Homework 09 Solutions
Math 21a
Spring, 2014
1. (Stewart 9.2 #30 ) See the image in the textbook. Ropes 3 m and 5 m in length are fastened to a holiday decoration that
is suspended over a town square. The decoration has a mass of 5 kg. The ropes, fastened at different heights, make
angles of 52◦ and 40◦ with the horizonatal. Find the tension in each wire and the magnitude of each tension. (See §9.2,
Example 7 for tips.)
Solution: The idea is to break everything into horizontal and vertical components. We let T1 denote the tension in
the 3 meter rope and T2 denote the tension in the 5 meter rope. The horizontal part of T1 is −|T1 | cos(52◦ ) and the
vertical part is |T1 | sin(52◦ ); thus T1 = h−|T1 | cos(52◦ , |T1 | sin(52◦ i. Similarly, T2 = h|T2 | cos(40◦ , |T2 | sin(40◦ i. Since
the horizontal parts of T1 and T2 should balance out, we have
−|T1 | cos 52◦ + |T2 | cos 40◦ = 0,
◦
so |T1 | = |T2 | · cos 40◦ / cos 52
. In the vertical direction, the total of the components must balance out the force of the
2
mass: mg ≈ (5 kg) 9.8 m/s = 49 N. Thus we have
|T1 | sin 52◦ + |T2 | sin 40◦ ≈ 49.
Plugging in the first equation into the second equation, we get
cos 40◦
◦
|T2 | · sin 52◦ ·
+
sin
40
= 49.
cos 52◦
Solving, we find |T2 | ≈ 30 N and then |T1 | ≈ 38 N. Therefore, the vectors are T1 ≈ h−38 cos 52◦ , 38 sin 52◦ i ≈ h−23, 30i
and T2 ≈ h30 cos 40◦ , 30 sin 40◦ i ≈ h23, 19i.
5m
3m
52◦
40◦
5 kg
2. (Stewart 9.2 #34 )
(a) Find the two unit vectors tangent to the curve y = x2 at the point (a, a2 ).
Solution: The derivative
tangent vector in this same
√
|h1, 2ai| = 4a2 + 1. Hence
opposite direction: √ −1
of x2 is 2x, so at x = a the tangent line will have slope 2a. The vector h1, 2ai is a
2a
direction (with slope
1 = 2a). To find a unit tangent vector, we must divide it by
4a2 +1
√
1
, √ 2a2
4a +1
.
4a2 +1
, √−2a
2
4a +1
is a unit tangent vector. The other unit tangent vector is in the
(b) Find the two unit vectors normal to the curve y = x2 at the point (a, a2 ).
Solution:
Any normal vector n = (A, B) should be perpendicular to the tangent vector (1, 2a), so
hA, Bi · h1, 2ai = 0.
The vector
so it is anormal vector. As before, we find that the two unit normal
h−2a, 1i satisfiesthis condition,
and √ 2a2 , √ −12
.
vectors − √ 2a2 , √ 12
4a +1
4a +1
4a +1
4a +1
(c) Draw a simple sketch of the curve y = x2 with the four unit vectors from parts (a) and (b) for at least two choices
of the constant a.
Solution: Here is a sketch of y = x2 (in blue), together with with tangent vectors (in red) and normal vectors
(in green) at four points: (x, y) = (−2, 4), (−1/2, 1/4), (1, 1) and (7/4, 49/16).
y = x2
y
4
3
2
1
−2
−1
1
2
x
3. (Based on Stewart 9.3 #28 ) Consider the line L1 given by x + 2y = 7 and the line L2 given by 5x − y = 2.
(a) There are two unit vectors that are parallel to L1 . What are they?
Solution: If we write the equation of the line L1 as y = 7/2 − x/2, we see that L√1 has slope −1/2. Thus the
vector h−2, 1i is parallel
to L1 , but is not√ a unit
√
√ vector since the length |h−2, 1i| = 5. To obtain a unit vector,
we must divide by 5. Thus
v
=
h−2/
5,
1/
5i is a unit vector parallel to L1 . The other unit vector parallel
1
√
√
to L1 is v2 = −v1 = h2/ 5, −1/ 5i.
(b) There are two unit vectors that are perpendicular to L1 . What are they?
Solution: A vector n is normal
L1 if v1 · n = 0. It follows that h1, 2i is a normal vector, and √
the associated
√ to √
√
unit normal vector is n1 = h1/ 5, 2/ 5i. The other unit normal vector to L1 is n2 = −n1 = h−1/ 5, −2/ 5i.
(c) Find the acute angle between the lines L1 and L2 .
√
√
Solution: Similarly to part (a), we see that a unit vector parallel to L2 is v2 = h1/ 26, 5/ 26i. Let θ be the
angle between L1 and L2 . Since v1 · v2 = cos θ · |v1 | · |v2 | and both v1 and v2 are unit vectors, we see that
2
1
1
5
3
cos θ = v1 · v2 = − √ · √ + √ · √ = √
.
5
26
5
26
130
√
Thus, the acute angle between L1 and L2 is arccos(3/ 130).
4.
(a) To the right is a diagram of seven vectors labeled a, b, c, d, e, f , g.
For each of the following vectors, write one of a, b, c, d, e, f , g or 0 to
indicate which vector it is equal to. You should not use any vector more
than once.
(i) projd f
(ii) d + c
(iii) proja f
(iv) d − c
(v) projf b
(vi) −g
(vii) 2c
(viii) d − f
Solution:
(i) d
(v) 0
The answers are
(ii) f
(vi) a
(iii) c
(iv) b
(vii) g
(viii) e
(b) What is a + b + c + d + e + f + g? Write your answer in as simple a form as possible, and then write a few complete
English sentences that might convince a classmate that your answer is right.
Solution:
In the sum, a + g cancels out. Both b + c and f + e both add up to d. Thus, the sum is 3d.
5. Suppose you have two non-zero vectors a and b in R3 .
(a) Is it possible that |b|a + |a|b = 0? If so, under what circumstances does that happen? If not, explain why not.
Solution: For the sum of two vectors to be 0, they must be collinear, point in opposite directions and have
equal magnitude. Clearly, the magnitude of |b|a is |b| · |a|. Similarly, the magnitude of |a|b is |a| · |b|. Since the
magnitudes of |b|a and |a|b are always the same, the sum |b|a + |a|b = 0 if and only if |b|a and |a|b point in
opposite directions. Since rescaling a vector by a constant does not change its direction, this happens precisely
when a and b point in opposite directions.
(b) (Stewart 9.3 #46 ) Suppose that you let c = |b|a + |a|b. And suppose that c 6= 0. Show that c bisects the angle
between a and b.
Solution:
The cosine of the angle between the vectors a and c is
cos θa,c =
|b||a|2 + |a| (b · a)
|b| |a| + b · a
a·c
=
=
.
|a| |c|
|a| |c|
|c|
Similarly, the cosine of the angle between the vectors b and c is
cos θb,c =
|b| (b · a) + |a||b|2
a · b + |b| |a|
b·c
=
=
|b| |c|
|b| |c|
|c|
Since these two expressions are exactly the same, the angle between a and c equals to the angle between b and
c. This tells us either that c bisects the angle between a and c or that a and b are collinear. But if a and b are
collinear, than c = |b|a + |a|b is also collinear, so c bisects a and b as well.