − k.
Answer.
a · b = (2i − k) · (4j − k) = 2(0) + 0(4) + (−1)(−1) = 1.
ä
2. Compute the angle between the vectors.
(2.1) a = 〈 2, 0, −2 〉 and b = 〈 0, −2, 4 〉.
Answer.
cos θ =
〈 2, 0, −2 〉 · 〈 0, −2, 4 〉
a·b
2
=
= −p ,
kakkbk k 〈 2, 0, −2 〉 kk 〈 0, −2, 4 〉 k
10
µ
θ = cos
−1
¶
2
≈ 2.25552
−p
10
ä
(2.2) a = 3i + j − 4k and b = −2i + 2j + k.
Answer.
cos θ =
a·b
−8
(3i + j − 4k) · (−2i + 2j + k)
=
= p ,
kakkbk k3i + j − 4kk − 2i + 2j + kk 3 26
−1
θ = cos
µ
¶
8
− p
≈ 2.12114
3 26
ä
3. Determine whether the vectors are orthogonal.
(3.1) a = 〈 4, −1, 1 〉 and b = 〈 2, 4, 4 〉.
Answer.
a · b = 〈 4, −1, 1 〉 · 〈 2, 4, 4 〉 = 8 6= 0.
ä
So a and b are not orthogonal.
(3.2) a = 6i + 2j and b = −i + 3j .
Answer.
a · b = (6i + 2j ) · (−i + 3j ) = 0.
ä
So a and b are orthogonal.
4. Find a vector perpendicular to the given vector.
(4.1) 〈 4, −1, 1 〉.
Page 1 of 4
Calculus II for Engineering
2ND HOMEWORK – SOLUTION
Fall, 2008
Answer. Let v = 〈 a, b, c 〉 be a vector perpendicular to u = 〈 4, −1, 1 〉. Then we should
have v · u = 0, i.e.,
0 = v · u = 〈 a, b, c 〉 · 〈 4, −1, 1 〉 = 4a − b + c,
i.e.,
4a − b + c = 0.
There are so many numbers, a, b, and c satisfying the equation. We choose just
one, a = 1, b = 4 and c = 0. That is, v = 〈 1, 4, 0 〉 is one vector perpendicular to u =
〈 4, −1, 1 〉.
ä
(4.2) 6i + 2j − k.
Answer. Let v = 〈 a, b, c 〉 = ai + bj + ck be a vector perpendicular to w = 6i + 2j − k.
Then by the same argument as above, we deduce
0 = v · w = (ai + bj + ck) · (6i + 2j − k) = 6a + 2b − c,
i.e.,
6a + 2b − c = 0.
There are so many numbers, a, b, and c satisfying the equation. We choose just one,
a = 1, b = −1 and c = 4. That is, v = 〈 1, −1, 4 〉 = i − j + 4k is one vector perpendicular
to w = 6i + 2j − k.
ä
5. Find Compb a and Projb a.
(5.1) a = 3i + j and b = 4i − 3j .
Answer.
a · b (3i + j ) · (4i − 3j ) 9
=
=
kbk
k4i − 3j k
5
¡
¢ b
9 4i − 3j
9
=
=
Projb a = Compb a
(4i − 3j )
kbk 5 k4i − 3j k 25
Compb a =
ä
(5.2) a = 〈 3, 2, 0 〉 and b = 〈 −2, 2, 1 〉.
Answer.
a · b 〈 3, 2, 0 〉 · 〈 −2, 2, 1 〉
2
=
=−
kbk
k 〈 −2, 2, 1 〉 k
3
¡
¢ b
2 〈 −2, 2, 1 〉
2
Projb a = Compb a
=−
= − 〈 −2, 2, 1 〉
kbk
3 k 〈 −2, 2, 1 〉 k
9
Compb a =
ä
6. A constant force of 〈 60, −30 〉 pounds moves an object in a straight line from the point
(0, 0) to the point (10, −10). Compute the work done.
Answer. The displacement vector is d = 〈 10 − 0, −10 − 0 〉 = 〈 10, −10 〉. The force is F =
〈 60, −30 〉. By the formula, the work W done is obtained by
W = F · d = 〈 10, −10 〉 · 〈 60, −30 〉 = 10(60) + (−10)(−30) = 900.
ä
7. Label each statement as true or false. If it is true, briefly explain why; if it is false, give a
counterexample.
Page 2 of 4
Calculus II for Engineering
2ND HOMEWORK – SOLUTION
Fall, 2008
(7.1) If a · b = a · c, then b = c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. a = 〈 1, 0, 0 〉, b = 〈 0, 1, 0 〉 and c = 〈 0, 0, 1 〉 satisfies a · b = 0 = a · c. But, obviously, b 6= c.
ä
(7.2) If b = c, then a · b = a · c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T
Answer. If b = c, then b − c = 0 and so
a · (b − c) = a · 0 = 0,
i.e.,
a · b − a · c = 0,
i.e.,
a · b = a · c.
(7.3) a · a = kak2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ä
T
Answer. The formula a · b = kakkbk cos θ , where θ is the angle between a and b, implies
a · a = kakkak cos 0 = kak2 ,
i.e.,
a · a = kak2 .
One may compute the dot product with a = 〈 a, b, c 〉 and prove the equality.
ä
(7.4) If kak > kbk, then a · c > b · c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. We observe a = 〈 2, 0 〉 and b = 〈 0, 1 〉 satisfy the inequality kak = 2 > 1 = kbk.
However, with c = 〈 0, 3 〉, we get a · c = 0 < 3 = b · c.
ä
(7.5) If kak = kbk, then a = b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. We observe i = 〈 1, 0, 0 〉, j = 〈 0, 1, 0 〉 and k = 〈 0, 0, 1 〉 have kik = 1 = kj k =
kkk. However, obviously, i 6= j 6= k.
ä
8. By the Cauchy–Schwartz Inequality, |a · b| ≤ kakkbk. What relationship must exist between a and b to have the equality |a · b| = kakkbk?
Answer. (1) a = 0 or b = 0. (2) The cosine of the angle between the vectors is ±1. This
happens exactly when the vectors point in the same or opposite directions. In other words,
when a = sb for some scalar s.
ä
9. By the Triangle Inequality, ka + bk ≤ kak + kbk. What relationship must exist between
a and b to have the equality ka + bk = kak + kbk?
Answer. (1) a = 0 or b = 0. (2) Vectors a and b must be parallel with the same direction so
that a = sb for some positive scalar s.
ä
10. The orthogonal projection of vector a along vector b is defined as Orthb a = a − Projb a.
Sketch a picture showing vectors a, b, Projb a and Orthb a, and explain what is orthogonal
about Orthb a.
Answer. Orthb a is the component of a that is orthogonal to b:
¡
¢
b · Orthb a = b · a − Projb a = b · a − b · Projb a
(a · b)b
(a · b)(b · b)
= b·a−b·
=
b
·
a
−
= b · a − a · b = 0,
kbk2
kbk2
Page 3 of 4
Calculus II for Engineering
2ND HOMEWORK – SOLUTION
Fall, 2008
ä
where b · b = kbk2 is used above.
11. A car makes a turn on a banked road. If the road is banked at 15◦ , show that a vector
parallel to the road is 〈 cos 15◦ , sin 15◦ 〉. If the car has weight 2500 pounds, find the component of the weight vector along the road vector. This component of weight provides a
force that helps the car turn.
Answer. The vector b = 〈 cos 15◦ , sin 15◦ 〉 represents the direction of the banked road. The
weight of the car is w = 〈 0, −2500 〉. The component of the weight in the direction of the
bank is
w·b
Compb w =
= −2500 sin 15◦ ≈ −647.0 lbs
kbk
ä
toward the inside of the curve.
Page 4 of 4
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 3 – SOLUTION
Section 10.4 The Cross Product
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Monday, October 6, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
3RD HOMEWORK – SOLUTION
1. Compute the determinant:
Answer.
Fall, 2008
¯
¯
¯ 0 2 −1 ¯
¯
¯
¯
¯
¯ 1 −1 2 ¯
¯
¯
¯ 1 1
2 ¯
¯
¯
¯ 0 2 −1 ¯
¯
¯
¯
¯
¯ 1 −1 2 ¯ = 0(−2 − 2) − 2(2 − 2) − (1 + 1) = −2.
¯
¯
¯ 1 1
2 ¯
ä
2. Compute the cross product a × b.
(2.1) a = 〈 2, −2, 0 〉 and b = 〈 3, 0, 1 〉.
Answer.
¯
¯
¯ i j k ¯
¯
¯
¯
¯
a × b = 〈 2, −2, 0 〉 × 〈 3, 0, 1 〉 = ¯ 2 −2 0 ¯ = −2i − 2j + 6k = −2 〈 1, 1, −3 〉
¯
¯
¯ 3 0 1 ¯
ä
(2.2) a = −2i + j − 3k and b = 2j − k.
Answer.
¯
¯
¯ i j k ¯
¯
¯
¯
¯
a × b = 〈 2, 1, −3 〉 × 〈 0, 2, −1 〉 = ¯ −2 1 −3 ¯ = 5i − 2j − 4k = 〈 5, −2, −4 〉
¯
¯
¯ 0 2 −1 ¯
ä
3. Find two unit vectors orthogonal to the two given vectors.
(3.1) a = 〈 0, 2, 1 〉 and b = 〈 1, 0, −1 〉.
Answer. We recall
Orthogonal Vector
a × b is orthogonal to both a and b. Thus, ±
to both a and b.
a×b
are two unit vectors orthogonal
ka × bk
a × b = 〈 0, 2, 1 〉 × 〈 1, 0, −1 〉 = 〈 −2, 1, −2 〉 ,
±
a×b
1
= ± 〈 −2, 1, −2 〉
ka × bk
3
ä
(3.2) a = −2i + 3j − 3k and b = 2i − k.
Answer.
a × b = 〈 −2, 3, −3 〉 × 〈 2, 0, −1 〉 = 〈 −3, −8, −6 〉 ,
Page 1 of 5
±
a×b
1
〈 −3, −8, −6 〉
= ±p
ka × bk
109
ä
Calculus II for Engineering
3RD HOMEWORK – SOLUTION
Fall, 2008
4. Use the cross product to determine the angle between the vectors, assuming that 0 ≤ θ ≤
π/2.
(4.1) a = 〈 2, 2, 1 〉 and b = 〈 0, 0, 2 〉.
Answer. We recall
Angle Between Two Vectors
ka × bk = kak kbk sin θ ,
ka × bk
sin θ =
,
kak kbk
i.e.,
µ
i.e.,
θ = sin
−1
ka × bk
kak kbk
¶
where 0 ≤ θ ≤ π is the angle between a and b.
a × b = 4 〈 1, −1, 0 〉 ,
kak = 3,
kbk = 2,
p
ka × bk = 4 2,
à p !
4 2
θ = sin−1
≈ 1.23096 rad ≈ 70.5288◦
6
ä
(4.2) a = i + 3j + 3k and b = 2i + j .
Answer.
a × b = 〈 −3, 6, −5 〉 ,
kak =
p
19,
kbk =
p
5,
Ãp !
70
θ = sin−1 p
≈ 1.03213 rad ≈ 59.1369◦
95
ka × bk =
p
70,
ä
5. Find the distance from the point Q = (1, 3, 1) to the line through (1, 3, −2) and (1, 0, −2).
Answer. The distance d from the point Q to the line through the points P and R is obtained by the following formula.
Distance from a Point to a Line
°−−→ −−→°
°
°
°PQ × PR °
°−−→°
d=
°
°
°PR °
Letting P = (1, 3, −2) and R = (1, 0, −2), we have
−−→
PQ = 3 〈 0, 0, 1 〉 ,
−−→
PR = −3 〈 0, 1, 0 〉 ,
−−→ −−→
PQ × PR = 9 〈 1, 0, 0 〉 ,
d=
9 k〈 1, 0, 0 〉k
= 3.
3 k〈 0, 1, 0 〉k
ä
6. If you apply a force of magnitude 30 pounds at the end of an 8–inch–long wrench at an
angle of π/3 to the wrench, find the magnitude of the torque applied to the bolt.
Answer. We recall
Page 2 of 5
Calculus II for Engineering
3RD HOMEWORK – SOLUTION
Fall, 2008
Torque
Torque τ is defined to be the cross product of the position vector r and force vector F ,
i.e.,
kτ k = kr k kF k sin θ
τ = r×F ,
where θ is the angle between r and F .
Given information: kF k = 30 and θ = π/3 and kr k = 8 inch = 8/12 = 2/3 feet. Thus we have
p
2
π
kτ k = (30) sin = 10 3 ft–lbs
3
3
ä
7. Label each statement as true or false. If it is true, briefly explain why. If it is false, give a
counterexample.
(7.1) If a × b = a × c, then b = c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. We observe that for a = 〈 1, 0, 0 〉, b = 〈 1, 1, 1 〉 and c = 〈 2, 1, 1 〉,
a × b = 〈 0, −1, 1 〉 = a × c,
but
b 6= c.
(7.2) a × b = −b × a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ä
T
Answer. We observe that for a = 〈 a 1 , a 2 , a 3 〉 and b = 〈 b 1 , b 2 , b 3 〉,
a × b = 〈 a 2 b3 − a 3 b2 , a 1 b3 − a 3 b1 , a 1 b2 − a 2 b1 〉 = −b × a.
(7.3) a × a = kak2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ä
F
Answer. a × a = 〈 0, 0, 0 〉 is a vector, while kak is a scalar. As a counterexample, for
a = i, we get i × i = 0 = 〈 0, 0, 0 〉 6= 1 = kik2 .
ä
(7.4) a · (b × c) = (a · b) × c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. (a · b) × c is not possible because a · b is a scalar. A cross product must involve
two vectors.
ä
(7.5) If the force is doubled, the torque doubles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T
Answer. Torque is the cross product of direction and force.
r × (2F ) = 2(r × F ) = 2τ .
ä
8. Find the area of the parallelogram with two adjacent sides formed by 〈 −2, 1 〉 and 〈 1, −3 〉.
Answer. We recall
Page 3 of 5
Calculus II for Engineering
3RD HOMEWORK – SOLUTION
Fall, 2008
Area
The area of the parallelogram with two adjacent sides formed by a and b is the magnitude of their cross product:
A = ka × bk = kak kbk sin θ
where 0 ≤ θ ≤ π is the angle between a and b.
The vectors 〈 −2, 1 〉 and 〈 1, −3 〉 are in the plane (R2 ). The cross product is defined only
for the vectors in the space (R3 ). But we observe 〈 −2, 1, 0 〉 and 〈 1, −3, 0 〉 in the space can
correspond to those vectors in the plane, respectively. So we compute the area with these
vectors:
A = k〈 −2, 1, 0 〉 × 〈 1, −3, 0 〉k = k5 〈 0, 0, 1 〉k = 5.
ä
9. Find the area of the triangle with vertices (0, 0, 0), (0, −2, 1) and (1, −3, 0).
Answer. Letting P(0, 0, 0), Q(0, −2, 1) and R(1, −3, 0), by the formula, the area of the par−−→
−−→
allelogram with two adjacent sides formed by PQ and PR is the magnitude of their cross
product:
°−−→ −−→°
p
°
°
A parallelogram = °PQ × PR ° = k〈 0, −2, 1 〉 × 〈 1, −3, 0 〉k = k〈 3, 2, 1 〉k = 14.
Since the area of the triangle is the half of the area of the parallelogram, hence the desired
area is
°−−→ −−→°
°
° p
°PQ × PR °
14
=
.
ä
A triangle =
2
2
10. Find the volume of the parallelepiped with three adjacent edges formed by 〈 0, −1, 0 〉,
〈 0, 2, −1 〉, and 〈 1, 0, 2 〉.
Answer. We recall
Volume
The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product:
V = |c · (a × b)|
Letting a = 〈 0, −1, 0 〉, b = 〈 0, 2, −1 〉 and c = 〈 1, 0, 2 〉, the formula implies the volume
V = 1.
ä
11. Use geometry to identify the cross product. (Do not compute!)
(11.1) j × (j × k).
Answer. Geometry implies
Page 4 of 5
Calculus II for Engineering
3RD HOMEWORK – SOLUTION
Fall, 2008
Cross Product on Standard Basis Vectors
i×j = k
j × i = −k
j ×k = i
k × j = −i
k×i = j
i × k = −j
Hence, we deduce
j × (j × k ) = j × i = − k
ä
(11.2) (j × i) × k.
Answer. By the same argument as above, we get
(j × i) × k = −k × k = 0
ä
12. Use the parallelepiped volume formula to determine whether the vectors 〈 1, 1, 2 〉 and
〈 0, −1, 0 〉 and 〈 3, 2, 4 〉 are coplanar.
Answer. Letting a = 〈 1, 1, 2 〉, b = 〈 0, −1, 0 〉 and c = 〈 3, 2, 4 〉, the formula above implies
the volume V = 2. Since the volume of the parallelepiped is not zero, it means those three
vectors do not lie on the same plane. That is, they are not coplanar.
ä
Page 5 of 5
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 4 – SOLUTION
Section 10.5 Lines and Planes in Space
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Monday, October 13, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
4TH HOMEWORK – SOLUTION
Fall, 2008
1. Find (a) parametric equations and (b) symmetric equations of the line.
(1.1) The line through (3, −2, 4) and parallel to 〈 3, 2, −1 〉
Answer.
(a)
x = 3 + 3t,
y = −2 + 2t,
z = 4− t
x−3 y+2 z−4
=
=
3
2
−1
(b)
(1.2) The line through (−1, 0, 0) and parallel to the line
ä
x+1 y
= = z−2
−2
3
Answer. The direction comes from the given line, i.e., 〈 −2, 3, 1 〉.
(a)
x = −1 − 2t,
y = 0 + 3t,
z = 0+ t
x+1 y z
= =
−2
3 1
(b)
ä
(1.3) The line through (−3, 1, 0) and perpendicular to both 〈 0, −3, 1 〉 and 〈 4, 2, −1 〉
Answer. 〈 0, −3, 1 〉 × 〈 4, 2, −1 〉 = 〈 1, 4, 12 〉 is in the direction perpendicular to both
vectors.
(a)
x = −3 + t,
y = 1 + 4t,
z = 0 + 12t
x+3 y−1
z
=
=
1
4
12
(b)
ä
(1.4) The line through (0, −2, 1) and normal to the plane y + 3z = 4
Answer. 〈 0, 1, 3 〉 is normal to the plane.
(a)
x = 0 + 0t = 0,
y = −2 + t,
z = 1 + 3t
(b)
x = 0,
y+2 z−1
=
1
3
ä
2. State whether the lines are parallel or perpendicular and find the angle between the lines.
L : x = 4 − 2t,
M : x = 4 + s,
y = 3t,
y = −2s,
z = −1 + 2t
z = −1 + 3s
Answer. The vectors parallel to lines L and M are respectively vL = 〈 −2, 3, 2 〉 and v M =
〈 1, −2, 3 〉. Since there is no constant c satisfying vL = 〈 −2, 3, 2 〉 = c 〈 1, −2, 3 〉 = cv M , so
the lines L and M are not parallel.
By the formula on the dot product, we have
vL · v M
−2 − 6 + 6
2
cos θ =
= p p
= −p
,
kv L k k v M k
238
17 14
−1
θ = cos
µ
¶
2
−p
≈ 1.7 rad,
238
which is not π/2. Thus, the lines L and M are not perpendicular and the angle between
the lines L and M is about 1.7 rad.
ä
3. Determine whether the lines are parallel, skew or intersect.
L : x = 3 + t,
y = 3 + 3t,
Page 1 of 5
z = 4− t
Calculus II for Engineering
4TH HOMEWORK – SOLUTION
M : x = 2 − s,
y = 1 − 2s,
Fall, 2008
z = 6 + 2s
Answer. The vectors parallel to lines L and M are respectively vL = 〈 1, 3, −1 〉 and v M =
〈 −1, −2, 2 〉. Since there is no constant c satisfying vL = 〈 1, 3, −1 〉 = c 〈 −1, −2, 2 〉 = cv M ,
so the lines L and M are not parallel. To determine whether or not the lines intersect, we
set the x–, y– and z–values equal:
3 + t = 2 − s,
simpl y,
(i)
3 + 3t = 1 − 2s,
t + s = −1,
(ii) 3t + 2s = −2,
4 − t = 6 + 2s,
(iii)
t + 2s = −2.
From (i) and (ii), we get t = 0 and s = −1. Putting t = 0 and s = −1 into (iii), the equation
(iii) holds. This implies that the lines L and M intersect when t = 0 and s = −1:
x = 3,
y = 3,
z = 4.
ä
Hence, the lines intersect at the point (3, 3, 4).
4. Find an equation of the given plane.
(4.1) The plane containing the point (−2, 1, 0) with normal vector 〈 −3, 0, 2 〉
Answer.
−3(x + 2) + 0(y − 1) + 2(z − 0) = 0,
simpl y,
3x − 2z + 6 = 0.
ä
(4.2) The plane containing the points (−2, 2, 0), (−2, 3, 2) and (1, 2, 2)
−−→
Answer. Letting P(−2, 2, 0), Q(−2, 3, 2) and R(1, 2, 2), we deduce PQ = 〈 0, 1, 2 〉 and
−−→ −−→
−−→
PR = 〈 3, 0, 2 〉 and PQ × PR = 〈 2, 6, −3 〉 is normal to the plane. Hence, using the
point P(−2, 2, 0) (one can use Q or R), the equation of the plane is obtained as
2(x + 2) + 6(y − 2) − 3(z − 0) = 0,
simpl y,
2x + 6y − 3z − 8 = 0.
ä
(4.3) The plane containing the point (3, −2, 1) and parallel to the plane x + 3y − 4z = 2
Answer. The vector normal to the plane is 〈 1, 3, −4 〉. Hence, the equation of the
plane is
(x − 3) + 3(y + 2) − 4(z − 1) = 0,
simpl y,
x + 3y − 4z + 7 = 0.
ä
(4.4) The plane containing the point (3, 0, −1) and perpendicular to the planes x + 2y − z = 2
and 2x − z = 1
Answer. Let v1 and v2 be the vectors normal to the planes: v1 = 〈 1, 2, −1 〉 and v2 =
〈 2, 0, −1 〉. Normal vector must be perpendicular to the normal vectors of both planes.
So, the plane that we are looking for has the normal vector v1 × v2 = 〈 −2, −1, −4 〉.
Hence, the equation of the plane is
−2(x − 3) − (y − 0) − 4(z + 1) = 0,
Page 2 of 5
simpl y,
2x + y + 4z = 2.
ä
Calculus II for Engineering
4TH HOMEWORK – SOLUTION
Fall, 2008
5. Sketch the given plane.
(5.1) 2x − y + 4z = 4
Answer. The equation represents a plane which has the normal vector 〈 2, −1, 4 〉 and
passes through the point P(2, 0, 0), Q(0, −4, 0) and R(0, 0, 1). We connect those three
points and sketch the plane roughly.
ä
Y-axis
4
2
0
-2
Normal Vector
-4
4
2
R
P
Z-axis 0
Q
-2
-4
-4
-2
0
2
X-axis
4
(5.2) x + y = 1
Answer. The equation represents a plane such that
(i) it is parallel to the z–axis and perpendicular to the x y–plane,
(ii) it has the normal vector 〈 1, 1, 0 〉,
ä
(iii) it passes through the point P(1, 0, 0), Q(0, 1, 0).
Y-axis
0
-4
4
4
2
-2
2
Normal Vector
Q
P
Z-axis 0
-2
-4
-4
-2
0
X-axis
2
4
6. Find the intersection of the planes 3x + y − z = 2 and 2x − 3y + z = −1.
Answer. Solve the equations for z and equate:
3x + y − 2 = z = −2x − 3y − 1,
i.e.,
Page 3 of 5
3x + y − 2 = −2x − 3y − 1,
Calculus II for Engineering
i.e.,
4TH HOMEWORK – SOLUTION
5x + 4y = 1,
i.e.,
y=
Fall, 2008
−5x + 1
.
4
Putting it into the first equation, we get
2 = 3x +
−5x + 1
− z,
4
i.e.,
z=
7x − 7
.
4
Using x = t as a parameter, we deduce the line:
x = t,
y=
1 5
− t,
4 4
7 7
z = − + t.
4 4
If one use y = s as a parameter, one can get the line:
x=
1 2
+ s,
5 5
7 11
z=− +
s.
5 5
y = s,
One can use z = u as a parameter and get the equation of the line.
ä
7. Find the distance between the given objects.
(7.1) The point (1, 3, 0) and the plane 3x + y − 5z = 2
Answer. The distance formula implies
d=
|3(1) + 1(3) − 5(0) − 2|
4
=p .
p
35
32 + 12 + (−5)2
ä
(7.2) The planes x + 3y − 2z = 3 and x + 3y − 2z = 1
Answer. We choose a point P(1, 0, 0) on the plane x + 3y − 2z = 1. Applying the distance
formula to the point P and the other plane, we get
d=
|1(1) + 3(0) − 2(0) − 3|
2
=p .
p
14
32 + 12 + (−2)2
ä
8. Find an equation of the plane containing the lines
L : x = 4 + t,
M : x = 2 + 2s,
y = 2,
y = 2s,
z = 3 + 2t
z = −1 + 4s
Answer. The vectors parallel to the lines are respectively vL = 〈 1, 0, 2 〉 and v M = 〈 2, 2, 4 〉.
The normal vector to the plane is vL × v M = 〈 −4, 0, 2 〉. We choose one point (4, 2, 3) from
L. (One can choose any point from either line.) Hence, the equation of the plane is
−4(x − 4) + 0(y − 2) + 2(z − 3) = 0,
simpl y,
2x − z = 5.
ä
9. State whether the statement is true or false (not always true).
(9.1) Two planes either are parallel or intersect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Page 4 of 5
T
Calculus II for Engineering
4TH HOMEWORK – SOLUTION
Fall, 2008
Answer. Two planes can be even both parallel and intersect if the planes coincide.
ä
(9.2) The intersection of two planes is a line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer. It can be a plane if the planes coincide, or can be empty.
(9.3) The intersection of three planes is a point. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer. It can be a point, a line, or a plane, or can be empty.
(9.4) Lines that lie in parallel planes are always skew. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer. It is true, unless the parallel planes coincide.
(9.5) The set of all lines perpendicular to a given line forms a plane. . . . . . . . . . . . . . . . . . .
F
ä
F
ä
T
ä
F
Answer. It is false. However, it is true if we take all lines perpendicular to a given
line through a given point.
ä
(9.6) There is one line perpendicular to a given plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. It is false. There is one line perpendicular to a given plane through each
point of the plane.
ä
(9.7) The set of all points equidistant from two given points forms a plane. . . . . . . . . . . .
T
10. Determine whether the given lines are the same:
L : x = 1 + 4t,
y = 2 − 2t,
M : x = 9 − 2s,
y = −2 + s,
z = 2 + 6t
z = 8 − 3s
Answer. The vectors parallel to the lines are vL = 〈 4, −2, 6 〉 and v M = 〈 −2, 1, −3 〉, respectively. Since vL = −2v M , the lines L and M are parallel. The point (9, −2, 8) lies on the
line M when s = 0. We solve for t in the x coordinate of the line L to see that 1 + 4t = 9
implies t = 2. It means when t = 2 and s = 0, the lines L and M have the same x coordinate
9. What about the y and z coordinates when t = 2 and s = 0? At this time t = 2 and s = 0,
the line L has the y and z coordinates, y = 2 − 2(2) = −2 and z = 2 + 6(2) = 14, respectively.
However, the line M has the y and z coordinates, y = −2 and z = 8. That is,
t = −2 & s = 0
=⇒
L : (x, y, z) = (9, −2, 14),
It implies that these lines L and M are not the same.
Page 5 of 5
M : (x, y, z) = (9, −2, 8).
ä
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 5 – SOLUTION
Section 11.2 The Calculus of Vector–Valued Functions
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Monday, October 20, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
5TH HOMEWORK – SOLUTION
Fall, 2008
1. Find the derivative of the given vector–valued function.
(1.1) r (t) = 〈
t−3
, te2t , t3 〉
t+1
Answer.
µ
¶
t − 3 0 ¡ 2t ¢0 ¡ 3 ¢0
4
r (t) = 〈
, te
, t 〉=〈
, (1 + 2t) e2t , 3t2 〉 .
2
t+1
(t + 1)
0
ä
(1.2) r (t) = 〈 cos(5t), tan t, 6 sin t 〉
Answer.
r 0 (t) = 〈 (cos(5t))0 , (tan t)0 , (6 sin t)0 〉 = 〈 −5 sin(5t), sec2 t, 6 cos t 〉 .
(1.3) r (t) = 〈
p
ä
t2 + 1, cos t, e−3t 〉
Answer.
0
r (t) = 〈
³p
t2 + 1
´0
¡
¢0
t
, (cos t)0 , e−3t 〉 = 〈 p
, − sin t, −3e−3t 〉 .
2
t +1
Page 1 of 4
ä
Calculus II for Engineering
5TH HOMEWORK – SOLUTION
Fall, 2008
2. Evaluate the given indefinite or definite integral.
Z
(2.1)
〈
3 4
, 〉 dt
t2 t
Answer.
Z
3 4
〈 2 , 〉 dt = 〈
t t
Z
3
dt,
t2
Z
4
3
dt 〉 = 〈 − , 4 ln t 〉 + c,
t
t
ä
where c is an arbitrary constant vector.
Z
〈 e−3t , sin(5t), t3/2 〉 dt
(2.2)
Answer.
Z
Z
〈e
−3t
, sin(5t), t
3/2
Z
〉 dt = 〈
e
=〈 −
−3t
dt,
Z
t3/2 dt 〉
sin(5t) dt,
e−3t
cos(5t) 2 5/2
,−
, t 〉 + c,
3
5
5
ä
where c is an arbitrary constant vector.
Z
〈 e−3t , t2 cos t3 , t cos t 〉 dt
(2.3)
Answer. Using u = t3 to integrate the second component of the vector and integration
by parts where u = t and dv = cos t dt to integrate the third component of the vector
gives
Z
Z
〈e
−3t
2
Z
3
, t cos t , t cos t 〉 dt = 〈
e
=〈 −
−3t
Z
2
dt,
3
t cos t dt 〉
t cos t dt,
e−3t sin t3
,
, t sin t + cos t 〉 + c,
3
3
ä
where c is an arbitrary constant vector.
Z 4
p
〈 t, 5 〉 dt
(2.4)
1
Answer.
Z
4
〈
p
Z
t, 5 〉 dt = 〈
1
Z
(2.5)
4
〈 2te4t , t2 − 1,
0
4t
t2 + 1
4p
Z
4
5 dt 〉 = 〈
t dt,
1
1
14
, 15 〉 .
3
ä
〉 dt
Answer. Using integration by parts where u = t and dv = e4t dt to integrate the first component and using u = t2 + 1 to integrate the third component of the vector gives
Z
0
4
4t
〈 2te , t − 1, 2
〉 dt = 〈
t +1
4t
Z
Z
4
2
4
4t
2te dt,
0
Z
2
(t − 1) dt,
0
15 16 1 52
e + ,
, 2 ln 17 〉 .
=〈
8
8 3
Page 2 of 4
0
4
4t
dt 〉
t2 + 1
ä
Calculus II for Engineering
5TH HOMEWORK – SOLUTION
Fall, 2008
3. Find t such that r (t) and r 0 (t) are perpendicular.
(3.1) r (t) = 〈 2 cos t, sin t 〉.
Answer. It is easy to get r 0 (t) = 〈 −2 sin t, cos t 〉.
r (t) · r 0 (t) = 〈 2 cos t, sin t 〉 · 〈 −2 sin t, cos t 〉 = −3 sin t cos t = 0
⇐⇒
π
3π
t = 0, ± , ±π, ± , . . . .
2
2
Therefore, r (t) and r 0 (t) are perpendicular when t =
nπ
for any integer n.
2
ä
(3.2) r (t) = 〈 t2 , t, t2 − 5 〉
Answer. It is easy to get r 0 (t) = 〈 2t, 1, 2t 〉.
r (t) · r 0 (t) = 〈 t2 , t, t2 − 5 〉 · 〈 2t, 1, 2t 〉 = 4t3 − 9t = 0
t = 0,
⇐⇒
3
Therefore, r (t) and r 0 (t) are perpendicular when t = 0, t = ± .
2
3
± .
2
ä
4. Find all values of t such that r 0 (t) is parallel to the x y–plane.
(4.1) r (t) = 〈 t2 , t, sin(t2 ) 〉
Answer. r 0 (t) is parallel to the x y–plane when the third component is 0.
¢
d ¡
sin(t2 ) = 2t cos(t2 ) = 0
dt
r
⇐⇒
t = 0,
±
nπ
2
ä
for any odd integer n.
(4.2) r (t) = 〈
p
t + 1, cos t, t4 − 8t3 〉
Answer. r 0 (t) is parallel to the x y–plane when the third component is 0.
¢
d ¡ 4
t − 8t3 = 4t3 − 24t2 = 4t2 (t − 6) = 0
dt
Page 3 of 4
⇐⇒
t = 0,
6.
ä
Calculus II for Engineering
5TH HOMEWORK – SOLUTION
Fall, 2008
5. Label as true or false and explain why.
r (t)
(5.1) If u(t) =
and u(t) · u0 (t) = 0, then r (t) · r 0 (t) = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
kr (t)k
F
Answer. False. For any function r (t), u(t) is a unit vector for all t, and ku(t)k = 1 is
constant. Recall the theorem
Relation between Magnitude and Orthogonality
kw(t)k = constant
⇐⇒
⇐⇒
w(t) and w0 (t) are orthogonal, for all t
w(t) · w0 (t) = 0 for all t
It implies u(t) · u0 (t) = 0 for any function r (t), but there are clearly functions r (t) with
kr (t)k 6= constant, so for these functions r (t) · r 0 (t) will not be equal 0.
ä
(5.2) If r (t 0 ) · r 0 (t 0 ) = 0 for some t 0 , then kv (t)k is constant. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. False. By the Theorem in the solution to the previous problem, kr (t)k is
constant if and only if r (t) and r 0 (t) are perpendicular for all t. It is not enough that
they are perpendicular for some particular value of t = t 0 .
ä
Page 4 of 4
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 6 – SOLUTION
Section 12.3 Partial Derivatives & Section 12.5 Chain Rule
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Monday, November 3, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Fall, 2008
Section 12.3 Partial Derivatives
1. Find all first–order partial derivatives.
(1.1) f (x, y) = x2 y3 − 3x
Answer.
f x = 2x y3 − 3,
(1.2) f (x, y) = 3e x
2y
−
p
ä
f y = 3x2 y2 .
x−1
Answer.
¢
2
1
1
x2 y − p
= 6x ye x y − p
,
∂x
2 x−1
2 x−1
2
2
∂ ¡ 2 ¢
f y = 3e x y
x y = 3x2 e x y .
∂y
f x = 3e x
(1.3) f (x, y) =
2y
∂ ¡
ä
x−3
+ x2 tan y
y
Answer.
fx =
(1.4) f (x, y, z) = p
1
+ 2x tan y,
y
fy = −
x−3
+ x2 sec2 y.
2
y
ä
2
x2 + y2 + z2
Answer.
fx = −¡
2x
x2 + y2 + z2
¢3/2 ,
fy = −¡
2y
x 2 + y2 + z 2
Page 1 of 8
¢3/2 ,
fz = −¡
2z
x2 + y2 + z2
¢3/2 .
ä
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Fall, 2008
2. Find the indicated partial derivatives.
∂2 f ∂2 f
∂2 f
(2.1)
,
and
of f (x, y) = x2 y − 4x + 3 sin y
2
2
∂ y∂ x
∂x ∂ y
Answer.
f x = 2x y − 4,
∂2 f
∂ y∂ x
= fxy =
∂
∂y
f y = x2 + 3 cos y,
f xx = 2y
f yy = −3 sin y,
ä
(2x y − 4) = 2x.
p
(2.2) f xx , f x y and f yyx of f (x, y) = e4x − sin y2 − x y
Answer.
y
f x = 4e4x − p ,
2 xy
f xx = 16e4x +
y2
,
4 (x y)3/2
x
x2
f y = − p − 2y cos(y2 ),
f yy =
− 2 cos(y2 ) + 4y2 sin(y2 ),
3/2
2 xy
4 (x y)
µ
¶
2
∂ f
∂
y
1
= fxy =
4e4x − p
=− p .
∂ y∂ x
∂y
2 xy
4 xy
(2.3) f xx , f yy and f yyzz of f (x, y, z) = e2x y −
ä
z2
+ xz sin y
y
Answer.
f x = 2ye2x y + z sin y,
f y = 2xe2x y +
f yyz = −
f xx = 4y2 e2x y ,
z2
+ xz cos y,
y2
4z
− x sin y,
y3
f yy = 4x2 e2x y −
f yyzz = −
(2.4) f xx , f yy and f wx yz of f (w, x, y, z) =
p
2z2
− xz sin y,
y3
4
.
y3
ä
w yz − x3 sin w
Answer.
f x = −3x2 sin w,
f xx = −6x sin w,
yz
fw = p
− x3 cos w,
2 w yz
wz
fy = p
,
2 w yz
f wx = −3x2 cos w,
Page 2 of 8
f wx y = 0,
f yy = −
w2 z 2
4 (w yz)3/2
f wx yz = 0.
,
ä
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Fall, 2008
3. For van der Waals’ equation,
µ
¶
an2
P + 2 (V − nb) = nRT
V
Here, P is the pressure of the gas, V is the volume of the gas, T is the temperature (in
degrees Kelvin), n is the number of moles of gas, R is the universal gas constant and a
and b are constants. Show that
∂ T ∂ P ∂V
= −1.
∂ P ∂V ∂ T
If you misunderstood the chain rule, why might you expect this product to equal 1?
Answer. We recall: the Chain Rule says that the partial derivative of F(u, v), where u =
f (x, y) and v = g(x, y), with respect to x is obtained by
∂
∂x
F(u, v) =
∂F ∂ u
∂u ∂ x
+
∂F ∂v
∂v ∂ x
=
∂F ∂ f
+
∂u ∂ x
∂F ∂ g
∂v ∂ x
.
We will use this rule. First, let us introduce two functions as follows:
F(P, V ) = P +
an2
,
V2
G(V , T) =
nRT
.
V − nb
Before we go further, we find the derivatives of the functions:
∂
F(P, V ) =
∂P
∂
∂V
∂
∂V
∂
∂T
F(P, V ) =
G(V , T) =
G(V , T) =
∂F ∂P
∂P ∂P
∂F ∂P
∂ P ∂V
∂G ∂V
∂V ∂V
∂G ∂V
∂V ∂ T
+
+
+
+
∂ F ∂V
∂V ∂P
∂ F ∂V
∂V ∂V
∂G ∂T
∂ T ∂V
∂G ∂T
∂T ∂T
= FP + FV VP
= F P P V + FV
= G V + G T TV
= G V VT + G T .
Now, we are ready to deal with the given equation. Using the functions, the given van der
Waals’ equation is simplified as follows:
µ
¶
an2
P + 2 (V − nb) = nRT,
V
i.e.,
P+
an2
nRT
=
,
2
V − nb
V
i.e.,
F(P, V ) = G(V , T).
I. We differentiate the whole equation with respect to V . Then we get
F P P V + FV = G V + G T T V .
( F)
Next step is one of the key parts in this solution. What do we want to find from the
equation?
1. If we want to find PV , then we deduce TV = 0, because T is regarded as a constant with
Page 3 of 8
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Fall, 2008
respect to V . So in this case, the equation (F) becomes
FP PV + FV = G V + 0,
i.e.,
F P P V = G V − FV ,
i.e.,
G V − FV
.
FP
PV =
(i)
2. If we want to find TV , then we deduce PV = 0, because P is regarded as a constant with
respect to V . So in this case, the equation (F) becomes
0 + FV = G V + G T T V ,
i.e.,
Moreover, we observe
VT =
∂V
∂T
G T T V = FV − G V ,
=
1
∂T
∂V
=
i.e.,
TV =
FV − G V
.
GT
1
GT
=
.
T V FV − G V
(ii)
II. We differentiate the whole equation F(P, V ) = G(V , T) with respect to P. Then we get
FP + FV VP = G V VP + G T T P .
Since we want to find T P , we get VP = 0, because V is regarded as a constant with respect
to P. So in this case, the equation becomes
FP + FV VP = G V VP + G T T P ,
i.e.,
FP + 0 = 0 + G T T P ,
i.e.,
TP =
FP
. (iii)
GT
Collecting (i), (ii), and (iii), we finally conclude
GT
FP
,
TP =
,
FV − G V
GT
∂ T ∂ P ∂V
F P G V − FV
GT
G V − FV
= T P PV VT =
=
= −1.
∂ P ∂V ∂ T
GT
FP
FV − G V FV − G V
PV =
G V − FV
,
FP
VT =
In this way, we don’t have to find the derivatives explicitly. All we need is the relations
between the derivatives – the results (i), (ii) and (iii).
ä
4. For the specific case of van der Waals’ equation given by
µ
use the partial derivative
volume of one unit.
∂T
∂V
¶
14
P + 2 (V − 0.004) = 12T
V
to estimate the change in temperature due to an increase in
Answer. We will use the solution of the previous problem:
F(P, V ) = P +
14
,
V2
and
Page 4 of 8
G(V , T) =
12T
.
V − 0.004
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Fall, 2008
We compute the derivatives:
28
12T
12
,
G
=
−
,
G
=
V
T
V − 0.004
V3
(V − 0.004)2
FV − G V
7(V − 0.004)
T
=−
+
,
TV =
GT
V − 0.004
3V 3
FV = −
from the result (ii) in the solution of the previous problem. Suppose V is sufficiently large.
Then our result becomes
7(V − 0.004)
T
T
7(V − 0)
+
+
≈
−
V − 0.004
V −0
3V 3
3V 3
T
7
T T
∂T T
+ ≈ 0+ = ,
=−
i.e.,
≈ .
V V
∂V V
3V 2 V
TV = −
Let 4V and 4T be small changes of V and T, respectively. Then for sufficiently large V ,
we have
4T ∂T T
4 T 4V
≈
≈ ,
i.e.,
≈
,
4V ∂ V V
T
V
which implies that an increase in volume of one unit increases the change in temperature
of one degree for sufficiently large V .
ä
5. The ideal gas law relating pressure, temperature and volume is
P=
for some constant C. Show that
T
CT
,
V
∂ P ∂V
∂T ∂T
= C.
Answer. Again, we use the solution to the problem 3 above to the functions:
F(P, V ) = P,
and
G(V , T) =
CT
.
V
From the solution (explicitly the results (ii) and (iii)), we recall
PT =
∂P
∂T
=
1
∂T
∂P
=
1
1
GT
= F =
,
P
TP
FP
GT
.
FV − G V
and
VT =
CT
,
V2
GT =
GT
For our functions F and G, we have
FP = 1,
FV = 0,
GV = −
C
.
V
Putting them into the equation on P T and VT , we deduce
C
C
GT GT
V
P T VT =
=V
=
F P FV − G V
1 0 + CT2
V
C2
V2
CT
V2
Page 5 of 8
=
C
,
T
i.e.,
TP T VT = C.
ä
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Fall, 2008
Section 12.5 Chain Rule
6. Use the chain rule to find the indicated derivative(s).
(6.1) g0 (t), where g(t) = f (x(t), y(t)),
q
f (x, y) =
x2 + y2 ,
x(t) = sin t,
y(t) = t2 + 2.
Answer. The Chain Rule implies
d
y
∂ f dx ∂ f d y
x
cos t + p
f (x, y) =
+
=p
(2t)
dt
∂ x dt ∂ y dt
x2 + y2
x 2 + y2
t2 + 2
sin t cos t + 2t(t2 + 2)
=p
cos t + p
(2t) = p
sin2 t + (t2 + 2)2
sin2 t + (t2 + 2)2
sin2 t + (t2 + 2)2
sin t
(6.2)
∂g
∂u
and
∂g
∂v
ä
, where g(u, v) = f (x(u, v), y(u, v)),
p
2
f (x, y) = x y3 − 4x2 ,
x(u, v) = e u ,
y(u, v) =
v2 + 1 sin u.
Answer. The Chain Rule implies
∂
∂u
∂
∂v
f (x, y) =
∂ f ∂x
+
∂f ∂y
¡
¢
¡
¢
2
= y3 − 8x 2ue u + 3x y2 cos u
∂ x ∂u ∂ y ∂u
i
h¡
¢2/3
2
2
2
3
u2
sin u − 8e
2ue u + 3e u (v2 + 1) sin2 u cos u
= v +1
¡
¢
¡
¢ v sin u
= y3 − 8x 0 + 3x y2 p
∂ x ∂v ∂ y ∂v
v2 + 1
p
2
2
v sin u
= 3e u (v2 + 1) sin2 u p
= 3v v2 + 1e u sin3 u
v2 + 1
f (x, y) =
∂ f ∂x
+
∂f ∂y
ä
7. Suppose the production of a firm is modeled by
P(k, l) = 16k1/3 l 2/3
where k measures capital (in millions of dollars) and l measures the labor force (in thousands of workers). Suppose that l = 2 and k = 5, the labor force is increasing at the rate of
40 workers per year and capital is decreasing at the rate of $100, 000 per year. Determine
the rate of change of production.
Answer. From the given hypotheses, we deduce
dl(t)
= 0.04 (thousand)
dt
and
dk(t)
= −0.1 (mill ion)
dt
Since l and t are functions of time t, the production P is the function of time t. The Chain
Page 6 of 8
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Rule implies
Fall, 2008
d
∂P dk ∂P dl
P(k, l) =
+
.
dt
∂ k dt
∂ l dt
We compute
Pk =
16 −2/3 2/3
k
l ,
3
and
Pl =
(♠ )
32 1/3 −1/3
k l
.
3
Putting them and l 0 (t) = 0.04 and k0 (t) = −0.1 into the equation (♠) above, we get
16 −2/3 2/3
32
d
P(k, l) =
k
l (−0.1) + k1/3 l −1/3 (0.4),
dt
3
3
d
16
32
P(5, 2) = 5−2/3 22/3 (−0.1) + 51/3 2−1/3 (0.4) = 0.2895
dt
3
3
8. Use implicit differentiation to find
∂z
∂x
and
∂z
∂y
ä
.
(8.1) x yz − 4y2 z2 + cos(x y) = 0
Answer. Letting F(x, y, z) = x yz − 4y2 z2 + cos(x y) = 0, we deduce
F x = yz − y sin(x y) = y (z − sin(x y)) ,
F y = xz − 8yz2 − x sin(x y),
F z = x y − 8y2 z = y (x − 8yz) .
Therefore, by the formula, we conclude
∂z
Fx
y (z − sin(x y))
z − sin(x y)
=−
=−
∂x
Fz
y (x − 8yz)
x − 8yz
2
Fy
∂z
xz − 8yz − x sin(x y)
=−
=−
∂y
Fz
y (x − 8yz)
=−
ä
(8.2) 3yz2 − e4x cos(4z) − 3y2 = 4
Answer. Letting F(x, y, z) = 3yz2 − e4x cos(4z) − 3y2 − 4 = 0, we deduce
F x = −4e4x cos(4z),
F y = 3z2 − 6y,
F z = 6yz + 4e4x sin(4z).
Therefore, by the formula, we conclude
∂z
∂x
=−
Fx
4e4x cos(4z)
=
,
F z 6yz + 4e4x sin(4z)
and
∂z
∂y
=−
Fy
Fz
=−
3z2 − 6y
6yz + 4e4x sin(4z)
9. For a differentiable function f (x, y) with x = r cos θ and y = r sin θ, show that
f θθ = f xx r 2 sin2 θ − 2 f x y r 2 cos θ sin θ + f yy r 2 cos2 θ − f x r cos θ − f y r sin θ .
Page 7 of 8
ä
Calculus II for Engineering
6TH HOMEWORK – SOLUTION
Fall, 2008
Answer. The Chain Rule implies
fθ =
∂
∂ f ∂x
∂f ∂y
= f x (− r sin θ ) + f y (r cos θ )
∂ x ∂θ ∂ y ∂θ
¤
∂ £
f θθ =
f θ (x, y) =
f x (− r sin θ ) + f y (r cos θ )
∂θ
∂θ
∂fy
∂fx
∂
∂
=
(− r sin θ ) + f x (− r sin θ ) +
(r cos θ ) + f y (r cos θ )
∂θ
∂θ
∂θ
∂θ
∂fy
∂fx
=
(− r sin θ ) + f x (− r cos θ ) +
(r cos θ ) + f y (− r sin θ )
∂θ
∂θ
∂θ
∂
f (x, y) =
All we have to do is to find
∂
∂θ
∂
∂θ
fx =
fy =
+
∂fx
∂θ
and
∂ f x ∂x
∂ x ∂θ
∂ f y ∂x
∂ x ∂θ
+
+
∂fy
∂θ
(Product Rule)
(♣ )
. Again, by the Chain Rule, we have
∂fx ∂y
∂ y ∂θ
∂f y ∂y
∂ y ∂θ
= f xx (− r sin θ ) + f x y (r cos θ )
= f yx (− r sin θ ) + f yy (r cos θ ) .
Putting them into the equation (♣) above, we conclude
f θθ =
∂fx
(− r sin θ ) + f x (− r cos θ ) +
∂fy
(r cos θ ) + f y (− r sin θ )
∂θ
∂θ
¤
£
= f xx (− r sin θ ) + f x y (r cos θ ) (− r sin θ ) + f x (− r cos θ )
£
¤
+ f yx (− r sin θ ) + f yy (r cos θ ) (r cos θ ) + f y (− r sin θ )
= f xx r 2 sin2 θ − 2 f x y r 2 cos θ sin θ + f yy r 2 cos2 θ − f x r cos θ − f y r sin θ
Page 8 of 8
ä
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 7 – SOLUTION
Section 12.6 The Gradient and Directional Derivatives
Section 12.7 Extrema of Functions of Several Variables
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Wednesday, November 19, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
Section 12.6 The Gradient and Directional Derivatives
1. Find the gradient of the given function.
(1.1) f (x, y) = x3 e3y − y4
Answer.
Grad f (x, y) = ∇ f (x, y) = 〈 f x (x, y), f y (x, y) 〉 = 〈 3x2 e3y , 3x3 e3y − 4y3 〉 .
ä
(1.2) f (x, y) = e3y/x − x2 y3
Answer.
Grad f (x, y) = ∇ f (x, y) = 〈 f x (x, y), f y (x, y) 〉 = 〈 −
3y 3y/x
3 3 3y/x
e
−
2x
y
, e
− 3x2 y2 〉
2
x
x
3y 3y/x 3 3y/x
〉 + 〈 −2x y3 , −3x2 y2 〉
e
, e
2
x
x
3 3y/x
〈 y, − x 〉 − x y2 〈 2y, 3x 〉 .
=− 2 e
x
=〈 −
ä
2. Find the gradient of the given function at the indicated point.
(2.1) f (x, y) = sin(3x y) + y2 at (π, 1)
Answer.
Grad f (x, y) = ∇ f (x, y) = 〈 3y cos(3x y), 3x cos(3x y) + 2y 〉 = 3 cos(3x y) 〈 y, x 〉 + 2y 〈 0, 1 〉
Grad f (π, 1) = ∇ f (π, 1) = 3 cos(3π) 〈 1, π 〉 + 2(1) 〈 0, 1 〉 = 〈 −3, 2 − 3π 〉 .
ä
(2.2) f (x, y, z) = z2 e2x− y − 4xz2 at (1, 2, 2)
Answer.
Grad f (x, y, z) = ∇ f (x, y, z) = 〈 2z2 e2x− y − 4z2 , − z2 e2x− y , 2ze2x− y − 8xz 〉
= ze2x− y 〈 2z, − z, 2 〉 − 4z 〈 z, 0, 2x 〉
Grad f (1, 2, 2) = ∇ f (1, 2, 2) = 2e2−2 〈 2(2), −2, 2 〉 − 4(2) 〈 2, 0, 2(1) 〉 = −4 〈 2, 1, 3 〉 .ä
Page 1 of 11
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
3. Compute the directional derivative of f at the given point in the direction of the indicated
vector.
1 1
(3.1) f (x, y) = x3 y − 4y2 at (2, −1) with u = 〈 p , p 〉
2 2
Answer.
1 1
D u f (x, y) = ∇ f (x, y) · u = 〈 f x (x, y), f y (x, y) 〉 · 〈 p , p 〉
2 2
2
1 1
3x y + x3 − 8y
= 〈 3x2 y, x3 − 8y 〉 · 〈 p , p 〉 =
p
2 2
2
2
3
p
3(2 )(−1) + 2 − 8(−1)
4
D u f (2, −1) =
= p = 2 2.
p
2
2
(3.2) f (x, y) = e4x
2− y
ä
at (1, 4), u in the direction of 〈 −2, −1 〉
Answer. The unit vector u in the direction 〈 −2, −1 〉 is obtained by
u=
〈 −2, −1 〉
1
= − p 〈 2, 1 〉 .
k〈 −2, −1 〉k
5
Using this unit vector u, we get
1
D u f (x, y) = ∇ f (x, y) · u = 〈 f x (x, y), f y (x, y) 〉 · − p 〈 2, 1 〉
5
1
(16x − 1)e4x
〉 · − p 〈 2, 1 〉 = −
= 〈 8xe
, −e
p
5
5
4(12 )−4
p
(16(1) − 1)e
D u f (1, 4) = −
= −3 5.
p
5
4x2 − y
4x2 − y
2− y
ä
(3.3) f (x, y) = x2 sin(4y) at (−2, π/8), u in the direction from (−2, π/8) to (0, 0)
Answer. The vector in the direction from (−2, π/8) to (0, 0) is 〈 2, −π/8 〉. So the unit
vector u in the direction from (−2, π/8) to (0, 0) is obtained by
u=
〈 2, −π/8 〉
1
〈 16, −π 〉 .
=p
k〈 2, −π/8 〉k
π2 + 256
Using this unit vector u, we get
1
〈 16, −π 〉
D u f (x, y) = ∇ f (x, y) · u = 〈 f x (x, y), f y (x, y) 〉 · p
π2 + 256
1
〈 16, −π 〉
= 〈 2x sin(4y), 4x2 cos(4y) 〉 · p
π2 + 256
32x sin(4y) − 4π x2 cos(4y)
=
p
π2 + 256
Page 2 of 11
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
D u f (−2, π/8) =
Fall, 2008
32(−2) sin (4(π/8)) − 4π(−2)2 cos (4(π/8))
64
= −p
.
p
π2 + 256
π2 + 256
ä
(3.4) f (x, y) = y2 + 2ye4x at (0, −2), u in the direction from (0, −2) to (−4, 4)
Answer. The vector in the direction from (0, −2) to (−4, 4) is 〈 −4, 6 〉. So the unit
vector u in the direction from (0, −2) to (−4, 4) is obtained by
u=
〈 −4, 6 〉
1
= p 〈 −2, 3 〉 .
k〈 −4, 6 〉k
13
Using this unit vector u, we get
1
D u f (x, y) = ∇ f (x, y) · u = 〈 f x (x, y), f y (x, y) 〉 · p 〈 −2, 3 〉
13
1
(6 − 16y)e4x + 6y
= 〈 8ye4x , 2y + 2e4x 〉 · p 〈 −2, 3 〉 =
p
13
13
4(0)
p
(6 − 16(−2))e
+ 6(−2)
D u f (0, −2) =
= 2 13.
p
13
ä
q
(3.5) f (x, y, z) =
x2 + y2 + z2 at (1, −4, 8), u in the direction of 〈 1, 1, −2 〉
Answer. The unit vector u in the direction 〈 1, 1, −2 〉 is obtained by
u=
〈 1, 1, −2 〉
1
= p 〈 1, 1, −2 〉 .
k〈 1, 1, −2 〉k
6
Using this unit vector u, we get
1
D u f (x, y, z) = ∇ f (x, y, z) · u = 〈 f x (x, y, z), f y (x, y, z), f z (x, y, z) 〉 · p 〈 1, 1, −2 〉
6
1
1
x + y − 2z
〈 x, y, z 〉 · p 〈 1, 1, −2 〉 = p
=p
6
x2 + y2 + z2
6(x2 + y2 + z2 )
1 − 4 − 2(8)
19
=− p .
D u f (1, −4, 8) = p
9 6
6(12 + (−4)2 + 82 )
Page 3 of 11
ä
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
4. Find the directions of maximum and minimum change of f at the given point, and the
values of the maximum and minimum rates of change.
(4.1) f (x, y) = x2 − y3 at (−1, −2)
Answer.
Grad f (x, y) = ∇ f (x, y) = 〈 2x, −3y2 〉 ,
Grad f (−1, −2) = 〈 −2, −12 〉 = −2 〈 1, 6 〉 .
It implies
(a) Direction of maximum change of f is −2 〈 1, 6 〉,
(b) Direction of minimum change of f is 2 〈 1, 6 〉,
p
(c) Value of maximum rate of change is k−2 〈 1, 6 〉k = 2 37,
p
(d) Value of minimum rate of change is − k−2 〈 1, 6 〉k = −2 37.
ä
(4.2) f (x, y) = y2 e4x at (3, −1)
Answer.
Grad f (x, y) = ∇ f (x, y) = 〈 4y2 e4x , 2ye4x 〉 = 2ye4x 〈 2y, 1 〉 ,
Grad f (3, −1) = −2e12 〈 −2, 1 〉 = −2e12 〈 −2, 1 〉 .
It implies
(a) Direction of maximum change of f is −2e12 〈 −2, 1 〉,
(b) Direction of minimum change of f is 2e12 〈 −2, 1 〉,
p
°
°
(c) Value of maximum rate of change is °−2e12 〈 −2, 1 〉° = 2e12 5,
p
°
°
(d) Value of minimum rate of change is − °−2e12 〈 −2, 1 〉° = −2e12 5.
ä
(4.3) f (x, y) = x cos(3y) at (−2, π)
Answer.
Grad f (x, y) = ∇ f (x, y) = 〈 cos(3y), −3x sin(3y) 〉 ,
Grad f (−2, π) = 〈 cos(3π), −3(−2) sin(3π) 〉 = 〈 −1, 0 〉 .
It implies
(a) Direction of maximum change of f is 〈 −1, 0 〉,
(b) Direction of minimum change of f is − 〈 −1, 0 〉,
(c) Value of maximum rate of change is k〈 −1, 0 〉k = 1,
ä
(d) Value of minimum rate of change is − k〈 −1, 0 〉k = −1.
q
(4.4) f (x, y) = x2 + y2 at (3, −4)
Answer.
Grad f (x, y) = ∇ f (x, y) = p
1
x2 + y2
〈 x, y 〉 ,
Page 4 of 11
Grad f (3, −4) =
1
〈 3, −4 〉 .
5
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
It implies
1
〈 3, −4 〉,
5
1
(b) Direction of minimum change of f is − 〈 3, −4 〉,
° 5
°
°1
°
°
(c) Value of maximum rate of change is ° 〈 3, −4 〉°
° = 1,
5°
°
°1
°
°
(d) Value of minimum rate of change is − ° 〈 3, −4 〉°
° = −1.
5
q
(4.5) f (x, y, z) = x2 + y2 + z2 at (1, 2, −2)
(a) Direction of maximum change of f is
ä
Answer.
Grad f (x, y, z) = ∇ f (x, y, z) = p
Grad f (1, 2, −2) =
1
x2 + y2 + z2
〈 x, y, z 〉 ,
1
〈 1, 2, −2 〉 .
3
It implies
1
〈 1, 2, −2 〉,
3
1
(b) Direction of minimum change of f is − 〈 1, 2, −2 〉,
° 3
°
°1
°
° = 1,
〈
〉
(c) Value of maximum rate of change is °
1,
2,
−
2
°3
°
°
°
°1
°
° = −1.
〈
〉
(d) Value of minimum rate of change is − °
1,
2,
−
2
°3
°
(a) Direction of maximum change of f is
Page 5 of 11
ä
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
Section 12.7 Extrema of Functions of Several Variables
5. Locate all critical points and classify them using the Second Derivative Test.
(5.1) f (x, y) = cos2 x + y2
Answer. Step 1 Critical Points We compute the partial derivatives and find the critical
points:
π
3π
f x = −2 sin x cos x = 0 =⇒ sin x = 0 or cos x = 0 =⇒ x = 0, ± , ±π, ± , ±2π, . . .
2
2
f y = 2y = 0 =⇒ y = 0.
Hence, all critical points are as follows:
(x, y) = (0, 0),
³ nπ
2
´
,0 ,
and
(mπ, 0) ,
where n = ±1, ±3, ±5, . . . and m = ±1, ±2, ±3, . . . .
Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the
critical point.
¡
¢
¡
¢
¡
¢
f xx = −2 cos2 x − sin2 x = −2 1 − 2 sin2 x = 2 2 sin2 x − 1 ,
¡
¢
D(x, y) = f xx f yy − f x2y = 4 2 sin2 x − 1 .
f yy = 2,
f x y = 0,
Putting each critical point into the determinant D(x, y), we get
D(0, 0) = −4 < 0,
³ nπ ´
D
, 0 = 4 > 0,
2
D (mπ, 0) = −4 < 0,
³ nπ ´
and
f xx
, 0 = 2 > 0.
2
Therefore, we conclude (0, 0) and (mπ, 0) are saddle points and (nπ/2, 0) gives the local
minimum.
ä
(5.2) f (x, y) = 4x y − x4 − y4 + 4
Answer. Step 1 Critical Points We compute the partial derivatives and find the critical
points:
f x = 4y − 4x3 = 4(y − x3 ) = 0
=⇒
y = x3 ,
(1)
f y = 4x − 4y3 = 4(x − y3 ) = 0
=⇒
x = y3 .
(2)
Solving the equations (1) and (2) for x and y, we get
y = x3 = (y3 )3 ,
i.e.,
y9 = y,
0 = y9 − y
0 = y(y8 − 1) = y(y4 + 1)(y4 − 1) = y(y4 + 1)(y2 + 1)(y2 − 1)
= y(y4 + 1)(y2 + 1)(y − 1)(y + 1)
=⇒
Page 6 of 11
y = 0,
or
y = −1,
or
y = 1.
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
By the same argument (or by putting into the equation (1) or (2)), we get x = 0, x = −1
or x = 1. Hence, all critical points are (x, y) = (0, 0), (−1, −1), and (1, 1).
Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the
critical point.
f xx = −12x2 ,
f yy = −12y2 ,
f x y = 4,
D(x, y) = f xx f yy − f x2y = (12x y)2 − 16 = 16 (3x y − 1) (3x y + 1) .
Putting each critical point into the determinant D(x, y), we get
D(0, 0) = −16 < 0,
D(−1, −1) = 96 > 0,
D(1, 1) = 96 > 0,
and
and
f xx (1, 1) = −12 < 0,
f xx (−1, −1) = −12 < 0.
Therefore, we conclude (0, 0) is a saddle point and (1, 1) and (−1, −1) give the local
maximum.
ä
(5.3) f (x, y) = 2x2 + y3 − x2 y − 3y
Answer. Step 1 Critical Points We compute the partial derivatives and find the critical
points:
f x = 4x − 2x y = 2x(2 − y) = 0
=⇒
x = 0 or y = 2,
f y = 3y2 − x2 − 3 = 0.
(3)
(4)
When x = 0 from the result (3), the equation (4) becomes 3y2 − 3 = 0, i.e., 0 = 3(y2 − 1) =
3(y − 1)(y + 1), i.e., y = ±1. So we have the critical points (0, ±1).
When y = 2 from the result (3), the equation (4) becomes 3(22 ) − x2 − 3 = 0, i.e., 0 =
x2 − 9 = (x − 3)(x + 3), i.e., x = ±3. So we have the critical points (±3, 2).
That is, all the critical points are (0, ±1) and (±3, 2).
Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the
critical point.
f xx = 4 − 2y = 2(2 − y),
f yy = 6y,
£
f x y = −2x,
¤
D(x, y) = f xx f yy − f x2y = 12y(2 − y) − 4x2 = 4 3y(2 − y) − x2 .
Putting each critical point into the determinant D(x, y), we get
D(0, 1) = 12 > 0,
D(0, −1) = −36 < 0,
and
and
f xx (0, 1) = 2 > 0,
D(−3, 2) = −36 < 0,
and
D(3, 2) = −36 < 0.
Therefore, we conclude (0, −1) and (±3, 2) are saddle points and (0, 1) gives the local
minimum.
ä
(5.4) f (x, y) = x sin y
Answer. Step 1 Critical Points We compute the partial derivatives and find the critical
Page 7 of 11
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
points:
f x = sin y = 0
y = 0, ±π, ±2π, . . .
=⇒
f y = x cos y = 0
=⇒
x = 0,
π 3π
y = ± ,± ,....
2
2
or
Hence, the critical points are (x, y) = (0, 0), and (0, nπ), where n is any integer.
Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the
critical point.
f xx = 0,
f yy = − x sin y,
f x y = cos y,
D(x, y) = f xx f yy − f x2y = − cos2 y.
Putting each critical point into the determinant D(x, y), we get
D(0, 0) = −1 < 0,
D(0, nπ) = −1 < 0
ä
Therefore, we conclude that all the critical points are saddle points.
Max
Max
Saddle
4
Saddle Min
Saddle Min
Z 2
-2
0
0
2
4
Z
2
Saddle Min
Saddle Min
Min
-1
Saddle
6
1
0
0Y
-5
Min
-2
Y
0
-1
X
-1
0
X
1
5
1
2 -2
2
Figure for Problem 5 (1)
Figure for Problem 5 (2)
Saddle
Min
Saddle
Saddle
Saddle
Saddle
Saddle
Saddle
Saddle
Saddle
Saddle
-5
5
Z
0 Z
X 0
Y
-5
5
X
5
0
-5
Figure for Problem 5 (3)
Y
Figure for Problem 5 (4)
Page 8 of 11
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
6. Find the absolute extrema of the function on the region.
(6.1) f (x, y) = x2 + y2 − 4x y, region bounded by y = x, y = −3 and x = 3.
Answer. Let R be the region bounded by the given graphs.
Step 1 Critical Points We find the critical points on the region R.
f x = 2x − 4y = 2(x − 2y) = 0
=⇒
x = 2y,
f y = 2y − 4x = 2(y − 2x) = 0
=⇒
y = 2x.
Solving two equations for x and y, we get
y = 2x = 2(2y),
i.e.,
y = 4y2 ,
y(4y − 1) = 0,
y=0
or
1
y= .
4
Putting into the equation x = 2y, we have the critical points (x, y) = (0, 0), (1/2, 1/4),
which are on the region R. At (x, y) = (0, 0) and (1/2, 1/4), we have f (0, 0) = 0 and
f (1/2, 1/4) = −3/16.
Step 2 Local Extrema on Boundary of R The region R has three boundaries: (a) y = x
for −3 ≤ x ≤ 3; (b) y = −3 for −3 ≤ x ≤ 3; (c) x = 3 for −3 ≤ y ≤ 3. We find the local
extrema of f (x, y) on each boundary.
(a) y = x for −3 ≤ x ≤ 3: On this boundary, f (x, y) becomes
g(x) = f (x, x) = x2 + x2 − 4x2 = −2x2 ,
−3 ≤ x ≤ 3,
which has the absolute maximum value g(0) = 0 at x = 0 and the absolute minimum
value g(−3) = −18 = g(−3) at x = ±3. That is, on this boundary, f (x, y) has the absolute
maximum value 0 at (x, y) = (0, 0) and the absolute minimum value −18 at (x, y) =
(−3, −3).
(b) y = −3 for −3 ≤ x ≤ 3: On this boundary, f (x, y) becomes
g(x) = f (x, −3) = x2 + 12x + 9,
−3 ≤ x ≤ 3,
which has the absolute maximum value g(3) = 54 at x = 3 and the absolute minimum value g(−3) = −18 at x = −3. That is, on this boundary, f (x, y) has the absolute
maximum value 54 at (x, y) = (3, −3) and the absolute minimum value −18 at (−3, −3).
(c) x = 3 for −3 ≤ y ≤ 3: On this boundary, f (x, y) becomes
g(x) = f (3, y) = y2 − 12y + 9,
−3 ≤ y ≤ 3,
which has the absolute maximum value g(−3) = 54 at y = −3 and the absolute minimum value g(3) = −18 at y = 3. That is, on this boundary, f (x, y) has the absolute
maximum value 54 at (x, y) = (3, −3) and the absolute minimum value −18 at (3, 3).
Step 2 Conclusion Collecting all the results above, we deduce finally, f (x, y) has the
following table:
(x, y) (0, 0) (1/2, 1/4) (3, −3) (3, 3) (−3, −3)
f (x, y)
0
−3/16
54
−18
−18
Page 9 of 11
Calculus II for Engineering
7TH HOMEWORK – SOLUTION
Fall, 2008
Hence, we conclude that the surface of f (x, y) over the given region has the absolute maximum value f (3, −3) = 54 at (x, y) = (3, −3) and the absolute minimum value
f (−3, −3) = −18 at (x, y) = (−3, −3) and (3, 3).
ä
(6.2) f (x, y) = x2 + y2 − 2x − 4y, region bounded by y = x, y = 3 and x = 0.
Answer. Let R be the region bounded by the given graphs.
Step 1 Critical Points We find the critical points on the region R.
f x = 2x − 2 = 2(x − 1) = 0
=⇒
x = 1,
f y = 2y − 4 = 2(y − 2) = 0
=⇒
y = 2.
So we have the critical point (x, y) = (1, 2), which is on the region R and f (1, 2) = −5.
Step 2 Local Extrema on Boundary of R The region R has three boundaries: (a) y = x
for 0 ≤ x ≤ 3; (b) y = 3 for 0 ≤ x ≤ 3; (c) x = 0 for 0 ≤ y ≤ 3. We find the local extrema of
f (x, y) on each boundary.
(a) y = x for 0 ≤ x ≤ 3: On this boundary, f (x, y) becomes
g(x) = f (x, x) = x2 + x2 − 6x = 2x2 − 6x,
which has the absolute minimum value g(3/2) = −9/2 at x = 3/2 and the absolute
maximum value g(0) = 0 = g(3) at x = 3 and x = 0. That is, on this boundary, f (x, y)
has the absolute minimum value −9/2 at (x, y) = (3/2, 3/2) and the absolute maximum
value 0 at (x, y) = (3, 3) and (x, y) = (0, 0)
(b) y = 3 for 0 ≤ x ≤ 3: On this boundary, f (x, y) becomes
g(x) = f (x, 3) = x2 − 2x − 3,
which has the absolute minimum value g(1) = −4 at x = 1 and the absolute maximum
value g(3) = 0 at x = 3. That is, on this boundary, f (x, y) has the absolute maximum
value 0 at (x, y) = (3, 3) and the absolute minimum value −4 at (1, 3).
(c) x = 0 for 0 ≤ y ≤ 3: On this boundary, f (x, y) becomes
g(x) = f (0, y) = y2 − 4y,
which has the absolute minimum value g(2) = −4 at y = 2 and the absolute maximum
value g(0) = 0 at y = 0. That is, on this boundary, f (x, y) has the absolute maximum
value 0 at (x, y) = (0, 0) and the absolute minimum value −4 at (0, 2).
Step 2 Conclusion Collecting all the results above, we deduce finally, f (x, y) has the
following table:
(x, y) (1, 2) (3/2, 3/2) (0, 0) (3, 3) (1, 3) (0, 2)
f (x, y) −5
−9/2
0
0
−4
−4
Hence, we conclude that the surface of f (x, y) over the given region has the absolute
maximum value f (3, 3) = 0 at (x, y) = (3, 3) or (0, 0) and the absolute minimum value
f (1, 2) = −5 at (x, y) = (1, 2).
ä
Page 10 of 11
Calculus II for Engineering
Y
0
7TH HOMEWORK – SOLUTION
Fall, 2008
2
-2
Abs. Max
0 Abs. Max
40
Abs. Max
-2
20
Z
Z
Abs. Min
-4
3
0
0
Abs. Min
Abs. Min
2
1
Y
-2
1
X
0
X
2
2
3
Figure for Problem 6 (1)
0
Figure for Problem 6 (2)
Page 11 of 11
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 8 – SOLUTION
Section 13.1 Double Integrals
Section 13.2 Area, Volume and Center of Mass
Calculus II for Engineering
MATH 1120 SECTION 4 CRN 23510
11:00 – 12:50 on Monday & Wednesday
Due Date: Sunday, November 30, 2008
ID No: Solution
Name: Solution
Score: Solution
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
Section 13.1 Double Integrals
1. Evaluate the double integral.
Ï
(1.1)
R
4xe2y d A, where R = { (x, y) | 2 ≤ x ≤ 4, 0 ≤ y ≤ 1 }.
Answer. Fubini’s Theorem implies
Ï
Z
4xe
2y
dA =
R
4xe
(1.2)
R
2y
·
y=0
¸
x
2 =4 · 2y ¸ y=1
x=2
e
2
¸ ·Z
x=4
d ydx = 4
x =2
x
=4
2
Ï
·Z
x=4 Z y=1
x dx
x =2
·
y=1
42 − 22
=4
2
y=0
¸·
e
2y
¸
dy
y=0
¸
¡
¢
e2 − e0
= 12 e2 − 1
2
ä
¡
p ¢
3x − 4x x y d A, where R = { (x, y) | 0 ≤ x ≤ 4, 0 ≤ y ≤ 9 }.
Answer. Fubini’s Theorem implies
Ï
x=4 ·
¸ y=9
8 p
3x y − x y x y
3x − 4x x y d ydx =
dx
3
x=0 y=0
x =0
y=0
¸
·
Z x =4 h
i
3528
27 2 144 5/2 x=4
3/2
x −
x
ä
=
=−
27x − 72x
dx =
2
5
5
x =0
x=0
¡
p ¢
3x − 4x x y d A =
R
Z
x=4 Z y=9 ¡
p
Z
¢
2. (Solve ONLY ONE Problem.) Sketch the solid whose volume is given by the iterated
integral.
Z 2Z
(2.1)
0
1
(2 + x + 2y) d ydx
−1
Answer. The full expression of the double integral is
Z
2Z 1
Z
x=2 Z y=1
(2 + x + 2y) d ydx =
0
(2 + x + 2y) d ydx.
−1
x=0
y=−1
It implies the following representation of the region R of the integration:
R = { (x, y) | 0 ≤ x ≤ 2, −1 ≤ y ≤ 1 } ,
which is a square on the x y–plane. Further, the integrand z = f (x, y) = 2 + x + 2y is a
plane passing through three points (−2, 0, 0), (0, −1, 0) and (0, 0, 2). So the figure of the
solid is as given below.
ä
Z 1Z 1
(2.2)
(4 − x2 − y2 ) d ydx
−1 −1
Answer. The full expression of the double integral is
Z
1
Z
1
Z
2
x=1
2
Z
y=1
(4 − x − y ) d ydx =
−1 −1
x=−1 y=−1
Page 1 of 18
(4 − x2 − y2 ) d ydx.
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
It implies the following representation of the region R of the integration:
R = { (x, y) | −1 ≤ x ≤ 1, −1 ≤ y ≤ 1 } ,
which is a square on the x y–plane. Further, the integrand z = f (x, y) = 4 − x2 − y2 is
a paraboloid. Hence, the solid has the base the square R and formed by the curve
z = 4 − x2 − y2 . The figure of the solid is given below.
ä
Solid Bounded by z=4-x 2 -y 2 ,-1£x£1,-1£y£1
Solid Bounded by z=2+x+2y,0<=x<=2,-1£y£1
6
4
4
3
Z
Z
2
2
1
1
1
0
-1
0
0
0Y
0Y
1
X
X0
2 -1
1 -1
Figure for 2 (1)
Z
2Z 1
Z
(2 + x + 2y) d ydx
0
−1
1
Figure for 2 (2)
Z 1
(4 − x2 − y2 ) d ydx
−1 −1
Page 2 of 18
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
3. (Solve ONLY TWO Problems.) Evaluate the iterated integral.
Z
(3.1)
2 Z x2
(x + 3) d ydx
0
0
Answer.
Z
x=2 Z y= x2
Z
x=2
(x + 3) d ydx =
x=0
y=0
Z
x =0
x=2 £
=
y= x 2
[x y + 3y] y=0 dx
3
x + 3x
2
¤
x =0
Z
(3.2)
·
x4
+ x3
dx =
4
¸ x =2
= 12.
ä
x =0
πZ 2
y sin(x y) dxd y
0
0
Answer.
Z
y=π Z x=2
Z
y=π
y sin(x y) dxd y =
y=0
x=0
Z
y=0
y=π
=
y=0
Z
(3.3)
1
2 Z 2/x
2
[− cos(x y)] xx=
=0 d y
¸ y=π
sin(2y)
= π.
+y
[− cos(2y) + 1] d y = −
2
y=0
·
ä
e x y d ydx
0
Answer.
Z
x=2 Z y=2/x
Z
e
x=1
Z
(3.4)
0
1 Z y2
0
y=0
xy
d ydx =
x=2 · e x y ¸ y=2/x
Z
x=2 · e2 − 1 ¸
dx =
x y=0
x
x=1
£¡
¢
¤ x =2 ¡
¢
= e2 − 1 ln x x=1 = e2 − 1 ln 2.
dx
x=1
ä
3
dxd y
4 + y3
Answer.
Z
y=1 Z x= y2
y=0
x=0
¸ 2
¸
Z y=1 ·
3x x= y
3y2
dy
dy =
4 + y3 x=0
4 + y3
y=0
y=0
£ ¡
¢¤ y=1
= ln 4 + y3 y=0 = ln 5 − ln 4.
3
dxd y =
4 + y3
Z
y=1 ·
Page 3 of 18
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
4. (Solve ONLY TWO Problems.) Setup, but do not evaluate, an integral equal to the
volume of the solid bounded by the given surfaces.
(4.1) z = 3x2 + 2y, z = 0, y = 0, y = 1, x = 1, x = 3
Answer. Let the region R in the x y–plane be the base of the solid. Then the region R
has the representation:
R = { (x, y) | 1 ≤ x ≤ 3, 0 ≤ y ≤ 1 } .
Hence, the volume V of the solid is computed by
Ï
¡ 2
¢
3x + 2y d A =
V=
R
x=3 £
Z
=
Z
x=3 Z y=1 ¡
¢
3x2 + 2y d ydx
x =1
3x2 y + y2
x=1
¤ y=1
Z
y=0 dx =
y=0
x =3 ¡
¢
£
¤ x=3
3x2 + 1 dx = x3 + x x=1 = 28.
x =1
ä
(4.2) z = 3x2 + 2y, z = 0, y = 1 − x2 , y = 0
Answer. Let the region R in the x y–plane be the base of the solid. Then the region R
has the representation:
ª
©
R = (x, y) | −1 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 .
Hence, the volume V of the solid is computed by
Ï
V=
¡
Z
¢
x=1
Z
3x + 2y d A =
Z
R
x =1
Z
x=−1
x =1 ¡
=
=
2
x=−1
y=1− x2 ¡
¢
3x2 + 2y d ydx
x=−1 y=0
¤ y=1− x2
£ 2
3x y + y2 y=0
dx =
·
Z
x =1
x=−1
5
¡
¢
3x2 (1 − x2 ) + (1 − x2 )2 dx
x3 2x
x − 2x + 1 dx =
−
+x
3
5
2
4
¢
¸ x =1
=
x=−1
28
.
15
ä
(4.3) z = 4 − 2y, z = 0, x = y4 , x = 1
Answer. Let the region R in the x y–plane be the base of the solid. Then the region R
has the representation:
©
ª
R = (x, y) | −1 ≤ y ≤ 1, y4 ≤ x ≤ 1 .
Hence, the volume V of the solid is computed by
Ï
V=
Z
(4 − 2y) d A =
y=1
Z
x=1
(4 − 2y) dxd y
y=−1 x= y4
Z
Z y=1
x=1
=
(4 − 2y)(1 − y4 ) d y
[(4 − 2y)x] x= y4 d y =
y=−1
y=−1
R
y=1
Page 4 of 18
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
·
¸ y=1
¡
¢
4y5
y6
32
4
5
2
−y +
.
=
4 − 4y − 2y + 2y d y = 4y −
=
5
3 y=−1 5
y=−1
Z
y=1
ä
(4.4) z = x2 , z = 0, y = x, y = 4, x = 0
Answer. Let the region R in the x y–plane be the base of the solid. Then the region R
has the representation:
R = { (x, y) | 0 ≤ x ≤ 4, x ≤ y ≤ 4 } .
Hence, the volume V of the solid is computed by
Ï
Z
V=
x dA =
Z
=
x=4 Z y=4
2
R
x =4 £
x =0
x =0
2
x y
¤ y=4
y= x
x2 d ydx
y= x
Z
x =4 ¡
dx =
2
4x − x
x =0
3
¢
·
4x3 x4
−
dx =
3
4
¸ x=4
=
x=0
64
.
3
ä
Region R: -1£x£1, 0£y£1-x 2
Region R:1£x£3,0£y£1
Y
1
Y
1
X
X
1
2
3
1
-1
Figure for 4 (1) z = 3x2 + 2y, z = 0,
y = 0, y = 1, x = 1, x = 3
Figure for 4 (2) z = 3x2 + 2y, z = 0,
y = 1 − x2 , y = 0
Region R:0£x£4,x£y£4
Region R: y 4£x£1,-1£y£1
Y
4
Y
1
3
X
2
1
1
X
1
-1
Figure for 4 (3) z = 4 − 2y, z = 0,
x = y4 , x = 1
Page 5 of 18
2
3
4
Figure for 4 (4) z = x2 , z = 0, y = x,
y = 4, x = 0
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
5. (Solve ONLY TWO Problems.) Sketch the region and change the order of integration.
Z
(5.1)
1Z 2
f (x, y) d ydx
0
2x
Answer. The full expression of the double integral is
Z
1Z 2
Z
x=1 Z y=2
f (x, y) d ydx =
0
f (x, y) d ydx.
2x
x=0
y=2x
It implies the following representation of the region R of the integration:
R = { (x, y) | 0 ≤ x ≤ 1, 2x ≤ y ≤ 2 } ,
which is obtained by cutting the region vertically. When we cut the region horizontally, we deduce the representation
n
yo
.
R = { (x, y) | 0 ≤ x ≤ 1, 2x ≤ y ≤ 2 } = (x, y) | 0 ≤ y ≤ 2, 0 ≤ x ≤
2
Using this representation, the double integral becomes
Z
x=1 Z y=2
Z
y=2 Z x= y/2
f (x, y) d ydx =
x =0
Z
(5.2)
f (x, y) dxd y.
y=2x
y=0
x=0
ä
1 Z 2y
f (x, y) dxd y
0
0
Answer. The full expression of the double integral is
Z
1 Z 2y
Z
y=1 Z x=2y
f (x, y) dxd y =
0
0
f (x, y) dxd y.
y=0
x=0
It implies the following representation of the region R of the integration:
R = { (x, y) | 0 ≤ y ≤ 1, 0 ≤ x ≤ 2y } ,
which is obtained by cutting the region horizontally. When we cut the region vertically, we deduce the representation
o
x
R = { (x, y) | 0 ≤ y ≤ 1, 0 ≤ x ≤ 2y } = (x, y) | 0 ≤ x ≤ 2, ≤ y ≤ 1 .
2
n
Using this representation, the double integral becomes
Z
y=1 Z x=2y
Z
x=2 Z y=1
f (x, y) dxd y =
y=0
Z
(5.3)
x =0
2 Z ln y
f (x, y) dxd y
1
f (x, y) d ydx.
x=0
0
Page 6 of 18
y= x/2
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
Answer. The full expression of the double integral is
Z
2 Z ln y
Z
y=2 Z x=ln y
f (x, y) dxd y =
1
0
f (x, y) dxd y.
y=1
x=0
It implies the following representation of the region R of the integration:
R = { (x, y) | 1 ≤ y ≤ 2, 0 ≤ x ≤ ln y } ,
which is obtained by cutting the region horizontally. When we cut the region vertically, we deduce the representation
©
ª
R = { (x, y) | 1 ≤ y ≤ 2, 0 ≤ x ≤ ln y } = (x, y) | 0 ≤ x ≤ ln 2, e x ≤ y ≤ 2 .
Using this representation, the double integral becomes
Z
y=2 Z x=ln y
Z
x=ln 2 Z y=2
f (x, y) dxd y =
y=1
x=0
x =0
Region R:0£x£1, 2x£y£2
y= e x
ä
f (x, y) d ydx.
Region R:0£x£2y, 0£y£1
Y
2
Y
1
1
X
Z
1Z
X
1
1
Z
2
Figure for 5 (1)
f (x, y) d ydx
0
2y
Figure for 5 (2)
2x
f (x, y) dxd y
0
Region R:0£x£Log@yD,1£y£2
Y
2
1
0
2
1Z
Z
X
logH2L
2Z
ln y
f (x, y) dxd y
Figure for 5 (3)
1
0
Page 7 of 18
0
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
6. Evaluate the iterated integral by first changing the order of integration.
Z
(6.1)
0
1Z 1
p
3
d ydx
3
x 4+ y
Answer. The region R of integration has the representation
©
ª
p
R = (x, y) | 0 ≤ x ≤ 1, x ≤ y ≤ 1 ,
which is obtained by cutting the region vertically. When we cut the region horizontally, we deduce the representation
ª ©
ª
©
p
R = (x, y) | 0 ≤ x ≤ 1, x ≤ y ≤ 1 = (x, y) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y2 .
Using this representation, the double integral becomes
Z
¸ 2
Z y=1 ·
3
3x x= y
dxd y =
dy
3
4 + y3 x=0
y=0 x=0 4 + y
y=0
Z y=1
¡
¢ y=1
3y2
=
d y = ln 4 + y3 y=0 = ln 5 − ln 4.
3
y=0 4 + y
x=1 Z y=1
x=0
3
d ydx =
p
3
y= x 4 + y
Z
y=1 Z x= y2
ä
See the problem # 3 (4) above.
Z 1Z 1
¡ 3¢
(6.2)
cos
x dxd y
p
0
y
Answer. The region R of integration has the representation
©
ª
p
R = (x, y) | 0 ≤ y ≤ 1, y ≤ x ≤ 1 ,
which is obtained by cutting the region horizontally. When we cut the region vertically, we deduce the representation
©
ª
ª ©
p
R = (x, y) | 0 ≤ y ≤ 1, y ≤ x ≤ 1 = (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x2 .
Using this representation, the double integral becomes
Z
y=1 Z x=1
y=0
Z
=
p cos
x= y
x =1 £
¡
y cos x
¡
3
x
3
¢
Z
dxd y =
¢¤ y= x2
x =0
x =1 Z y= x 2
y=0
Z
x=0
x =1
¡ ¢
cos x3 d ydx
y=0
2
¡
x cos x
dx =
x=0
3
¢
·
¸
¡ 3 ¢ x=1 1
1
dx =
sin x
= sin (1)
3
3
x=0
7. Sketch the solid whose volume is described by the given iterated integral.
Z
(7.1)
4 Z 4− x
(4 − x − y) d ydx
0
0
Answer. The base region R of the solid is represented as follows:
R = { (x, y) | 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 − x }
Page 8 of 18
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
and the solid has the roof or is covered by the surface z = 4 − x − y, which is a plane
passing through (4, 0, 0), (0, 4, 0) and (0, 0, 4). Hence, the sold is sketched as given
figure below.
ä
Z 1 Z p1 − x 2
(7.2)
(x2 + y2 ) d ydx
0
0
Answer. The base region R of the solid is represented as follows:
n
o
p
R = (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 ,
which is the part of the disk x2 + y2 ≤ 1 in the first quadrant of the x y–plane. The
solid has the roof or is covered by the surface z = x2 + y2 , which is a paraboloid passing
through (0, 0, 0). Hence, the sold is sketched as given figure below.
ä
Solid Bounded by z=4-x-y over R
Region R:0£x£4,0£y£4-x
Y
4
4
3
3
2
Z
2
1
0
4
0
1
3
1
1
2
3
2
2
X
0
4
X
1
3
Y
40
Z
4 Z 4− x
Figure for 7 (1)
(4 − x − y) d ydx
0
Region R:0£x£1, 0£y£
0
1 - x2
Solid Bounded by z=x 2 +y 2 over R
Y
1
X
0
1
1
Z
0
1
0
X
0
1
Z
1Z
p
1− x 2
Figure for 7 (2)
0
0
Page 9 of 18
(x2 + y2 ) d ydx
Y
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
Section 13.2 Area, Volume and Center of Mass
8. (Solve ONLY TWO Problems.) Use a double integral to compute the area of the region
bounded by the curves.
(8.1) y = x2 , y = x + 2
Answer. The intersection points of two curves are obtained as follows:
x2 = y = x + 2,
x2 = x + 2 0 = x2 − x − 2 = (x − 2)(x + 1),
x = −1 or x = 2.
That is, the intersection points are (x, y) = (−1, 1) and (2, 4). We represent the region
R bounded by the curves by cutting the region vertically:
©
ª
R = (x, y) | −1 ≤ x ≤ 2, x2 ≤ y ≤ x + 2 .
Thus, the area of the region R is
Ï
Z
A=
1dA =
Z
=
R
x =2
Z
y= x+2
1 d ydx
x=−1 y= x2
· 2
Z x=2
¤
£
x
y= x+2
2
x + 2 − x dx =
[y] y= x2 dx =
2
x=−1
x=−1
x =2
x3
+ 2x −
3
¸ x=2
=
x=−1
9
2
ä
(8.2) y = 3x, y = 5 − 2x, y = 0
Answer. The intersection point of two curves y = 3x and y = 5 − 2x is obtained as
follows:
3x = y = 5 − 2x, 3x = 5 − 2x 0 = 5x − 5 = 5(x − 1), x = 1.
That is, the intersection point is (x, y) = (1, 3). We represent the region R bounded by
the curves by cutting the region horizontally:
½
R=
y
5− y
(x, y) | 0 ≤ y ≤ 3, ≤ x ≤
3
2
¾
.
Thus, the area of the region R is
Z
Ï
y=3 Z x= x+2
1 dxd y
1dA =
A=
Z
=
R
y=0
x= y/3
· 2
Z y=3
£
¤
x
y= x+2
2
x + 2 − x dx =
[y] y= x2 dx =
2
y=0
y=0
y=3
x2
+ 2x −
3
¸ x=2
=
x=−1
(8.3) y = x3 , y = x2
Answer. The intersection points of two curves are obtained as follows:
x3 = y = x2 ,
x3 = x2
0 = x3 − x2 = x2 (x − 1),
Page 10 of 18
x = 0 or x = 1.
13
2
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
That is, the intersection points are (x, y) = (0, 0) and (1, 1). We represent the region R
bounded by the curves by cutting the region vertically:
©
ª
R = (x, y) | 0 ≤ x ≤ 1, x3 ≤ y ≤ x2 .
Thus, the area of the region R is
Ï
Z
A=
1dA =
Z
=
x=1 Z y= x2
R
y= x 3
x=0
1 d ydx
· 3
Z x =1
£ 2
¤
x
y= x 2
3
x − x dx =
[y] y= x3 dx =
3
x=0
x=0
x=1
x4
−
4
¸ x=1
x=0
1
12
y
5-y
=
Region R: 0£y£3,
3
Region R:-1£x£2,x 2£y£x+2
£x£
ä
2
Y
3
Y
4
3
2
2
1
1
X
5
X
-1
1
0
2
2
Figure for 8 (1) y = x , y = x + 2
1
2
Figure for 8 (2) y = 3x, y = 5 − 2x,
y=0
Region R: 0£x£1, x 3£y£x 2
Y
1
X
1
Figure for 8 (3) y = x3 , y = x2
Page 11 of 18
2
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
9. (Solve ONLY THREE Problems.) Compute the volume of the solid bounded by the
given surfaces.
(9.1) x + 2y − 3z = 6 and the three coordinate planes
Answer. First, we determine the base of the solid. We take the projection R of the
solid onto the x y–plane as the base, where the region R has the representation from
x + 2y ≤ 6:
n
xo
.
R = (x, y) | 0 ≤ x ≤ 6, 0 ≤ y ≤ 3 −
2
Now, we determine the upper and lower boundaries of the solid. The surface of x +
2y − 3z = 6 is a plane passing through the points (6, 0, 0), (0, 3, 0) and (0, 0, −2). When
we connect those three points, we observe that the solid is under the x y–plane (i.e.,
z = 0), because of (0, 0, −2). It implies the upper boundary of the solid is the x y–plane,
x 2y
while the lower boundary is the plane x + 2y − 3z = 6 (i.e., z = +
− 2). Hence, the
3 3
double integral for the volume is obtained as follows:
¶¸
¸ y=3− x/2
Z x=6 ·
x y y2
x 2y
− −
dx
0− +
− 2 d ydx =
+ 2y
V=
3 3
3
3
x =0
x=0 y=0
y=0
¸
· 3
¸ x=6
Z x=6 · 2
x
x
x2
=
− x + 3 dx =
− + 3x
=6
12
36 2
x=0
x=0
Z
µ
x=6 Z y=3− x/2 ·
ä
(9.2) z = x2 + y2 , z = 0, x = 0, x = 1, y = 0, y = 1
Answer. First, we determine the base of the solid. We take the projection R of the
solid onto the x y–plane as the base, where the region R has the representation:
R = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 } .
Now, we determine the upper and lower boundaries of the solid. The surface of z =
x2 + y2 is a paraboloid passing through the point (0, 0, 0). We observe that the solid is
under the surface of z = x2 + y2 and above the x y–plane (i.e., z = 0). Hence, the double
integral for the volume is obtained as follows:
Z
V=
x=1 Z y=1 £
x =0
Z
x =1 ·
=
x =0
2
2
x +y
¤
Z
x =1 ·
d ydx =
y=0
¸
· 3
1
x
x
x +
dx =
+
3
3 3
x =0
¸ x=1
2
Page 12 of 18
=
x=0
y3
x y+
3
2
3
¸ y=1
2
dx
y=0
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
x
Solid Bounded by z=
3
2y
+
Solid Bounded by z=x 2 +y 2 ,z=0,0£x£1,0£y£1
1
Y
x
-2,0£x£6,0£y£3-
3
Fall, 2008
2
Y
0
3
2
1
0
0
2
Z
Z
0
0
-2
0
2
X
4
6
X
1
Figure for 9 (2) z = x2 + y2 , z = 0,
x = 0, x = 1, y = 0, y = 1
Figure for 9 (1) x + 2y − 3z = 6 and
the three coordinate planes
(9.3) z = 2 + x, z = 0, x = 0, y = 0, y = 1
Answer. We observe z = 2 + x is a plane parallel to y–axis and passing through (−2, 0)
and (0, 2) in the xz–plane. z = 0 and x = 0 imply the x y–plane and yz–plane, respectively. y = 0 and y = 1 imply that y moves between 0 and 1. The figure is given
below.
We take the region R on the x y–plane as the base of the solid:
R = { (x, z) | −2 ≤ x ≤ 0, 0 ≤ y ≤ 1 }
and take the surface of z = 2 + x as the height of the solid. Then, the double integral
for the volume is obtained as follows:
Z
x =0
Z
V=
Z
Z
y=1
x=−2 y=0
[2 + x] d ydx =
·
x =0
=
x=−2
[2 + x] dx = 2x +
x=0
x=−2
y=1
[(2 + x)y] y=0 dx
2 ¸ x =0
x
2
ä
=2
x=−2
(9.4) z = 1 − x2 − y2 , x + y = 1 and the three coordinate planes
Answer. We take the region R formed by x + y ≤ 1 as the base of the solid and take the
surface of z = 1 − x2 − y2 as the height of the solid. Then, R has the representation:
R = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x } .
Hence, the double integral for the volume is obtained as follows:
Z
x=1 Z y=1− x £
Z
x =1 ·
y3
(1 − x )y −
1 − x − y d ydx =
V=
3
x =0
x=0 y=0
·
¸
·
¸
Z x=1
x=1
4x3
x4 2x3 2x
1
2 2
− 2x +
dx =
−
+
=
=
3
3
3
3
3 x=0 3
x=0
2
2
¤
Page 13 of 18
¸ y=1− x
2
dx
y=0
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
Solid Bounded by z=1-x 2 -y 2 ,x+y=1
1
Y
Solid Bounded by z=2+x,z=0,x=0,y=0,y=1
0
1
2
Z
1
Z
1
0
0
-2
0
Y
-1
X
X
1
0
0
Figure for 9 (4) z = 1 − x2 − y2 ,
x + y = 1 and the three coordinate
planes
Figure for 9 (3) z = 2 + x, z = 0, x = 0,
y = 0, y = 1
(9.5) z = x2 + y2 + 1, z = −1, y = x2 , y = 2x + 3
Answer. We take the region R formed by y = x2 and y = 2x + 3 as the base of the solid
and take the surface of z = x2 + y2 + 1 as the upper boundary and z = −1 as the lower
boundary of the solid. Then, R has the representation:
©
ª
R = (x, y) | −1 ≤ x ≤ 3, x2 ≤ y ≤ 2x + 3 .
Hence, the double integral for the volume is obtained as follows:
Z
V=
Z
x=3
Z
y=2x+3 £
x=−1 y= x2
x=3 ·
=
x=−1
·
= −
2
x + y + 1 − (−1) d ydx =
3 ¸ y=2x+3
y
(x + 2)y +
3
2
Z
¤
2
Z
x =3 ·
dx =
y= x2
x=3
y=2x+3 £
x=−1 y= x2
¤
x2 + y2 + 2 d ydx
¸
3
x6
4 14x
2
− −x +
+ 13x + 22x + 15 dx
3
3
x=−1
¸ x =3
x7 x5 7x4 13x3
− +
+
+ 11x2 + 15x
21 5
6
3
Z
=
x=−1
22016
105
ä
(9.6) z = 2x + y + 1, z = −2x, x = y2 , x = 1
Answer. This problem actually defines two solids. The figure is given below. We will
compute the total volume of this solid. The solid described is bounded above and
below by z = −2x and z = 2x + y + 1, respectively.
The region of integration is the region R in the x y–plane. Note that the planes z = −2x
and z = 2x + y − 1 intersect when y = −4x + 1. So we separate the region R by R 1 and
R 2 , i.e.,
R = R1 ∪ R2,
where R 1 and R 2 are the left and right regions, respectively.
On the left region R 1 , the top of the solid is z = −2x and the bottom is z = 2x + y − 1.
In this case, we will integrate the function −2x − (2x + y − 1) = −4x − y + 1.
Page 14 of 18
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
On the right region R 2 , the top of the solid is z = 2x + y − 1 and the bottom is z = −2x.
In this case, we will integrate the function (2x + y − 1) − (−2x) = 4x + y − 1.
Note that the intersection points of y = 4x + 1 and x = y2 are
p
1
17
,
c=− −
8
8
p
1
17
d=− +
.
8
8
It implies the following representation of R 1 :
½
1− y
(x, y) | c ≤ y ≤ d, y ≤ x ≤
4
¾
2
R1 =
.
To get the representation of R 2 , it need to work more. We separate R 2 into three
regions R 2A and R 2B as given in figure below. Then R 2 = R 2A ∪ R 2B and we can express
each region as follows:
©
p ª
R 2A = (x, y) | d 2 ≤ x ≤ c2 , −4x + 1 ≤ y ≤ x
©
p
p ª
R 2B = (x, y) | c2 ≤ x ≤ 1, − x ≤ y ≤ x .
Hence, the double integral for the volume is obtained as follows:
Ï
Ï
V=
Ï
R1
=
R1
Z
(−4x − y + 1) d A +
Ï
(−4x − y + 1) d A +
y= d Z x=(1− y)/4
=
y= c
Z
x = y2
R2
R 2A
(4x + y − 1) d A
Ï
(4x + y − 1) d A +
p
x = c 2 Z y= x
x= d 2
(4x + y − 1) d A
(−4x − y + 1) dxd y
Z
p
x=1 Z y= x
(4x + y − 1) d ydx +
x= c 2
p
p
p
289 17 344 + 185 17 3541 + 13 17
+
+
=
15360 p
15360
1920
28 289 17
=
+
≈ 2.02182
15
7680
+
R 2B
y=−4x+1
Page 15 of 18
p (4x + y − 1)
y=− x
d ydx
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
Solid Bounded by z=x 2 +y 2 +1 over -1£x£3,x 2 £y£2x+3
Region R: -1£x£3, x 2£y£2x+3
Y
9
90
8
60
6
Z
30
4
0
-1
2
9
-1
0
6
X
1
-1
2
1
X
3
Y
3
2
30
2
2
2
Figure for 9 (5) z = x + y + 1, z = −1, y = x , y = 2x + 3
t ˆ› G j–” —–šŒ™ G XU XU \
› › —a V Vž ž ž U ” ˆ› Š–” —–šŒ™ U Š–”
Region R: R1 HLeftL and R2 HRightL
Y
1
y
0.8
0.5
0.4
A
R2
R1
0.2
X
0.2
0.4
0.6
0.8
1
0.4
0.6
B
R2
-0.4
-0.5
-0.8
-1
Solid Bounded by z=2x+y+1, z=-2x, x=y 2 , x=1
1
Y
0
-1
4
3
Z 2
1
0
0
X
1
Figure for 9 (6) z = 2x + y + 1, z = −2x, x = y2 , x = 1
Page 16 of 18
0.8
x
1.0
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
10. (Solve ONLY ONE Problem.) Set up, but do not evaluate, a double integral for the
volume bounded by the given surfaces.
q
(10.1) z = 4 − x2 − y2 , inside x2 + y2 = 1, first octant
Answer. The base region R in the x y–plane is inside the circle x2 + y2 = 1 and in the
first octant, i.e., x ≥ 0, y ≥ 0 and z ≥ 0. So we have the representation
n
o
p
R = (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 .
Therefore, the volume of the solid is
Z
Ï q
4 − x2 − y2 d A =
V=
R
(10.2) z = e x
2 + y2
p
4
π
3
4 − x2 − y2 d ydx =
−
≈ 1.4681
3
2
p
x=1 Z y= 1− x2 q
x=0
y=0
ä
, z = 0 and x2 + y2 = 4
Answer. The base region R in the x y–plane is inside the circle x2 + y2 = 4. So we have
the representation
o
n
p
p
R = (x, y) | −2 ≤ x ≤ 2, − 4 − x2 ≤ y ≤ 4 − x2 .
Therefore, the volume of the solid is
Ï
V=
e
x 2 + y2
Z
dA =
R
x=2
Z
p
y = 4− x 2
p
x=−2 y=− 4− x2
ex
2 + y2
ä
d ydx ≈ 168.384
11. (Solve ONLY ONE Problem.) Find the mass and center of mass of the lamina with the
given density.
(11.1) Lamina bounded by y = x4 and y = x2 , ρ (x, y) = 4
Answer. The region R has the representation
©
ª
R = (x, y) | 0 ≤ x ≤ 1, x4 ≤ y ≤ x2 .
Using this, the definitions imply
Z
x=1 Z y= x2
m=
Z
My =
x̄ =
x =0
x =1 Z
x =0
My
y= x4
y= x2
y= x4
5
= ,
m
8
4 d ydx =
8
15
1
4x d ydx = ,
3
Mx 1
ȳ =
= .
m
3
Z
Mx =
x=1 Z y= x2
x =0
y= x4
4y d ydx =
µ
¶
8
5 1
Therefore, the mass is m =
and the center of mass is ( x̄, ȳ) = , .
15
8 3
Page 17 of 18
8
45
ä
Calculus II for Engineering
8TH HOMEWORK – SOLUTION
Fall, 2008
(11.2) Lamina bounded by x = y2 and x = 4, ρ (x, y) = y + 3
Answer. The region R has the representation
©
ª
R = (x, y) | −2 ≤ y ≤ 2, y2 ≤ x ≤ 4 .
Using this, the definitions imply
Z
m=
y=2
Z
x =4
y=−2 x= y2
Z y=2 Z x=4
(y + 3) dxd y = 32
384
My =
,
x(y + 3) dxd y =
5
y=−2 x= y2
M y 12
Mx
4
x̄ =
=
,
ȳ =
=
.
m
5
m
15
Z
Mx =
y=2
Z
x=4
y=−2 x= y2
y(y + 3) dxd y =
128
15
µ
¶
12 4
Therefore, the mass is m = 32 and the center of mass is ( x̄, ȳ) =
.
,
5 15
Page 18 of 18
ä