AMATH 231 Solution ASSIGNMENT # 3: Arclength and Curvature Fall 2014

1. Solution: [2 marks]

We compute the derivative and its length,

~g(t) = et(cos t sin t, sin t+ cos t, 1), and ~g(t) =

3et.

Then we evaluate the integral,

s =

t0

3et dt =

3[et]30

=

3(e3 1).

2. Solution: [3 marks]

We compute the derivative,

~g() = (f () cos f() sin , f () sin + f() cos ),

now its length,

g() =

(f () cos f() sin )2 + (f () sin + f() cos )2,

=

(f ())2 + f 2().

Therefore, the arc length is,

s =

21

f 2() +

(d

df()

)2d.

3. Solution: [4 marks + 2 bonus]

In a span of 25 minutes the LP turns 33 25 = 825 times. If the spacing betweengroove is constant then it is reasonable to assume that the radial distance dependslinearly on the angle, i.e.

r = a, for 1 2,for constants a, 1, 2 that must be determined.

Since the record turns 825 times in total and each turn is 2pi then the difference inphase must be

2 1 = 825 2pi.Furthermore, we must have

6 = a1, and 14.5 = a2.

We can substitute the second two equations into the first equation multiplied by a,

a =14.5 6825 2pi 0.00164.

This then implies that 1 = 3658.5 and 2 = 8841.5. Therefore, our formula becomes,

~g() = 0.00164(cos , sin ).

To compute the arc length we must first compute the derivative and its length,

~g() = 0.00164(cos sin , sin + cos ),and

~g() = 0.00164

1 + 2.

The arc length is then set up,

s = 0.00164

8841.53658.5

1 + 2 d.

Bonus: To evaluate this we need to use a substitution = sinh and d = coshd,

s = 0.00164

9.78048.8980

1 + sinh2 cosh d,

= 0.00164

9.78048.8980

cosh2 d,

= 0.00164

9.78048.8980

1

2[1 + cosh 2] d,

= 0.00082

[+

1

2sinh 2

]9.78048.8980

53035.

4. Solution: [2 + 6 + 3 marks]

a) Because the unit tangent vector has a unit length, it is orthogonal to its derivativeand therefore the unit normal vector. Also, the cross product of two vectors isnecessarily orthogonal to those vectors.

b) As computed in class, the unit tangent vector is

~T (t) =~g(t)~g(t) =

(a sin t, a cos t, b)a2 + b2

=1

a2 + b2(a sin t, a cos t, b).

The unit normal vector is similarly computed,

~N(t) =~T (t)~T (t) = ( cos t, sin t, 0)

The unit bi-normal is,

~B =1

a2 + b2(b sin t,b cos t, a).

The curvature is

(s) =

~T (t)~g(t) =

(a cos t,a sin t, 0)a2 + b2

=a

a2 + b2.

To compute the torsion we first need the following vector,

~g(t) ~g(t) = (a sin t, a cos t, b) (a cos t,a sin t, 0),= (ab sin t,ab cos t, a2),

and ~g(t) ~g(t)2 = a2(a2 + b2). Therefore, the formula yields,

=(~g(t) ~g(t)) ~g(t)~g(t) ~g(t)2 ,

=(ab sin t,ab cos t, a2) (a sin t,a cos t, 0)

a2(a2 + b2),

=a2b

a2(a2 + b2)=

b

a2 + b2.

c) The identities can be checked one at a time:

d~T

dt=ds

dt

[ ~N],

aa2 + b2

( cos t, sin t, 0) = aa2 + b2

( cos t, sin t, 0),

Figure 1: Field lines for F(x, y) = (x, 1).

d ~N

dt=ds

dt

[~T + ~B

],

(sin t, cos t, 0) =[ aa2 + b2

(a sin t, a cos t, b) + ba2 + b2

(b sin t,b cos t, a)],

(sin t, cos t, 0) = ( sin t, cos t, 0).

d ~B

dt=ds

dt

[ ~N

],

ba2 + b2

(cos t, sin t, 0) =b

a2 + b2(cos t, sin t, 0).

5. Solution:

The field lines of F(x, y) = (x, 1) are determined as follows: setting g(t) = F(g(t))implies

g(t) = (x(t), y(t))

= F(x(t), y(t))

= (x(t), 1).

It follows thatx(t) = x(t) and y(t) = 1.

Figure 2: Vector field and field lines (in red) for F(x, y) = (x, 1).

These are two uncoupled ODEs and hence are readily solved as x(t) = c1et and y = t+c2

where c1, c2 are constants. It follows that t = ln |x/c1| for x 6= 0. This means the fieldlines of F(x, y) = (x, 1) are governed by

y = ln

xc1+ c2

which are depicted by the red coloured curves in Figure 1 for various values of c1 andc2. When x = 0, we have

x(t) = 0 and y(t) = 1

which means the yaxis is a field line as well (see Figure 1). Directions of the fieldlines are determined from F(x, y) = (x, 1).

Aside: as a point of interest, the vector field becomes stronger and stronger as we moveaway from the yaxis as indicated by the larger magnitude of the vectors in the fieldportrait (see Figure 2). This is because the magnitude of F,

||F|| =x2 + (1)2,

increases when |x| increases.