Winter 2012
Math 255
Problem Set 2
1) Prove the chain rule ( − δ proofs are welcome)
dF (f (t)) dF df
=
dt
df dt
Solution
Method 1:
F (f (t + δt)) − F (f (t))
δt→0
δt
F (f (t + δt)) − F (f (t)) f (t + δt) − f (t)
= lim
δt→0
f (t + δt) − f (t)
δt
f (t + δt) − f (t)
F (f (t + δt)) − F (f (t))
lim
= lim
δt→0
δt→0
f (t + δt) − f (t)
δt
dF df
=
df dt
lim
where we have used the continuity of f (t) and the
fact that the limit of a product of functions is equal
to the product of the limits of two functions.
Method 2:
We recall that if limx→a y(x) = b then for every > 0
such that |x − z| < there exists a δ() such that
|y(x) − b| < δ(). Since F (f ) and f (t) are differentiable, we know that
F (f + ∆f ) − F (f ) dF
=
∆f →0
∆f
df
lim
and
f (t + ∆t) − f (t) df
=
.
∆t→0
∆t
dt
lim
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Thus for every F > 0 there exists δF > 0 such that
if |∆f | < F , then
F (f + ∆f ) − F (f ) dF < δF
−
∆f
df and f > 0 there exists δf > 0 such that if |∆t| < f ,
then
f (t + ∆t) − f (t) df < δf .
−
∆t
dt We therefore need to show that for every F f > 0
there exists δF f (F f ) such that if
|∆t| < F f
then F (f (t + ∆t)) − F (f (t)) dF df < δF f .
−
∆t
df dt Now
F (f (t + ∆t)) − F (f (t)) dF df −
∆t
df dt F (f (t + ∆t)) − F (f (t)) f (t + ∆t) − f (t) dF df = −
f (t + ∆t) − f (t)
∆t
df dt F (f (t + ∆t)) − F (f (t)) dF f (t + ∆t) − f (t)
= −
f (t + ∆t) − f (t)
df
∆t
dF f (t + ∆t) − f (t) df +
−
df
∆t
dt F (f (t + ∆t)) − F (f (t)) dF f (t + ∆t) − f (t) −
≤ f (t + ∆t) − f (t)
df
∆t
dF f (t + ∆t) − f (t) df + −
df
∆t
dt df dF ≤ δF + δf + δf
dt
df
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Math 255
In the last line, we have used the fact that f is continuous, so for ∆t < F f there is a ∆f such that
f (t + ∆t) − f (t) = ∆f
and as ∆t
→ 0 then ∆f → 0.
df dF We assume that dt and df are bounded for the
values of interest, and taking their maxima, we deduce that we have found our δF f .
Section 11.3, Pages 705–715:
3) The parametric equations of the cycloid are given by
x = r(θ − sin θ), y = r(1 − cos θ). Eliminate θ from
these equations to get a cartesian equation in x and
y.
Solution
We get that θ = cos−1 (1 − y/r) and so
x = r cos−1 (1 − y/r) − sin cos−1 (1 − y/r) .
4) Find the asymptotes to the tractrix, which is given
by x = a sin t and y = a (cos t + log(tan t/2))
Solution
We observe that y → ∞ when t → 0, π, 2π..., for
which x → 0. Hence the curve has an asymptote
along the y-axis. The curve is sketched in Fig. 1.
5) (72) Use a graphing device to graph the polar curve
r = sin2 (4θ) + cos(4θ). Choose the parameter interval to make sure that you produce the entire curve.
Solution
The curve is sketched in Fig. 2.
Section 11.4, Pages 715–720:
6) (14) Sketch the curve r = 2 + cos(2θ) and find the
area it encloses.
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2.5
-2.5
0
2.5
5
Figure 1: The Tractrix
1.5
0.5π
1
0.75π
0.25π
0.5
-2.4
-2
-1.6
π
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
-0.5
1.75π
1.25π
-1
1.5π
-1.5
Figure 2: The curve r = sin2 (4θ) + cos(4θ)
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Winter 2012
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Solution
The curve is sketched
R 2 in Fig. 3. We recall that the
area is given by r dθ. In this case, we can use the
symmetry of the curve and find
Z π/2
9π
4
.
(2 + cos(2θ))2 dθ =
2
0
2.4
1.6
0.5π
0.8
π
-4.8
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
4.8
-0.8
1.5π
-1.6
-2.4
Figure 3: The curve r = 2 + cos(2θ)
7) (42) Find all the points of intersection of r2 = sin(2θ)
and r2 = cos(2θ).
Solution
We need to find the points where tan(2θ) = 1, we
find that θ = π/8, 5π/8, 9π/8, and so r2 = ±2−1/2 .
We require r to be positive and real, hence r = 2−1/2
and θ = π/8, 9π/8. We check our results by plotting
a graph. The graph shows that the curves also cross
at the origin, but this occurs for different values of
θ, thus we also need to include the point r = 0.
8) (48) Find the exact length of the polar curve r = θ
for 0 ≤ θ ≤ 2π.
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Winter 2012
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1.5
0.5π
1
0.75π
0.25π
0.5
-2.4
-2
-1.6
π
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
-0.5
1.25π
1.75π
-1
1.5π
-1.5
Figure 4: The curves r2 = cos(2θ) and r2 = sin(2θ)
Solution
We use the formula for arclength in polar coordinates
s
2
Z b
dr
S=
r2 +
dθ
dθ
a
to find that
Z 2π p
S=
θ2 + 1dθ
0
p
p
= 0.5θ 1 + θ2 + 0.5 ln(θ + 1 + θ2 )|2π
0
p
p
= π 1 + 4π 2 + 0.5 ln(2π + 1 + 4π 2 )
where we have used a table to find the integral. We
could also use trigonometric substitution tan α = θ
and then integrate sec3 α.
9) (56)
a) Find a formula for the area of the surface generated by rotating the polar curve r = f (θ), a ≤
θ ≤ b (where f’ is continuous and 0 ≤ a < b ≤ π),
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Winter 2012
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about the line θ = π/2.
Solution
We recall that the area generated by rotating a
curve around the x axis is given by
s 2
Z
2
dx
dy
S = 2π y
+
dt.
dt
dt
For a rotation around the y axis we get
s 2
Z
2
dx
dy
S = 2π x
+
dt.
dt
dt
We choose θ instead of t to be our parameter and
recall that x = r cos θ, y = r sin θ, and letting
r = f (θ), after differentiating and substituting
we get that
s Z
2
df
S = 2π f (θ) cos θ
+ (f )2 dt.
dθ
b) Find the surface area generated by rotating the
lemniscate r2 = cos 2θ about the line θ = π/2.
Solution
The lemniscate is defined for −π/4 ≤ θ ≤ π/4
and 3π/4 ≤ θ ≤ 5π/4. By symmetry, we can use
one of these curves to integrate with. The area
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is given by
Z
s
√
π/4
2π cos 2θ cos θ
−π/4
Z
sin2 (2θ)
dθ
cos(2θ) +
cos(2θ)
π/4
= 2π
cos(θ)dθ
−π/4
π = 4π sin
4
√
= 2 2π.
Read sections 11.5 and 11.6 on conic sections and
conic sections in polar coordinates because similar
material occurs in Chapter 13.5, 13.6 and 13.7. Sections 11.5 and 11.6 are not directly examinable, but
13.5, 13.6 and 13.7 are examinable.
Section 13.1, Pages 829–834:
10) (4) What are the projections of the point (2, 3, 5) on
the xy−, yz−, and xz−planes? Draw a rectangular
box with the origin and (2, 3, 5) as opposite vertices
and with its faces parallel to the coordinate planes.
Label all vertices of the box. Find the length of the
diagonal of the box.
Solution
On the xy plane we have (2, 3, 0), on the yz plane we
have (0, 3, 5) and on the xz plane we have (2, 0, 5).
The length of the diagonal is
p
√
√
22 + 32 + 52 = 4 + 9 + 25 = 38.
A sketch of the box is in fig. 5.
11) (42) Find the volume of the solid that lies inside both
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Winter 2012
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Figure 5: A picture of a box.
of the spheres
x2 + y 2 + z 2 + 4x − 2y + 4z + 5 = 0
and
x2 + y 2 + z 2 = 4.
Solution
We first rewrite these equations in standard form to
get
(x + 2)2 + (y − 1)2 + (z + 2)2 = 4
and
x2 + y 2 + z 2 = 4.
Thus the √
spheres both have radius of 2 and centers
that are 22 + 12 + 22 = 3 units apart. To make
the calculations easier, we can therefore consider the
volume between the spheres
(x − 1.5)2 + y 2 + z 2 = 4
and
(x + 1.5)2 + y 2 + z 2 = 4.
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This is a volume of revolution (see pg. 387 or Stewart) so we need to calculate
Z 0.5
2π
y 2 dx
0
Z 0.5
= 2π
4 − (x + 1.5)2 dx
Z0 0.5
= 2π
4 − 1.52 − x2 − 3xdx
0
x3 3x2
7
= 2π x −
−
4
3
2
11π
=
12
This agrees with problem 49 on pg. 392.
Section 13.2, Pages 834–842:
12) Find the sum of the vectors < −1, 3, 4 > and <
1, 3, 2 >. Illustrate this geometrically.
Solution
< −1, 3, 4 > + < 1, 3, 2 >=< 0, 6, 6 >
This is illustrated in fig. 6.
13) a) Find |a|, a + b, a − b, 2a and 3a + 4b for
a =< 1, 2, 3, 4 >
b =< 5, 6, 0, 1 >
Solution
p
√
2
2
2
|a| = 1 + 2 + 3 + 4 = 30
a + b =< 6, 8, 3, 5 >
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Winter 2012
Math 255
Figure 6: A picture of demonstrating the sum of two vectors.
a − b =< −4, −4, 3, 3 >
2a =< 2, 4, 6, 8 >
3a + 4b =< 23, 30, 9, 16 >
b) Describe a situation where you may want to use
vectors with more than 3 entries.
Solution
Many possible choices, for example, a shopping
list, exam scores, coordinates in space and time.
14) A clothesline is tied between two poles 7m apart.
The line is quite taut and has negligible sag. When
a wet shirt with a mass of 0.5Kg is hung at the middle of the line, the midpoint is pulled down 5cm.
Find the tension in each half of the clothesline.
Solution
We decompose the tension into vertical and horizontal components. The sum of the vertical components
of the tensions in the two halves of the string must be
equal to 0.5Kg×10m/s2 = 5N. Thus 2.5N= T sin θ,
where T is the tension in the string and θ is the angle
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Winter 2012
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between
√ the horizontal and the string.
√ Hence sin θ =
0.05/ 3.52 + 0.052 and so T = 2.5 3.52 + 0.052 /0.05N≈
175.0179N.
Bonus problem
15) (pg. 735 no. 4) Four bugs are placed at the four
corners of a square with side length a. The bugs
crawl counterclockwise at the same speed and each
bug crawls directly toward the next bug at all times.
They approach the center of the square along spiral
paths.
a) Find the polar equation of a bug’s path assuming the pole is the center of the square (Use the
fact that the line joining one bug to the next is
tangent to the bug’s path).
Solution
Let1 us label the bugs 1, 2, 3 and 4. By symmetry all the bugs each have the same distance
from the origin as a function of time and also the
angles between radii from the origin to the bugs
also remain at π/2 or 90◦ . We let ri denote the
position of bug i, so that the equations of motion
are given by
ṙi = α(rmod(i+1,4) − ri ).
We let ri = r(t) [cos θi (t), sin θi (t)] and convert to
polar coordinates. We then using the fact that
θ4 = θ3 + π/2 = θ2 + π = θ1 + 3π/2 and choose
α = − (rK√2 , where K is a constant, so that |ṙ| =
1 We
can, following Gardner (1957), assume 1 and 3 are female and 2 and 4 are male.
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Winter 2012
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K. Now
K
ṙ1 = √ (sin θ1 − cos θ1 , − cos θ1 − sin θ1 )
2
and also
ṙ1 = (ṙ cos θ1 − θ̇r sin θ, ṙ sin θ + rθ̇ cos θ).
After equating components, we get that θ˙1 =
− √K2 and so since r(t = 0) = √a2 , we get that
K
√ . We also get that r θ̇ = − √
and so
r(t) = a−Kt
2
2
choosing θ(t = 0) = 0, we get that
√ 2r Kt = log θ = log 1 −
.
a a
It is also possible to derive the equation for this
curve without calculus.
b) Find the distance traveled by a bug by the time
it meets the other bugs at the center.
Solution
We find that r = 0 when t = Ka . We recall that
the distance traversed is given by
s
s 2
2
Z
Z
2
dr
dθ
dr
r2 +
dθ =
r2
+
dt.
dθ
dt
dt
In our parameterization this becomes
Z ar 2
K
K
K2
+
dt = a.
2
2
0
c) Download the extension by Chapman, Lottes and
Trefethen Four bugs on a rectangle, Proc. Roy.
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Math 255
Soc. A, vol. 467 pg. 881-8962 . Who originally
came up with the problem of Four Bugs on a
Square? Write a one paragraph summary of what
you consider to be the most important point in
Four bugs on a rectangle.
Solution
According to the paper cyclic pursuit first appeared in a Cambridge University tripos exam of
1871. It is also possible to get points for stating; in a 1735 paper by Pierre Bouguer on the
problem of a pirate ship chasing a merchant vessel. Anything reasonable for the one paragraph
summary will receive credit.
2 Bing,
Google, Yahoo or any other search engine should locate this for you.
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