Solution to Quiz 1
Calculus II for Engineering, Section 03
UID No: Solution
Name: Solution
Date: Wednesday, February 20, 2008
Score:
30/30
1/6 (5 points) Suppose that there are two forces acting on a sky diver: gravity at 150
pounds down and air resistance at 140 pounds up and 20 pounds to the right. What
is the net force acting on the sky diver?
Answer. Two forces are acting on the sky diver: Gravity G = h0; 150i (Why?) and
Air Resistance a = h20; 140i. Hence, the net force is
G + a = h0;
150i + h20; 140i = h0 + 20; 150 + 140i = h20; 10i ;
which saying 20 pounds to the right and 10 pounds down. (Exercise 31 in Section
10.1)
2/6 (5 points) For the points P = (2; 3; 1), Q = (4; 2; 2) and R = (8; 0; 4), (1) nd vectors
!
!
P Q and QR and (2) are those three points on the same line (i.e., colinear)? Justify
your answer.
Answer.
!
P Q = h4
We observe
!
2; 2 3; 2 1i = h2; 1; 1i ;
QR = h8
!
!
QR = h4; 2; 2i = 2 h2; 1; 1i = 2P Q;
!
!
4; 0 2; 4 2i = h4; 2; 2i :
i:e:;
!
!
QR = 2P Q:
It implies that two vectors QR and P Q are parallel. Moreover, those three points are
on the same line. (Think about a line passing through P , Q, R having two vectors
!
!
P Q and QR parallel!) (Exercise 43 in Section 10.2)
1/4
(Continued)
3/6 (5 points) Compute a b: (1) a = h3; 2; 0i and b = h 2; 4; 3i and (2) a = 2i k and
b = 4j k, where i, j and k are standard basis vectors.
Answer. By the denition of the dot product,
(1) a b = h3; 2; 0i h 2; 4; 3i = 6 + 8 + 0 = 2;
(2) a b = h2; 0; 1i h0; 4; 1i = 0 + 0 + 1 = 1:
(Exercise 4 and 5 in Section 10.3)
4/6 (5 points) A constant force of h30; 20i pounds moves an object in a straight line from
the point (0; 0) to the point (24; 10). Compute the work done.
Answer. The work done is obtained by the dot product of the force F = h30; 20i and
!
the moving direction d = OP = h24 0; 10 0i = h24; 10i, where O = (0; 0) and
P = (24; 10). Thus, we deduce
W = F d = h30; 20i h24; 10i = 920:
(Exercise 29 in Section 10.3)
2/4
(Continued)
5/6 (5 points) Find two unit vectors orthogonal to the two given vectors a = h2; 1; 0i
and b = h1; 0; 3i.
Answer. Method 1 Using Dot Product : Corollary 3.1 says that two vectors c and
a are orthogonal if and only if c a = 0. Let c = hc1 ; c2 ; c3 i and we nd c which is
a unit vector and orthogonal to a and b.
c a = hc1 ; c2 ; c3 i h2;
1; 0i = 2c1 c2 = 0; i:e:; c2 = 2c1;
c b = hc1 ; c2 ; c3 i h1; 0; 3i = c1 + 3c3 = 0; i:e:; c1 = 3c3 :
From those two equations c2 = 2c1 and c1 = 3c3, we deduce
c1 = 3c3;
c2 = 2c1 = 2( 3c3) = 6c3;
i:e:; c = hc1; c2; c3i = h 3c3; 6c3; c3i = c3 h 3; 6; 1i :
p
kck = kc3 h 3; 6; 1ik = jc3j 46:
We are
p looking for c which is a unit vector. It suggests that we should choose c3 =
1= 46. Therefore,
c=
p1 h
46
3; 6; 1i ;
and
c=
p1 h
46
3; 6; 1i :
Method 2 Using Cross Product : Theorem 4.2 says that a b is orthogonal to both
a and b. So we compute a b.
ab=
i
j
k
2 1 0 = i ( 3 0) j (6 0) + k (0 + 1)
1 0 3
= 3i 6j + k = h 3; 6; 1i
p
ka bk = kh 3; 6; 1ik = 46:
Thus, when we dene c as follows,
c=
p1 h
46
3; 6; 1i ;
and
c=
p1 h
46
3; 6; 1i ;
we have two unit vectors which are orthogonal to both a and b. Check yourself.
(Exercise 13 in Section 10.4)
3/4
(Continued)
6/6 (5 points) If you apply a force of magnitude 30 pounds at the end of an 2=3-feet-long
wrench at an angle of =3 to the wrench, nd the magnitude of the torque applied to
the bolt.
Answer. Torque
vector r:
τ
is dened by the cross product of the force F and the position
τ
=F
r:
It implies the magnitude of the torque:
kτ k = kF rk = kF k krk sin ;
where is the angle between F and r.
Since kF k = 30, krk = 2=3 and = =3, we deduce
p
2! kτ k = 30 3 sin 3 = 10 3:
(Exercise 26 in Section 10.4)
4/4
(by Theorem 4.4)
Quiz 3
Calculus II for Engineering, Section 03
Name: Solution
UID No: Solution
Date: Monday, April 21, 2008
Score: 5/5
Find the directional derivative of the function f (x; y ) = y 2 e4x at the point (0; 2) in the direction of h3; 1i.
Please, ll up the blanks.
Solution. Let u be the unit vector in the direction of h3; 1i. Then, u is obtained as follows:
h3; 1i = h3p; 1i
k h3; 1i k
10
u=
Using this vector and the Theorem, we can deduce the directional derivative Du f (x; y ) as follows:
D
E
D
E
Du f (x; y ) = 4y 2 e4x ; 2ye4x
=
=
4y 2 e4x ; 2ye4x
12y 2 e4x
p
10
u
*
2ye4x
p :
10
p3 ;
10
p1
10
+
Therefore, putting the point (x; y ) = (0; 2) into the equation above, we conclude
Du f (0; 2) =
12( 2)2 e4(0)
p
10
2( 2)e4(0)
52
p
=p :
10
10
Extracted from Calculus II for Engineers, Final Exam, Spring 2007
1
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