Download LB 220 Homework 1 (due Monday, 01/14/13)

LB 220 Homework 1 (due Monday, 01/14/13)
Directions. Please solve the problems below. Your solutions must begin
with a clear statement (or re-statement in your own words) of the problem.
You solutions should be clear, legible, and demonstrate at minimum partial
progress towards a complete solution to the problem. Please refer to the
syllabus for the policy on grading (communication, completeness, and
correctness) and late homework (homework is due at the start of class, late
homework is assessed a 20% penalty if submitted within the next 48 hours)
Collaboration. I encourage you to discuss the homework problems with
your classmates. However, each student must write and submit his or her
own homework solutions.
Calculators. You can use calculators to determine a numerical
approximation to an answer to an application question, but you should use
exact values until the very last step in the problem. Calculators are not,
however, permitted on any quizzes or exams.
1. Write an equation for the locus of points which are equidistant from
the points A and B. What shape is this locus? Use the following
points: A(2, 4, 1), B(−1, 0, 3).
Solution: Let P (x, y, z) be a point belonging to this locus. Then the
square of the distance from P to A is equal to the square of the
distance from P to B:
(x − 2)2 + (y − 4)2 + (z − 1)2 = (x + 1)2 + y 2 + (z − 3)2 .
This simplifies to 6x + 8y − 4z = 11, which is a linear equation.
Therefore, the locus is a plane. An alternate solution is to observe
that the midpoint M (1/2, 2, 2) of AB belongs to the locus and that if
−−→
the locus is a plane, then BA = h3, 4, −2i is a normal vector to this
plane. Therefore, the plane has equation
3(x − 1/2) + 4(y − 2) + −2(z − 2) = 0, which is the same (up to a
multiple of two) as the equation given above.
2. Write an equation for the locus of points whose distance to A is twice
their distance to B. What is the shape of this locus? Use the
following points: A(2, 4, 1), B(−1, 0, 3).
Solution: Let P (x, y, z) be a point belonging to this locus. Then the
square of the distance from P to A is equal to four times the square
of the distance from P to B:
(x − 2)2 + (y − 4)2 + (z − 1)2 = 4((x + 1)2 + y 2 + (z − 3)2 ).
This simplifies to 3x2 + 3y 2 + 3z 2 + 12x + 8y − 22z + 19 = 0. Now
divide through by 3 and complete the squares to obtain the following:
4
11
116
(x + 2)2 + (y + )2 + (z − )2 =
.
3
3
9
√
Therefore, the locus is a sphere of radius 116/3 centered at the
point (−2, −4/3, 11/3).
3. A team of dogs pull a sled up a snow covered path with a 5◦ incline
with a force of 500 N at an angle of 10◦ above the inclined path.
Determine the horizontal and vertical components of the force.
Solution: The force vector is directed 15◦ above the horizontal. The
horizontal component is therefore 500 cos 15◦ ≈ 483 N and the
vertical component is 500 sin 15◦ ≈ 129 N.
4. Suppose that wind blows with a speed of 50 km/hr from the direction
N45◦ W (meaning from a direction which is 45◦ to the west of the
northerly direction). A pilot is piloting an aircraft in the direction
N60◦ E at an airspeed of 250 km/hr. Determine the true course and
ground speed of the aircraft. (The true course is the direction of the
resultant of the velocity vectors of the wind and the aircraft; the
ground speed is the magnitude of this resultant vector.)
Solution: Let North be represented by the positive y-axis and East
by the positive x-axis. Then the wind blows towards the fourth
quadrant. Let w be the velocity vector of the wind. Its magnitude is
50 km/hr and its direction is the same as the unit vector
√
1
2
◦
◦
hcos −45 , sin −45 i = √ h1, −1i =
h1, −1i.
2
2
Therefore,
√
w = 25 2h1, 1i.
Similarly, if v is the velocity vector of the aircraft, then its magnitude
is 250 km/hr and its direction is the same as the unit vector
1 √
hcos 30◦ , sin 30◦ i = h 3, 1i.
2
Therefore,
√
v = 125h 3, 1i.
From this, we see that the resultant vector is equal to
√
√
√
v + w = h125 3 + 25 2, 125 − 25 2i.
The ground speed is the magnitude of this vector which is
approximately equal to 267 km/hr. The true course can be
determined by first finding the tangent of the angle θ with which the
vector makes with the positive x-axis:
√
125 − 25 2
√
√ .
tan θ =
125 3 + 25 2
Solving, we find that θ ≈ 20◦ . Therefore, the true course is
approximately N70◦ E.
5. A clothesline is tied between two poles, 8 m apart. The line is taught
and has negligible sag. When a wet shirt with a mass of 0.8 kg is
hung at the middle of the line, the midpoint is pulled down a distance
of 8 cm. Determine the tension in each half of the clothesline.
Solution: Let T be one of the two tension vectors directed along the
clothesline from the center (where the shirt is located) towards one of
the ends. Suppose that T = Tx i + Ty j, where Tx and Ty are the as
yet to be determined components of the tension.
The goal of the
q
2
problem is compute the magnitude kTk = Tx + Ty2 of the tension.
Since there are two tension vectors, one to the left and one to the
right, which together offset the downward force of the shirt, we have
that
m
2Ty = (0.8)g =⇒ Ty = 0.4g, where g ≈ 9.8 2 .
s
The ratio of Ty to Tx is equal to the ratio of 8 cm = 0.08 m to 4 m
(half the distance of the clothesline) because of the similarity of
triangles. This implies that Tx = 50Ty . Therefore,
q
q
√
√
kTk = Tx2 + Ty2 = 2500Ty2 + Ty2 = Ty 2501 ≈ (0.4)(9.8) 2501 ≈ 196 N
Remark: In the English or Imperial system, units of pounds are used.
There are two kinds of “pounds” however. There is the pound-mass
which is approximately equal to 0.453 kg or 1 kg ≈ 2.2 lb. And there
is the pound-force which is approximately equal to 4.45 N.
There is a simple relation: suppose I have 0.453 kg of vegetables, i.e.
one pound-mass of vegetables; if I place one pound-mass of
vegetables on a scale at the grocery store, then a force of 4.45N, i.e.
one pound-force, is exerted on the scale. So, one pound-mass exerts a
force of one pound-force.
To distinguish between the two pound-units, the pound-mass may be
denoted by lbm and the pound-force by lbf . To obtain the extimates
given above, use the definition of the newton and the measurement of
the acceleration of gravity near the earth’s surface:
m
),
s2
m
ft
≈ 32 2 ,
s2
s
ft
m
1 lbf ≈ (1 lbm )(32 2 ) ≈ (0.453 kg)(9.8 2 ) ≈ 4.45 N.
s
s
1 N = (1 kg)(1
g ≈ 9.8
The answer to the problem above was 196 N ≈ 44 lbf . So, the force of
tension is about the same as the force exerted by 44 pounds of
vegetables on a scale in the grocery store. You can check this is
correct: 44/2.2 ≈ 20 kg and (20)(9.8) ≈ 196 N.
The pound-mass of the wet shirt was given to be 0.8 kg ≈ 1.76 lbm .
The wet shirt exerts a force of approximately 1.76 pound-force which
is approximately (1.76)(4.45) ≈ (0.8)(9.8) ≈ 7.8 N of force. So, most
of the tension in the clothesline is due to the fact that the clothesline
has negligible sag.
·ft
Another units: the poundal (unit of force) is 1 pdl = 1 lbm
. So
s2
1 lbf ≈ 32 pdl. And the slug (unit of mass) is 1 lbf = (1 slug)(1 sft2 ).
Crazy. What’s the average mass of a slug in slugs?
Here’s what you should keep in mind: the force in newtons exerted
by 1 pound-mass of vegetables on a scale due to the acceleration of
gravity (i.e. 1 pound-force) is approximately equal to 4.45 N. In
reverse, to exert one newton of force on a scale, you should place
approximately 0.225 pound-mass of vegetables on a scale.
The following link is useful for specific examples of forces we might
observe, measured in newtons:
http://en.wikipedia.org/wiki/Orders_of_magnitude_(force)
6. Use vectors to prove that the line joining the midpoints of two sides
of a triangle is parallel to the third side and is half its length.
Solution: Let A, B, and C be the vertices of a triangle. Let M be
the midpoint of AB and N the midpoint of AC. The following vector
equations hold:
−−→ −−→ −→
AB + BC + CA = 0,
−−→
−−→
AB = 2M B,
−−→
−−→
BC = 2BN .
Consider the triangle 4M BN . The following vector equation holds:
−−→ −−→ −−→
M B + BN + N M = 0.
(I am using the fact that vector addition v + w can be computed by
placing the tail of w at the head of v and representing v + w by the
directed line segment from the tail of v to the head of w.)
−−→
Multiply the above equation by two and re-write in terms of AB, and
−−→
BC:
−−→
−−→
−−→ −−→ −−→
−−→
2M B + 2BN + 2N M = AB + BC + 2N M = 0.
Combined with the very first equation we deduce that
−−→ −→
2N M = CA.
−−→
−→
Thus, N M is parallel to CA. And therefore, AC is parallel to M N .
In fact, we have proved more: the length of AC is twice that of M N .
7. Determine the measure of the acute angles at the points of
intersection of the curves y = x2 and y = x3 .
Solution: x2 = x3 =⇒ x2 (x − 1) = 0 =⇒ x = 0, 1. So, the points of
intersection are O(0, 0) and P (1, 1).
At O, the slope of the tangent line to y = x2 is y 0 (0) = 2(0) = 0. The
same is true for the tangent line to y = x3 at O. Therefore, the angle
between the curves at O is equal to zero.
At P , the slope of the tangent line to y = x2 is 2 and the slope of the
tangent line to y = x3 is y 0 (0) = 3(1)2 = 3. The vector a = h1, 2i has
slope 2 and the vector b = h1, 3i has slope 3. The cosine of the angle
between
these two vectors is equal to (a · b)/(kakkbk)
√ √
= 7/( 5 10) = θ. Therefore, the angle between the two curves at P
is equal to arccos θ ≈ 8.13◦ .
8. Determine the two unit vectors orthogonal to both h1, 2, 3i and
h−1, 0, 1i.
Solution: Let v be the vector
i
v = 1
−1
product of the above two vectors:
j k
2 3 = h2, −4, 2i
0 1
√
√
The length of v is equal to 24 = 2 6. Therefore, the unit vectors
√1 v = √1 h1, −2, 1i and its negative are the two unit vectors
24
6
orthogonal to both of the two given vectors.
9. Use the triple scalar product to determine whether the points
A(1, 3, 2), B(3, −1, 6), C(5, 2, 0), and D(3, 6, −4) are coplanar.
−−→
−→
−−→
Solution: AB = h2, −4, 4i, AC = h4, −1, −2i, AD = h2, 3, −6i. The
points are coplanar if and only if the following determinant
(representing the volume of the parallelepiped spanned by the above
three vectors) is equal to zero.
2 −4 4 4 −1 −2 = 2 −1 −2 − (−4) 4 −2 + 4 4 −1
2 3 3 −6
2 −6
2 3 −6
= 2(6 + 6) + 4(−24 + 4) + 4(12 + 2) = 24 − 80 + 56 = 0
So, the points are coplanar.
10. A wrench 30 cm long lies along the positive y-axis and grips a bolt at
the origin. A force is applied in the direction h0, 3, −4i at the end of
the wrench. Find the magnitude of the force needed to supply
100 N · m of torque to the bolt.
Let τ be the torque vector so that τ = r × F, where
r = h0, 30/100, 0i (in meters) is the position vector of the point where
the force is applied (relative to the point of rotation placed at the
origin) and where F is the force vector. We are given that F has the
same direction as v = h0, 3, −4i. Therefore,
F = kFk
v
.
kvk
We are to determine the value of α = kFk so that kτ k = 100. From
the above equation we have that
F=
α
v.
5
The easiest way to solve the problem is to make a sketch and
recognize that the vectors are parallel to the yz-plane and so the
problem can be reduced to a two-dimensional problem using a
(3, 4, 5)-triangle.
The problem can be solved using the vector product, but it is
somewhat
√ easier to use the scalar product and the identity
sin θ = 1 − cos2 θ for 0 ≤ θ ≤ π. For,
cos θ =
r·F
3
= .
krkkFk
5
(You will see that the factor of α cancels in the above computation.)
From this we have that sin θ = 54 . And therefore,
kτ k = 100 = krkkFk sin θ =
and so α =
1250
3
3 4
α
10 5
= 416.6 N of force needs to be applied.