1 question on the cycloid - university college...

Answers to question sheet 2, qs02.tex

1 question on the cycloidcycloid

The position c(t) of particle of mass m and electric charge e in combined electric and magnetic fields E and B (accordingto Newton and Lorenz) satisfies the differential equation

mc = e(E+ c×B)

Assuming q = e = m = 1 and normal ’crossed fields’ E = j and B = k show that the equation becomes xyz

=

010

+

y−x

0

Show that, subject to initial conditions c(0) = c(0), the particle execute a cycloid in the plane z = 0.

1.1 answer

mc = e(E+ c×B)⇔ 1 · c = 1 · (j+ c×k)

⇔

xyz

=

010

+

i j kx y z0 0 1

⇔

xyz

=

010

+

y−x

0

⇔

xyz

=

y1− x

0

We solve the trivial differential equation z = 0 to obtain z(t) = z(0)+ tz(0)t = 0 + 0t = 0 (using the initial conditions).The main differential equation now becomes 2 dimensional(

xy

)=(

y1− x

)Below we valiently solve this differential equation subject to the initial conditions but it suffices to check that a parame-terization of the cycloid satisfies both the D.E. and the I.C.

Recall, parameterization of the cycloid with velocity and acceleration is(xy

)=(

t− sin t1− cos t

),

(xy

)=(

1− cos tsin t

),

(xy

)=(

sin tcos t

)Clearly the I.C. are satisfied x(0) = y(0) = x(0) = y(0) = 0. Also the D.E is satisfied x = sin t = y and y = cos t =1− (1− cos t) = 1− x.

Next we give a second method. We directly solve the D.E with I.C. Write the D.E. in the form(xy

)+ A

(xy

)=(

01

)where A =

(0 −11 0

)

1

⇔ exp(At)(

xy

)+ Aexp(At)

(xy

)= exp(At)

(01

)where we multiply across by the ’integrating factor’exp(At) = exp

(0 −tt 0

)=(

cos(t) −sin(t)sin(t) cos(t)

)⇔ d

dt

[exp(At)

(xy

)]=(−sin(t)

cos(t)

)⇔

Z t

0

dd p

[exp(Ap)

(xy

)]d p =

Z t

0

(−sin(p)

cos(p)

)d p

⇔[

exp(Ap)(

x(p)y(p)

)]t

p=0=(

cos(p)sin(p)

)t

p=0

⇔ exp(At)(

x(t)y(t)

)=(

cos(t)−1sin(t)

)on the LHS we used I.C. x(0) = y(0) = 0

⇔(

x(t)y(t)

)= exp(−At)

(cos(t)−1

sin(t)

)⇔

(x(t)y(t)

)=(

cos(t) sin(t)−sin(t) cos(t)

)(cos(t)−1

sin(t)

)⇔

(x(t)y(t)

)=(

1− cos(t)sin(t)

)⇔

Z t

0

(x(p)y(p)

)d p =

Z t

0

(1− cos(p)

sin(p)

)d p

⇔(

x(p)y(p)

)t

p=0=(

p− sin(p)−cos(p)

)t

p=0

⇔(

x(t)y(t)

)=(

t− sin(t)1− cos(t)

)on the LHS we used the I.C. x(0) = y(0) = 0

The solution is the well known parameterization of the cycloid.

2 question on cardioidLet C ⊂ R2 be the cardioid curve obtained by rolling a circle of radius 1 around the fixed circle x2 + y2 = 1 of radius 1and following the initial point (i)of contact. The parameterization is

c : [0,2π] 3 t 7→ c(t) =(

2cos(t)− cos(2t)2sin(t)− sin(2t)

)∈ R2

1. Sketch the initial configuration. Sketch the cardioid (heart shaped) curve.

2. Explain how the parameterization is obtained from the physical description (rolling circle) above.

3. Compute the velocity and acceleration vectors and the speed scalar.

4. Compute the arc length function s(t),0≤ t ≤ 2π and the total arc length L.

5. Compute the Seret-Frenet frame.

6. Compute the curvature function. Prove that κ(t) = 3/[8sin(t/2) ].

7. Parameterize the involute curve (i.e. compute i(t)).

8. Parameterize the evolute curve (i.e. compute e(t)).

9. Prove that the involute is another cardioid.

10. Prove that the evolute is also a cardioid.

11. Sketch the cardioid, its ivolute and its evolute together on one diagram

2

2.1 answer

Figure 1: (i) initial configuration of wheel rolling on wheel , (ii) cardioid involute and evolute

See figure(??).

2.2 answerThe argument is similar to that for the nephroid which is given in full, see below. Since both circles have the same radiusone might assume that a spoke on the rolling circle turns with angular speed 1. That is wrong! The spoke turns withangular speed 2. See the detailed treatment of the nephroid below.

2.3 answervelocity

c(t)

=(

x(t)y(t)

)=

(−2sin(t)+2sin(2t)2cos(t)−2cos(2t)

)= 4sin(t/2)

(cos(3t/2)sin(3t/2)

)acceleration

c(t) =(

x(t)y(t)

)=(−2cos(t)+4cos(2t)−2sin(t)+4sin(2t)

)speed scalar

|| c(t) ||=

√x2 + y2

=√

(4sin2(t/2))(cos2(3t/2)+ sin2(3t/2))

=√

(4sin2(t/2))(1)

= 4sin(t/2)

3

Caution:- in the current parametrization domain 0 ≤ t ≤ 2π the speed is always positive (as it should be). In a largerparametrization domain we might have to put c(t) = |sin(t/2) |.

2.4 answerarc-length and total arc length

s(t)

=Z t

0

∣∣∣∣c′(p)∣∣∣∣ d p,

=Z t

04sin(p/2)d p

= −8cos(p2)∣∣∣t0

= −8cos(t/2)+8cos(0)= 8[1− cos(t/2)]

The total arc length isL = s(2π) = 8[1− cos(π)] = 8[1− (−1)] = 16

Warning:- the arc length must be ’adjusted’ for parameterization of the involute, see below.

2.5 answerSeret-Frenet Frame

t(t)

=1

|| c(t) ||c(t)

=1

4sin(t/2)4sin(t/2)

(cos(3t/2)sin(3t/2)

)=

(cos(3t/2)sin(3t/2)

)

n(t)

=(

0 −11 0

)t(t)c(t)

=(

0 −11 0

)t(t)(

cos(3t/2)sin(3t/2)

)=

(−sin(3t/2)

cos(3t/2)

)The Seret-Frenet Frame is

{t,n} ={(

cos(3t/2)sin(3t/2)

),

(−sin(3t/2)

cos(3t/2)

)}

2.6 answercurvature

κ(t)

=xy − yx(x2, y2)3/2

∣∣∣∣t

4

= (−2sin(t)+2sin(2t))(−2sin(t)+4sin(2t))−

(2cos(t)−2cos(2t))(−2cos(t)+4cos(2t))

(4sin(t/2))3

=

4

(−sin(t)+ sin(2t))(−sin(t)+2sin(2t))−

(cos(t)− cos(2t))(−cos(t)+2cos(2t))

64sin3(t/2)

=

4

sin2 t−3sin t sin(2t)+2sin2(2t)−

−cos2 t +3cos t cos(2t)−2cos2(2t)

64sin3(t/2)

=

4

sin2 t + cos2 t

−3(cos(2t)cos t + sin(2t)sin t)

+2sin2(2t)+2cos2(2t)

64sin3(t/2)

=

4

1−

3cos(2t− t)+

2

64sin3(t/2)

=4 ·3(1− cos(t))

64sin3(t/2)

=12(2sin2(t/2))

64sin3(t/2)

=24sin2(t/2)64sin3(t/2)

=3

8sin(t/2)

2.7 answer

Above we saw that || c(t) ||= 4 |sin(t/2) | and that (t) =Z t

0|| c(p) || d p = 8(1− cos(t/2)).

5

However use of these formulae in construction of i(t) leads to a bizarre and uninteresting involute curve. As in the caseof the cycloid we use s(t) =

R tπ|| c(p) || d p =−8cos(t/2)). In effect we measure arc length from the mid (not the start)

point of the cycloid.

i(t)= c(t)− s(t)t(t)

=(


)− (−8cos(t/2)

(cos(3t/2)sin(3t/2)

)=

(2cos(t)− cos(2t)2sin(t)− sin(2t)

)+ 4

(2cos(3t/2)cos(t/2)2sin(3t/2)cos(t/2)

)=


)+ 4

(cos(2t)+ cos(t)sin(2t)+ sin(t)

)= 3

(2cos(t)+ cos(2t)2sin(t)+ sin(2t)

)Comparing i(t) to the original c(t) we see that the involute is some kind(?) of cardioid scaled up by a factor 3. In factwe can manipulate the details and get

i(t)

= −3(

2cos(t−π)− cos(2(t−π))2sin(t−π)− sin(2(t−π)

)= −3c(t−π)

Thus the involute is the original cardioid but

• scaled by 3

• and reflected in the origin (to explain the ’−’)

• and time delayed by π

2.8 answerevolute

e(t)

= c(t)+n(t)κ(t)

=(


)+

13/8sin(t/2)

(−sin(3t/2)

cos(3t/2)

)=


)+

8sin(t/2)3

(−sin(3t/2)

cos(3t/2)

)=


)+

43

(−2sin(3t/2)sin(t/2)

2cos(3t/2)sin(t/2)

)=


)+

43

(cos(2t)− cos(t)sin(2t)− sin(t)

)=

13

(6cos(t)−3cos(2t)6sin(t)−3sin(2t)

)+

13

(4cos(2t)−4cos(t)4sin(2t)−4sin(t)

)=

13

(2cos(t)+ cos(2t)2sin(t)+ sin(2t)

)We see that e(t) = i(t)/9 = −1/3c(t − π). The description of the evolute is similar to that of the involute above butscaled,not by 3, but by 1/3.

6

2.9 answerSketch the involute,evolute and cardoid on one diagram.See figure(??)

3 question on nephroidLet C ⊂ R2 be the nephroid (kidney shaped) curve obtained by rolling a circle of radius 1 around (outside) the fixedcircle x2 + y2 = 4 of radius 2 and following the initial point (2i)of contact. The parameterization is

c : [0,2π] 3 t 7→ c(t) =(


)∈ R2

1. Sketch the circles x2 +y2 = 4,x2 +y2 = 16 and (x−3)2 +y2 = 1 and the nephroid curve together on one diagram.

2. Explain how the parameterization is obtained from the physical description (rolling circle) above.

3. Compute the velocity and acceleration vectors and the speed scalar.

4. Compute the arc length function s(t),0≤ t ≤ 2π and the total arc length L.

5. Compute the Seret-Frenet frame.

6. Compute the curvature function. Prove that κ(t) = 1/[3 |sin(t) | ].

7. Parameterize the involute curve (i.e. compute i(t)).

8. Parameterize the evolute curve (i.e. compute e(t)).

9. Prove that the involute is another nephroid.

10. Prove that the evolute is also a nephroid.

11. Sketch the nephroid, its involute and its evolute together in one diagram.

3.1 answer

Figure 2: (i) initial configuration of wheel rolling on wheel , (ii) nephroid involute and evolute

See figure(??).

7

3.2 answerThe center of the rolling circle travels on x2 + y2 = 9, the circle of radius 3 centered at the origin. Initially this center isat position 3i and at time t its position is

a(t) = S(t)(3i) =(

3cos t3sin t

).

One imagines a spoke on the rolling wheel extending from the center to the initial point of contact. Initially this spokeis represented by the vector b(0) =−i. The position of the initial point of contact at time t = 0 is c(0) = a(0)+b(0) =3i− i = 2i.

The small wheel rolls ACW, the crucial question is ’with what angular speed?’. Since the ratio of radii is 2:1 one mightthink that this angular speed is 2. But that is wrong! The angular speed is in fact 3, see below for proof. Thus theposition of the initial point of contact at time t is (center + spoke)

c(t) = a(t)+b(t) = S(t)(3i)+S(3t)(−i) =(

3cos t− cos 3t3sin t− sin 3t

)We now give two arguments each of which justifies the assertion that the small wheel turns with angular speed 3.

• If the small wheel were fixed to the larger at its initial point of contact and both wheels rotated as a unit thenthe angular velocity of the spoke would be 1. If instead the small wheel rolls on the larger wheel it turns withadditional angular speed 2. The total angular speed of the spoke is 1+2 = 3.

• Make a diagram showing the small wheel in two positions, with center both at a(0) and at a(π/2). Draw in thespoke at position b(0) = −i and position b(π/2) = j = S(3π/2)b(0). This shows that the spoke is turning at 3times the angular speed of the center of the rolling circle.

3.3 answervelocity

c(t)

=(

x(t)y(t)

)=

ddt


)= 3

(−sin(t)+ sin(3t)

cos(t)− cos(3t)

)use the trig formulae

sin A− sin B = 2cos(A+B

2)sin(

A−B2

)

cos A− cos B =−2sin(A+B

2)sin(

A−B2

)

= 6(

cos(2t)sin(t)sin(2t)sin(t)

)= 6sin(t)

(cos(2t)sin(2t)

)The acceleration

c(t) = 6cos(t)(

cos(2t)sin(2t)

)+ 12sin(t)

(−sin(2t)

cos(2t)

)There is no point in trying to simplify the acceleration as it will here be used only to compute a determinant en route tothe curvature, see below.

8

The speed scalar

|| c(t) ||=

√x2 + y2

=∣∣∣∣6sin(t)

√cos2(2t)+ sin2(2t)

∣∣∣∣= |6sin(t) ·1 |= 6 |sin(t) |

Caution:- speed is always positive so be careful since the norm of velocity always involves a square root.

3.4 answertotal arc length is

L

=Z 2π

0

∣∣∣∣c′(p)∣∣∣∣ d p,

=Z 2/pi

06 |sin(p) | d p

note that |sin |=±sinaccording to the value ofp.

=Z

π

06sin(p)d p +

Z 2π

π

−6sin(p)d p

= −6cos(p)|π0 + 6cos(p)|2/piπ

= −6[−1−1] + 6[1− (−1)]= 24

The following arclength function works only in the range 0≤ t ≤ π

s(t) =Z t

06sin(p)d p = 6cos(t)−6

3.5 answerSeret-Frenet Frame

t(t)

=1

|| c(t) ||c(t)

=sin t|sin t |

(cos(2t)sin(2t)

)The fraction introduces a factor ±1, 2nπ < t < (2n+1)π or (2n+1)π < t < (2n+2)π

n(t)

=(

0 −11 0

)t(t)

=(

0 −11 0

)sin t|sin t |

(cos(2t)sin(2t)

)=

sin t|sin t |

(−sin(2t)

cos(2t)

)The Seret-Frenet Frame is

{t,n} =sin t|sin t |

{(cos(2t)sin(2t)

),

(−sin(2t)

cos(2t)

)}

9

3.6 answerRecall the formula for curvature

κ(t)

=det(c, c)

< c, c >3/2

=

det(

x xy y

)< c, c >3/2

=

det(

6sin(t)cos(2t) 6cos(t)cos(2t)−12sin(t)sin(2t)6sin(t)sin(2t) 6cos(t)sin(2t)+12sin(t)cos(2t)

)|63 sin(t) |3

=

det(

6sin(t)cos(2t) −12sin(t)sin(2t)6sin(t)sin(2t) 12sin(t)cos(2t)

)63 |sin(t) |3

by an E.C.Operation, col2 -> col2 - cot(t)·col1

=

(6sin(t)) · (12sin(t))det(

cos(2t) −sin(2t)sin(2t) cos(2t)

)63 |sin(t) |3

=(6sin(t)) · (12sin(t)) ·1

63 |sin(t) |3

=1

3 |sin(t) |

3.7 answerBefore we compute the parameterization of the involute let us rearrange (for use below) the parameterization of thenephroid using the triple angle formulae.

cos(3t) = 4cos3(t)−3cos(t)sin(3t) = 3sin(t)−4sin3(t)

An alternative parameterization of the nephroid is

c(t)

=(


)=

(3cos(t)−4cos3(t)+3cos(t)3sin(t)−3sin(t)+4sin3(t)

)

10

=(

6cos(t)−4cos3(t)4sin3(t)

)Furthermore we measure distance along the nephroid from the point c(π/2): ie we take s(t) = 6cos(t). The followingis only strictly valid in the range 0 ≤ t ≤ π but yields an interesting ’wind on-off’ sort of involute if taken in the range−∞ < t < ∞.

i(t)= c(t)− s(t)t(t)

=(


)+ 6cos(t)

(cos(2t)sin(2t)

)=

(3cos(t)− cos(3t)+6cos(t)cos(2t)3sin(t)− sin(3t)+6cos(t)sin(2t)

)next express this in terms of cos t and sin t only

=(

3cos(t)− (4cos3(t)−3cos(t))+6cos(t)(2cos2(t)−1)3sin(t)− (3sin(t)−4sin3(t))+6cos(t)2sin(t)cos(t)

)=

(6cos(t)−4cos3(t)+(12cos3(t)−6cos(t)

4sin3(t)+12sin(t)cos2(t)

)=

(8cos3(t)

4sin3(t)+12sin(t)(1− sin2(t))

)=

(8cos3(t)

−8sin3(t)+12sin(t)

)Comparison of i(t) with the alternate formula for c(t) suggests that the involute is some sort of another nephroid. Wemake this remark precise as follows

i(t)

= 2(

4cos3(t)6sin3(t)−4sin3(t)

)= 2

(4sin3(t−π/2)

6cos3(t−π/2)−4cos3(t−π/2)

)= 2

(0 −11 0

)(6cos3(t−π/2)−4cos3(t−π/2)

−4sin3(t−π/2)

)= 2

(0 −11 0

)(6cos3(−(t−π/2))−4cos3(−(t−π/2))

4sin3(−(t−π/2))

)= 2S(π/2)c(−(t−π/2))

We havei(t) = 2S

(π

2

)c(−(

t− π

2

))We interpret this as follows the involute is the original nephroid but

• scaled by 2.

• rotated by π/2

• traversed (parameterized) in the opposite sense and out of phase by π/2.

3.8 answerevolute

e(t)

= c(t)+n(t)κ(t)

11

=(


)+ (3 |sin(t) |) sin t

|sin t |

(−sin(2t)

cos(2t)

)=


)+ 3sin(t)

(−sin(2t)

cos(2t)

)=


)+ 3sin(t)

(−sin(2t)

cos(2t)

)=

(3cos(t)− cos(3t)−3sin(t)sin(2t)3sin(t)− sin(3t)+3sin(t)cos(2t)

)Observe that the first coord is an even and the second an odd function of t. It seems sensible to reduce these coords tobe in terms of cos and sin respectively. We will make use of the triple angle formulae


e(t)

=(

3cos(t)− [4cos3(t)−3cos(t)]−3sin(t)[2sin(t)cos(t)]3sin(t)− [3sin(t)−4sin3(t)]+3sin(t)[1−2sin2(t)]

)=

(6cos(t)−4cos3(t)−6sin2(t)cos(t)

3sin(t)−2sin3(t)

)=

(6cos(t)−4cos3(t)−6(1− cos2(t))cos(t)

3sin(t)−2sin3(t)

)=

(2cos3(t)

3sin(t)−2sin3(t)

)Comparison with the alternate nephroid parameterisation (see above)

c(t) =(

6cos(t)−4cos3(t)4sin3(t)

)suggests that the evolute is itself a nephroid. The clinching observation is

2e(t)

=(

4cos3(t)6sin(t)−4sin3(t)

)=

(4sin3(t−π/2)

6cos(t−π/2)−4cos3(t−π/2)

)=

(−4sin3(−(t−π/2))

6cos(−(t−π/2))−4cos3(−(t−π/2))

)=

(0 −11 0

)(6cos(−(t−π/2))−4cos3(−(t−π/2))

4sin3(−(t−π/2))

)= S(π/2)c(−(t−π/2))

Thuse(t) =

12

S(π

2)c(−(

t− π

2

))We interpret this as follows the evolute is the original nephroid but

• scaled by 1/2.

• rotated by π/2

• traversed (parameterized) in the opposite sense and out of phase by π/2.

12

3.9 answerSketch the involute,evolute and cardoid on one diagram.See figure(??)

3.10 footnote

cos(3t)+ isin(3t)= exp(3it)= exp(it)3

= [ cos(t)+ isin(t) ]3

= cos3(t)+3icos2(t)sin(t)+3(i)2 cos(t)sin2(t)+(i)3 sin3(t)= cos3(t)+3icos2(t)sin(t)−3cos(t)sin2(t)− isin3(t)= [cos3(t)−3cos(t)sin2(t)] + i[3cos2(t)sin(t)− sin3(t)]= [cos3(t)−3cos(t)(1− cos2(t))] + i[3(1− sin2(t))sin(t)− sin3(t)]= [4cos3(t)−3cos(t)] + i[3sin(t)−4sin3(t)]

Taking real and imaginary parts


4 question on nephroid causticLet

c1(t) = 4(

cos(t)sin(t)

), c2(t) = 4

(cos(3t−π)sin(3t−π)

)and c(t) =

14[3c1(t)+ c2(t)]

1. Prove that the line segment l = c1(t),c2(t) meets the nephroid Γ (see q03) tangentially at the point c(t) ∈ Γ.

2. Prove that the angle between the horizontal i and the radial r = c1(t) equals the angle between the radial r and theline segment l.

3. Prove that the rays of a horizontal beam of light reflect off the semicircular mirror{

x∣∣ x2 + y2 = 4 , x≥ 0

}to

form a nephroid caustic.

4. See the nephroid formed by a bright light in your coffee cup. See figure ??.

4.1 answerBy definition

c(t) =34

c1(t)+14

c2(t)

and therefore is a point on the line segment joining the point c1(t) to the point c2(t), in fact it is the point dividing theline segment in parts with lengths in the ratio 3 : 1.

Since

c(t)

=34

c1(t)+14

c2(t)

=34·4(

cos(t)sin(t)

)+

14·4(

cos(3t−π)sin(3t−π)

)=

(3cos(t)+ cos(3t−π)3sin(t)+ sin(3t−π)

)

13

Figure 3: envelop of reflected horizontal rays is a nephroid

=(


)Thus c(t) is a point on the nephroid Γ (see the standard parameterization of the nephroid). Thus indeed the line segmentc1(t),c2(t) meets the nephroid, i.e. at the common point c(t).

Next we will show that the tangent line at the point c(t) on the nephroid is parrallel to the line segment. This proves that(as required) the line segment meets the nephroid tangentially at the the point c(t). The tangent vector has direction thatof

c′(t) =(−3sin(t)+3sin(3t)3cos(t)−3cos(3t)

)while the line segment has direction that of

c1(t)− c2(t) =(

4cos(t)−4cos(3t−π)4sin(t)−4sin(3t−π)

)=(

cos(t)+ cos(3t)sin(t)+ sin(3t)

)These are parallel iff

cos(t)− cos(3t)−sin(t)+ sin(3t)

=sin(t)+ sin(3t)cos(t)+ cos(3t)

iff (cos(t)− cos(3t)

)·(

cos(t)+ cos(3t))

=(

sin(t)+ sin(3t))·(− sin(t)+ sin(3t)

)iff

cos2(t)− cos2(3t) = −sin2(t)+ sin2(3t)

iffcos2(t)+ sin2(t) = cos2(3t)+ sin2(3t)

iff1 = 1

which is true.

14

4.2 equiangles. The angle θ between the horizontal and the radial vector from 0 to the point c1(t) is of course θ = t. The angle φ

between the radial and the line segment is given by

cos(φ)

=< c1(t),c1(t)− c2(t) >

||c1(t) || · ||( ||c1(t))− c2(t)

= <

(4cos(t)4sin(t)

),

(4cos(t)−4cos(3t−π)4sin(t)−4sin(3t−π)

)>

/4 ·∣∣∣∣∣∣∣∣( 4cos(t)−4cos(3t−π)

4sin(t)−4sin(3t−π)

)∣∣∣∣∣∣∣∣= <

(cos(t)sin(t)

),

(cos(t)+ cos(3t)sin(t)+ sin(3t)

)>

/∣∣∣∣∣∣∣∣( cos(t)+ cos(3t)sin(t)+ sin(3t)

)∣∣∣∣∣∣∣∣=

1+ cos(3t)cos(t)+ sin(3t)sin(t)√2+2cos(3t)cos(t)+2sin(3t)sin(t)

=1+ cos(3t− t)√2+2cos(3t− t)

=1+ cos(2t)√2+2cos(2t

=1√2

√1+ cos2t

=1√2

√1+2cos2 t−1

= cos(t)

Thus θ = φ = t.

4.3 answer, nephroid from circular mirrorSee figure ??.

Figure 4: envelop of reflected horizontal rays is a nephroid

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5 question and answer, conical helixConical Helix with Darboux vector

Let Γ be the conical helix curve with parametrization c(t) = x(t)i+ y(t)j+ z(t)k = t cos(t)i+ t sin(t)j+ tk.

1. Sketch Γ = K∩S⊂ R3 showing the curve as the intersection of cone and screw surfaces.

2. Compute t(t), n(t), b(t).[Hint:- The computation is brutal here but: if you write v(t) = cos(t)i + sin(t) and w(t) = v(t) and note that{v,w,k} is a P.O.N. triple (with easily computed dot and cross products), then life becomes a lot easier. Forexample c′ is simply v+ tw+k.]

3. Prove that κ(t) =(t4 +5t2 +8)1/2

(t2 +2)3/2 , τ(t) =t2 +6

t4 +5t2 +8.

4. Let d : R 3 t 7→ d(t) ∈ R3 be the Darboux vector at the point c(t) ∈ Γ; i.e. d(t) is the eigenvector (associatedwith eigenvalue 0) of the curvature matrix, i.e. the angular velocity of the S.F. frame.Express d(t) in terms of the S.F. basis (t,n,b).

5. Express d(t) in terms of the standard basis (i, j,k).

6. Compute e(t) = Sk(−t)d(t). [ Sk, of course, denotes standard rotation about the z-axis].

7. Compute | |d(t) | |= | |e(t) | |, the angular speed of the S.F. frame. Prove that limt→∞| |d(t) | | = 0.

8. Prove that limt→∞

d(t)| |d(t) | |

= k.

9. Explain, in your own words, what all this means.

5.1 answer

Figure 5: (i) conical helix curve , (ii) intersection of cone and screw surfaces

See figures ?? and ??.

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5.2 answerFirst write, for convenience,

v(t) = cos ti+ sin tj and w(t) =−sin ti+ cos tj

Note that (v,w,k) is a positively oriented orthonormal triple: all inner products and all cross products amongst thesethree vectors are known (trivial). Also v′ = w and w′ =−v

c′

= (tv+ tk)′

= (v+ tv′+k)= v(t)+ tw+k

c′′

= (v+ tw+k)′

= v′+w+ tw′+k′

= w+w− tv+0= 2w− tv

c′′′

= (c′′)′

= (2w− tv)′

= 2w′−v− tv′

= −2v−v− t(w)= −3v− tw

We compute s′(t)

s′(t)

= < c′,c′ >1/2

= < v(t)+ tw(t)+k,v(t)+ tw(t)+k >1/2

=√

1+ t2 +1+0+0+0using orthonormality 6 times

=√

2+ t2

5.3 answer, S.F. FrameTo compute t,n,b we first need the following quantities < c′,c′ >,< c′,c′′ >,c′ × c′′,< c′ × c′′,c′ × c′′ > and <c′c′′,c′′′ >.

< c′,c′ >= < v(t)+ tw(t)+k,v(t)+ tw(t)+k >

= 2+ t2

< c′,c′′ >= < v(t)+ tw(t)+k,−tv(t)+2w >

= t

17

c′× c′′

= (v+ tw+k)× (2w− tv)= 2v×w− tv×v+2tw×w− t2w×v+2k×w− tk×v= 2k− t0+2t0+ t2k−2v− tw= 2k+ t2k−2v− tw= −2v− tw+(2+ t2)k

< c′× c′′,c′× c′′ >= <−2v− tw+(2+ t2)k,−2v− tw+(2+ t2)k >

= 4+ t2 +(2+ t2)2

= 8+5t2 + t4

< c′× c′′,c′′′ >= <−2v− tw+(2+ t2)k,−3v− tw >

= t2 +6

Now we are ready to compute t,n,b,κ,τ.

t

=c′

||c′ ||

=v+ tw+k√

2+ t2

=1√

2+ t2

cos t− t sin tsin t + t cos t

1

see expressions for v,w

The unit binormal vector is

b

=c′× c′′

||c′× c′′ ||

=−2v− tw+(t2 +2)k√

8+5t2 + t4

=1√

8+5t2 + t4

−2cos t + t sin t−2sin t− t cos t

t2 +2

see expressions for v,w

The unit normal is

n= b× t

=(−2v− tw+(t2 +2)k)× (v+ tw+ k)√

8+5t2 + t4 ·√

t2 +2

18

=−2v×v−2tv×w−2v×k− tw×v− t2w×w− tw×k+(t2 +2)k×v+(t2 +2)tk×w+(t2 +2)k×k√

8+5t2 + t4 ·√

t2 +2

=−20−2tk+2w+ tk− t20− tv+(t2 +2)w− (t2 +2)tv+(t2 +2)0√

8+5t2 + t4 ·√

t2 +2

=(−t3−3t)v + (t2 +4)w − tk√

8+5t2 + t4 ·√

t2 +2

=1√

8+5t2 + t4 ·√

t2 +2

(−t3−3t)cos t − (t2 +4)sin t(−t3−3t)sin t + (t2 +4)cos t

−t

Finally the S.F. frame is{

t,n,b}

= 1√2+ t2

cos t− t sin tsin t + t cos t

1

,1√

8+5t2 + t4 ·√

t2 +2

(−t3−3t)cos t − (t2 +4)sin t(−t3−3t)sin t + (t2 +4)cos t

−t

,1√

8+5t2 + t4


t2 +2

5.4 answer, curvature and torsion

κ =< c′× c′′,c′× c′′ >1/2

< c′,c′ >3/2=

(t4 +5t2 +8)1/2

(t2 +2)3/2

τ =< c′× c′′,c′′′ >

< c′× c′′,c′× c′′ >=

t2 +6t4 +5t2 +8

5.5 answer, Darboux vct w.r.t S.F.basis

d(t) is (by definition) the angular velocity converted from matrix

0 −κ 0κ 0 −τ

0 τ 0

form to vector form and is

d(t) =

τ(t)0

κ(t)

where κ and τ are as above. But this column vector is w.r.t. to the S.F frame. In other words

d(t) = τ(t)t(t) + κ(t)b(t)

5.6 answer, Darboux vct w.r.t. standard basis

d(t)

= τ(t)t(t) + κ(t)b(t)

=−6+ t2

8−3t2 + t41

8−3t2 + t4

2cos t− t sin t2sin t + t cos t

1

+1√

8+5t2 + t4


t2 +2

Note the following which we will use below

d(t)

=−6+ t2

8−3t2 + t41

8−3t2 + t4 Sk(t)

2t1

+1√

8+5t2 + t4Sk(t)

−2−t

t2 +2

= Sk(t)

−6+ t2

8−3t2 + t41

8−3t2 + t4

2t1

+1√

8+5t2 + t4

−2−t

t2 +2

19

5.7 answer, computing e(t)From the last remark, since Sk(−t)◦Sk(t) = Sk(0) = I .

e(t) =−6+ t2

8−3t2 + t41

8−3t2 + t4

2t1

+1√

8+5t2 + t4

−2−t

t2 +2

5.8 answer, ||d(t) ||Note that

||d(t) || = ||Sk(t)e(t) || = ||e(t) ||We can compute ||d(t) || with a hack slog from the preceeding explicit equation for e. But a clever way is to use Pythagoras theoremon the expression

d = τt+κb

So that

||d ||

=√

τ2 +κ2

=

√(t2 +6)2

(t4 +5t2 +8)2 +(t4 +5t2 +8)

(t2 +2)3

It is clear without working details that ||d(t) || → 0 like 1/t, ie that t ||d(t) || → 1 as t→ ∞

5.9 answer, computing limiting direction of Darboux vctIt is trivial to check that

limt→∞

tκ(t) = 1 , limt→∞

tτ(t) = 0

Thus

limt→∞

d(t)||d(t) ||

=

limt→∞

td(t)t ||d(t) ||

=limt→∞ td(t)

limt→∞ t ||d(t) ||

=limt→∞ td(t)

1

=

limt→∞

td(t)

=

limt→∞

(tτ(t)t(t) + tκ(t)b(t)

)=

limt→∞

(tκ(t)b(t)

)since lim tκ(t) = 0 and t is a unit vector

=

20

limt→∞

t(t4 +5t2 +8)1/2

(t2 +2)3/21√

8+5t2 + t4


t2 +2

=

limt→∞

t(t2 +2)3/2


t2 +2

= 0

01

=

k

The last step follows by taking the (straightforward) limit of each of 3 vector components; in particular

limt→∞

z(t) = limt→∞

t(t2 +2)(t2 +2)3/2

= 1

5.10 answer, explainThe speed with which the SF frame spins, i.e. ||d || tends to zero over time but the direction of spin d/ ||d || settles down to the verticalk, compare with the situation for the standard helix where d is constant at k.

The vector e could have been put to use in the above computations but proved unnecessary in the method used here.

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