Winter 2012 Math 255

Problem Set 2

1) Prove the chain rule (ε− δ proofs are welcome)

dF (f(t))

dt=

dF

df

df

dt

SolutionMethod 1:

limδt→0

F (f(t+ δt))− F (f(t))

δt

= limδt→0

F (f(t+ δt))− F (f(t))

f(t+ δt)− f(t)

f(t+ δt)− f(t)

δt

= limδt→0

F (f(t+ δt))− F (f(t))

f(t+ δt)− f(t)limδt→0

f(t+ δt)− f(t)

δt

=dF

df

df

dt

where we have used the continuity of f(t) and thefact that the limit of a product of functions is equalto the product of the limits of two functions.Method 2:We recall that if limx→a y(x) = b then for every ε > 0such that |x − z| < ε there exists a δ(ε) such that|y(x) − b| < δ(ε). Since F (f) and f(t) are differen-tiable, we know that

lim∆f→0

F (f + ∆f)− F (f)

∆f=

dF

df

and

lim∆t→0

f(t+ ∆t)− f(t)

∆t=

df

dt.

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Winter 2012 Math 255

Thus for every εF > 0 there exists δF > 0 such thatif |∆f | < εF , then∣∣∣∣F (f + ∆f)− F (f)

∆f− dF

df

∣∣∣∣ < δF

and εf > 0 there exists δf > 0 such that if |∆t| < εf ,then ∣∣∣∣f(t+ ∆t)− f(t)

∆t− df

dt

∣∣∣∣ < δf .

We therefore need to show that for every εFf > 0there exists δFf(εFf) such that if

|∆t| < εFf

then ∣∣∣∣F (f(t+ ∆t))− F (f(t))

∆t− dF

df

df

dt

∣∣∣∣ < δFf .

Now∣∣∣∣F (f(t+ ∆t))− F (f(t))

∆t− dF

df

df

dt

∣∣∣∣=

∣∣∣∣F (f(t+ ∆t))− F (f(t))

f(t+ ∆t)− f(t)

f(t+ ∆t)− f(t)

∆t− dF

df

df

dt

∣∣∣∣=

∣∣∣∣(F (f(t+ ∆t))− F (f(t))

f(t+ ∆t)− f(t)− dF

df

)f(t+ ∆t)− f(t)

∆t

+dF

df

(f(t+ ∆t)− f(t)

∆t− df

dt

)∣∣∣∣≤∣∣∣∣F (f(t+ ∆t))− F (f(t))

f(t+ ∆t)− f(t)− dF

df

∣∣∣∣ ∣∣∣∣f(t+ ∆t)− f(t)

∆t

∣∣∣∣+

∣∣∣∣dFdf∣∣∣∣ ∣∣∣∣f(t+ ∆t)− f(t)

∆t− df

dt

∣∣∣∣≤ δF

(∣∣∣∣dfdt∣∣∣∣+ δf

)+

∣∣∣∣dFdf∣∣∣∣ δf

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Winter 2012 Math 255

In the last line, we have used the fact that f is con-tinuous, so for ∆t < εFf there is a ∆f such thatf(t+ ∆t)− f(t) = ∆f and as ∆t→ 0 then ∆f → 0.

We assume that∣∣∣dfdt ∣∣∣ and

∣∣∣dFdf ∣∣∣ are bounded for the

values of interest, and taking their maxima, we de-duce that we have found our δFf .Section 11.3, Pages 705–715:

3) The parametric equations of the cycloid are given byx = r(θ − sin θ), y = r(1− cos θ). Eliminate θ fromthese equations to get a cartesian equation in x andy.SolutionWe 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 givenby x = a sin t and y = a (cos t+ log(tan t/2))SolutionWe observe that y → ∞ when t → 0, π, 2π..., forwhich x → 0. Hence the curve has an asymptotealong the y-axis. The curve is sketched in Fig. 1.

5) (72) Use a graphing device to graph the polar curver = sin2(4θ) + cos(4θ). Choose the parameter inter-val to make sure that you produce the entire curve.SolutionThe curve is sketched in Fig. 2.

Section 11.4, Pages 715–720:

6) (14) Sketch the curve r = 2 + cos(2θ) and find thearea it encloses.

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Winter 2012 Math 255

-2.5 0 2.5 5

2.5

Figure 1: The Tractrix

-2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4

-1.5

-1

-0.5

0.5

1

1.5

0.25π

0.5π

0.75π

π

1.25π

1.5π

1.75π

Figure 2: The curve r = sin2(4θ) + cos(4θ)

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Winter 2012 Math 255

SolutionThe curve is sketched in Fig. 3. We recall that thearea is given by

∫r2dθ. In this case, we can use the

symmetry of the curve and find

4

∫ π/2

0

(2 + cos(2θ))2dθ =9π

2.

-4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8

-2.4

-1.6

-0.8

0.8

1.6

2.4

0.5π

π

1.5π

Figure 3: The curve r = 2 + cos(2θ)

7) (42) Find all the points of intersection of r2 = sin(2θ)and r2 = cos(2θ).SolutionWe need to find the points where tan(2θ) = 1, wefind 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 plottinga graph. The graph shows that the curves also crossat the origin, but this occurs for different values ofθ, 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 Math 255

-2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4

-1.5

-1

-0.5

0.5

1

1.5

0.25π

0.5π

0.75π

π

1.25π

1.5π

1.75π

Figure 4: The curves r2 = cos(2θ) and r2 = sin(2θ)

SolutionWe use the formula for arclength in polar coordinates

S =

∫ b

a

√r2 +

(dr

dθ

)2

dθ

to find that

S =

∫ 2π

0

√θ2 + 1dθ

= 0.5θ√

1 + θ2 + 0.5 ln(θ +√

1 + θ2)|2π0= π

√1 + 4π2 + 0.5 ln(2π +

√1 + 4π2)

where we have used a table to find the integral. Wecould also use trigonometric substitution tanα = θand then integrate sec3 α.

9) (56)

a) Find a formula for the area of the surface gener-ated 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 Math 255

about the line θ = π/2.SolutionWe recall that the area generated by rotating acurve around the x axis is given by

S = 2π

∫y

√(dx

dt

)2

+

(dy

dt

)2

dt.

For a rotation around the y axis we get

S = 2π

∫x

√(dx

dt

)2

+

(dy

dt

)2

dt.

We choose θ instead of t to be our parameter andrecall that x = r cos θ, y = r sin θ, and lettingr = f(θ), after differentiating and substitutingwe get that

S = 2π

∫f(θ) cos θ

√(df

dθ

)2

+ (f)2dt.

b) Find the surface area generated by rotating thelemniscate r2 = cos 2θ about the line θ = π/2.SolutionThe lemniscate is defined for −π/4 ≤ θ ≤ π/4and 3π/4 ≤ θ ≤ 5π/4. By symmetry, we can useone of these curves to integrate with. The area

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Winter 2012 Math 255

is given by∫ π/4

−π/42π√

cos 2θ cos θ

√cos(2θ) +

sin2(2θ)

cos(2θ)dθ

= 2π

∫ π/4

−π/4cos(θ)dθ

= 4π sin(π

4

)= 2√

2π.

Read sections 11.5 and 11.6 on conic sections andconic sections in polar coordinates because similarmaterial occurs in Chapter 13.5, 13.6 and 13.7. Sec-tions 11.5 and 11.6 are not directly examinable, but13.5, 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) onthe xy−, yz−, and xz−planes? Draw a rectangularbox with the origin and (2, 3, 5) as opposite verticesand with its faces parallel to the coordinate planes.Label all vertices of the box. Find the length of thediagonal of the box.SolutionOn the xy plane we have (2, 3, 0), on the yz plane wehave (0, 3, 5) and on the xz plane we have (2, 0, 5).The length of the diagonal is√

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 Math 255

Figure 5: A picture of a box.

of the spheres

x2 + y2 + z2 + 4x− 2y + 4z + 5 = 0

andx2 + y2 + z2 = 4.

SolutionWe first rewrite these equations in standard form toget

(x+ 2)2 + (y − 1)2 + (z + 2)2 = 4

andx2 + y2 + z2 = 4.

Thus the spheres both have radius of 2 and centersthat are

√22 + 12 + 22 = 3 units apart. To make

the calculations easier, we can therefore consider thevolume between the spheres

(x− 1.5)2 + y2 + z2 = 4

and(x+ 1.5)2 + y2 + z2 = 4.

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Winter 2012 Math 255

This is a volume of revolution (see pg. 387 or Stew-art) so we need to calculate

2π

∫ 0.5

0

y2dx

= 2π

∫ 0.5

0

4− (x+ 1.5)2dx

= 2π

∫ 0.5

0

4− 1.52 − x2 − 3xdx

= 2π

[7

4x− x3

3− 3x2

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

|a| =√

1 + 22 + 32 + 42 =√

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 usevectors with more than 3 entries.SolutionMany possible choices, for example, a shoppinglist, 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. Whena wet shirt with a mass of 0.5Kg is hung at the mid-dle of the line, the midpoint is pulled down 5cm.Find the tension in each half of the clothesline.SolutionWe decompose the tension into vertical and horizon-tal components. The sum of the vertical componentsof the tensions in the two halves of the string must beequal 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 Math 255

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 fourcorners of a square with side length a. The bugscrawl counterclockwise at the same speed and eachbug crawls directly toward the next bug at all times.They approach the center of the square along spiralpaths.

a) Find the polar equation of a bug’s path assum-ing the pole is the center of the square (Use thefact that the line joining one bug to the next istangent to the bug’s path).SolutionLet1 us label the bugs 1, 2, 3 and 4. By sym-metry all the bugs each have the same distancefrom the origin as a function of time and also theangles between radii from the origin to the bugsalso remain at π/2 or 90◦. We let ri denote theposition of bug i, so that the equations of motionare given by

ri = α(rmod(i+1,4) − ri).

We let ri = r(t) [cos θi(t), sin θi(t)] and convert topolar coordinates. We then using the fact thatθ4 = θ3 + π/2 = θ2 + π = θ1 + 3π/2 and chooseα = − K

(r√

2, where K is a constant, so that |r| =

1We can, following Gardner (1957), assume 1 and 3 are female and 2 and 4 are male.

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Winter 2012 Math 255

K. Now

r1 =K√

2(sin θ1 − cos θ1,− cos θ1 − sin θ1)

and also

r1 = (r cos θ1 − θr sin θ, r sin θ + rθ cos θ).

After equating components, we get that θ1 =− K√

2and so since r(t = 0) = a√

2, we get that

r(t) = a−Kt√2

. We also get that rθ = − K√2

and so

choosing θ(t = 0) = 0, we get that

θ = log

∣∣∣∣1− Kt

a

∣∣∣∣ = log

∣∣∣∣∣√

2r

a

∣∣∣∣∣ .It is also possible to derive the equation for thiscurve without calculus.

b) Find the distance traveled by a bug by the timeit meets the other bugs at the center.SolutionWe find that r = 0 when t = a

K . We recall thatthe distance traversed is given by∫ √

r2 +

(dr

dθ

)2

dθ =

∫ √r2

(dθ

dt

)2

+

(dr

dt

)2

dt.

In our parameterization this becomes∫ aK

0

√K2

2+K2

2dt = a.

c) Download the extension by Chapman, Lottes andTrefethen Four bugs on a rectangle, Proc. Roy.

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Winter 2012 Math 255

Soc. A, vol. 467 pg. 881-8962. Who originallycame up with the problem of Four Bugs on aSquare? Write a one paragraph summary of whatyou consider to be the most important point inFour bugs on a rectangle.SolutionAccording to the paper cyclic pursuit first ap-peared in a Cambridge University tripos exam of1871. It is also possible to get points for stat-ing; in a 1735 paper by Pierre Bouguer on theproblem of a pirate ship chasing a merchant ves-sel. Anything reasonable for the one paragraphsummary will receive credit.

2Bing, Google, Yahoo or any other search engine should locate this for you.

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