Homework 1-5 Solutions Problems 1. (a) z = 1 + 0i ,

Homework 1-5 Solutions
Problems
1.
(a) z = 1 + 0i ,
(b) z = −7 + 24i .
2. f (z) = u(x, y) + iv(x, y) with u(x, y) = e2y cos(2x) and v(x, y) = −e2y sin(2x).
(a) To show f (z) is analytic, explicitly evaluate partials,
∂v ∂u
∂v
that they satisfy C-R equations: ∂u
= ∂y
, ∂y = − ∂x
.
∂x
(b) f 0 (z) =
∂v
∂u
+ i ∂x
∂x
−2iz
(c) f (z) = e
.
∂u
,
∂x
etc. and show
= −e2y 2 sin 2x − ie2y 2 cos 2x = −2if (z).
3. To prove u = sin x cosh y + 2 cos x sinh y + x2 − y 2 + 4xy is harmonic,
just
R
∂2u
∂2u
∂u
∂v
∂u
show that ∂x2 + ∂y2 = 0. To find v, integrate ∂x = ∂y to get v = dy ∂x =
cos x sinh y − 2 sin x cosh y + 2xy + 2y 2 + φ(x). To determine the function φ(x),
∂v
plug u and v into the second C-R equation ∂u
= − ∂x
, which gives φ0 (x) = −4x
∂y
so φ(x) = −2x2 + const. Finally, a little algebra can be used to show f (z) =
u + iv = (1 − 2i) [sin z + z 2 ] + i const.
4. u = ax3 + bx2 y + cxy 2 + dy 3 .
2
2
(a) Harmonic equation ∂∂xu2 + ∂∂yu2 = (6a+2c)x+(2b+6d)y = 0. This is satisfied
if c = −3a, b = −3d.
(b) Write u = ax3 − 3dx2 y − 3axy 2 + dy 3 . Then using C-R equations as in
problem 3, you can find v = 3ax2 y − ay 3 − 3dxy 2 + dx3 + const.
(c) A little algebra gives f (z) = (a + id)z 3 + i const.
5. We want to show that static Maxwell equations in free space imply that E ∗ (z) =
Ex − iEy is analytic as a function of the two dimensions x and y with z = x + iy.
• Writing E ∗ (z) = f (z) = u + iv in the standard notation, gives u = Ex ,
v = −Ey .
• Free space means ρ = 0, J = 0, and D = E, H = B. Static means timeindependent ( ∂E
= 0, ∂B
= 0). Thus, the relevant Maxwell equations
∂t
∂t
become ∇ · E = 0 and ∇ × E = 0.
• In 2-dimensions, ∇ · E = 0 implies
implies
∂Ex
∂y
−
∂Ey
∂x
=
∂u
∂y
+
∂v
∂x
= 0.
∂Ex
∂x
y
+ ∂E
=
∂y
∂u
∂v
− ∂y
∂x
= 0 and ∇ × E = 0
∂v
∂v
= ∂y
and ∂u
= − ∂x
are the Cauchy-Riemann equa• The two equations ∂u
∂x
∂y
tions. Assuming continuity of partial derivatives, leads to analyticity of
E ∗ (z).
6. Find Cauchy-Riemann equations in polar coordinates (z = reiθ ). There are
several ways to do this. Here’s one. ∆f = ∆u + i∆v and ∆z = eiθ ∆r + reiθ i∆θ.
We can take the derivative along a path with r constant or θ constant, then
take the limit in the other variable
• If we keep θ constant, ∆θ = 0. Then
• If we keep r constant, ∆r = 0. Then
df
dz
df
dz
=
=
∂u
+ i eiθ∂v∂r .
eiθ ∂r
∂u
+ i reiθ∂vi∂θ .
reiθ i∂θ
• Multiplying through by eiθ and equating real and imaginary parts of the
= 1r ∂v
, ∂v = − 1r ∂u
.
above expressions gives ∂u
∂r
∂θ ∂r
∂θ
7. The principle branch of the complex logarithm Log(z).
(a) The cut z plane associated with Log(z). There is a Branch point singularity
at the origin. There is a branch cut along the negative real axis. The phase
is θ = π on the negative real axis and approaching it from above. The phase
is θ = −π when approaching the negative real axis from below.
(b) The fact that the derivative of analytic function f (z) can be expressed as
+ i ∂v
) follows directly from the solution to problem 6.
f 0 (z) = e−iθ ( ∂u
∂r
∂r
(c)
dLog(z)
dz
= e−iθ 1r = z1 .
8. Evaluate the following.
(a) Log(−2i)=Log(2e−iπ/2 )=log 2 − i π2 .
(b) Tan−1 (−2i)= 2i [Log(−i)−Log(3i)]= 2i [Log(e−iπ/2 )−Log(3eiπ/2 )]
= 2i [−i π2 − (log 3 + i π2 )] = π2 − 2i log 3.
(c) sin(−2i) =
ei(−2i) −e−i(−2i)
2i
=
e2 −e−2
2i
= −i sinh 2.
1
9. The integral, −1
z ∗ dz, does not make sense. Since z ∗ is not an analytic function,
this integral will depend on the path. Since no path is given, the integral is
ambiguous.
Since sin 2z is an analytic function, the other integral has one unambiguous
answer, independent of path. It is
Ri
1
1
1
i
0 sin 2z dz = − 2 cos 2z|0 = − 2 (cos 2i − 1) = 2 (1 − cosh 2).
R
10. Use Cauchy’s integral formula to do the following contour integrals:
z
e dz
(a) I = C z−πi/2
, where C = boundary of a square with sides x = ±2, y = ±2.
The square encompasses the pole at z = πi/2, so we get
I = 2πi[ez ]z=πi/2 = 2πieπi/2 = −2πi.
H
(b) I = C z2dz+2 , where C = circle of radius 1, center i. Factoring the denomH
dz √
inator of the integral, I = C (z+√2i)(z−
, reveals that the integrand has
2i)
√
√
+ 2i is inside the contour,
two poles at z = ±h 2i. iOnly the pole at z = √
so we get I = 2πi z+1√2i √ = 2πi 2√12i = π/ 2.
H
z= 2i
11. Do the following integrals:
4
+2z+1
(a) I = C z(z−z
4 dz, where C is any closed contour containing z 0 . Using the
0)
derivatives
of
Cauchy’s integral
formula gives
i
h
1 d3
4
= 2πi 3!1 4 · 3 · 2 z0 = 8πiz0 .
I = 2πi 3! dz3 (z + 2z + 1)
H
z=z0
cosh z
C z n+1 dz,
(b) I =
where C is any closed contour containing the origin. Using
the derivatives
of
Cauchy’s
integral
formula givesi
h
i
h
1 dn 1 z
1 dn
= 2πi n! dzn 2 (e + e−z )
I = 2πi n! dzn cosh z
H
= 2πi
h
1 1 z
(e
n! 2
z=0
n −z
+ (−1) e )
i
z=0
=
2πi [1+(−1)n ]
.
n!
2
z=0
12. Find the order of the pole at the origin, the residue there, and the integral
around a path C enclosing the origin, but no other singularities, for each of the
following:
(a) I like to use Taylor expansions to find these.
1 2
1 2
z +...
z +...
1− 2!
1− 2!
z
=
= z1 + . . ..
=
cot z = cos
1
1 2
3
sin z
z +...)
z− 3! z +...
z(1− 3!
The pole is order 1. The Residue at the origin is 1. The integral around
the origin is 2πi.
−z− 1 z 2 −...
−z(1+ 1 z+...)
(b) csc2 z log (1 − z) = logsin(1−z)
= (z− 1 2z3 +...)2 = z2 (1− 1 2z2 +...)2 = − z1 + . . ..
2z
3!
3!
The pole is order 1. The Residue at the origin is −1. The integral around
the origin is −2πi.
(c)
z(1− 1 z 2 +...)
z cos z
= (z− 1 z3 +...)(1−2!1 z2 + 1 z4 +...−1)
sin z(cos z−1)
3!
2!
4!
1 2
z +...)
(1− 2!
0
2!
= − z2 + z + . . ..
1 2
1 2
2! 2
− 2!
z (1− 3!
z +...)(1− 4!
z +...)
z
sin z−tan z
=
=
1 2
(1− 2!
z +...)
1 2
1 4
1 2
(1− 3! z +...)(− 2!
z + 4!
z +...)
=
The pole is order 2. The Residue at the origin is 0. The integral around
the origin is 0. (Note that this function is even in z, so there are no odd
powers in the Laurent expansion.)
13. We want to evaluate I = 0∞ x3dx
by considering the contour integral
+a3
where the contour C consists of:
R
• C1 = a ray along θ = 0 from r = 0 to r = R,
• C2 = an arc at r = R from θ = 0 to θ = 2π/3,
• C3 = a ray along θ = 2π/3 from r = R to r = 0,
taking R → ∞. The various contributions to the contour integral give
dz
C z 3 +a3 ,
H
• C1 : Let z = x, dz = dx. Then
R
• C2 : Let z = Reiθ , dz = Reiθ idθ. Then
RR
dx
0 x3 +a3 → I as R
R
dz
3
C2 z 3 +a3 ∼ R/R → 0
dz
C1 z 3 +a3
=
→ ∞.
as R → ∞.
• C3 : Let
z = xe2πi/3
, dz = dxe2πi/3 .
R
R0
dz
e2πi/3 dx
→ −e2πi/3 I, where we used e2πi = 1.
Then C3 z3 +a3 = R 2πi/3
3
x) +a3
(e
= (1 − e2πi/3 )I. We can evaluate the
Thus, the contour integral is C z3dz
+a3
contour integral using the residue theorem. There are poles at −a, aeiπ/3 and
−iπ/3
ae
. Only aeiπ/3 is inside the hcontour,
h
h
i so
i
i
H
1
1
dz
1
2πi/3
=
2πi
=
2πi
)I = 2πiRes z3 +a3
.
2
2πi/3
2
C z 3 +a3 = (1−e
πi/3
πi/3
3z
3a e
H
h
z=ae
i
z=ae
P (z)
= QP0(z(z00)) if the polynomial Q(z) has only a simple
(I used the trick, Res Q(z)
z0
zero at z = z0 .) Solving for I gives
= 3√2π3a2 .
I = 3a2 e2πi/32πi
(1−e2πi/3 )
dθ
, let z = eiθ and dz = eiθ idθ. With this substi14. To evaluate I = 02π (a+cos
θ)2
tutionq, it becomes a contour integral around the unit circle, |z| = 1. Using
cos θH= (eiθ + e−iθ )/2 = (z
+ z −1 )/2, we get
dz
4 H
4 H
z dz
z dz
I = iz(a+(z+z −1 )/2)2 = i (z2 +2az+1)
,
2 = i
(z−z+ )2 (z−z− )2
√
2
where z± = −a ± a − 1. Assuming a > 1, then only z+ is inside the unit
circle. We can now use the Residue theorem (or the derivative of Cauchy’s
integral formula).
We
h
i find
+z− )
4
d
2πa
z
I = i 2πi dz (z−z− )2
= −8π (z(z++−z
2 = (a2 −1)3/2 .
−)
R
z=z+
ax
∞
e
15. We want to evaluate I = −∞
dx 1+e
x , where 0 < a < 1, by considering the
H
eaz
contour integral C dz 1+ez along the closed contour C connecting −ρ+0i, ρ+0i,
ρ + 2πi, −ρ + 2πi, and then letting ρ → ∞.
R
• Along Rthe segment
−ρ + 0i to ρ + 0i: Let z = x, dz = dx.
Rρ
eaz
eax
Then dz 1+ez = −ρ dx 1+e
x → I as ρ → ∞.
• Along Rthe segment
ρ + 0i to ρ + 2πi: Let z = ρ + yi, dz = idy.
R 2π
eaz
ea(ρ+iy)
(a−1)ρ
→ 0 as ρ → ∞.
Then dz 1+e
=
z
0 idy 1+e(ρ+iy) ∼ e
• Along Rthe segment
ρ + 2πi to −ρ + 2πi: Let z = x + 2πi, dz = dx.
R −ρ
eaz
eax e2πai
2πai
Then dz 1+e
=
I as ρ → ∞.
z
ρ dx 1+ex e2πi → −e
• Along Rthe segment
−ρ + 2πi to −ρ + 0i: Let z = −ρ + yi, dz = idy.
R0
ea(−ρ+iy)
eaz
−aρ
→ 0 as ρ → ∞.
Then dz 1+ez = 2π idy 1+e
(−ρ+iy) ∼ e
az
e
2πai
Thus, we get C dz 1+e
)I. The contour integral has only one pole
z = (1 − e
inside the contour, at z = πi. The residue can be found by Taylor expanding
eaπi ea(z−πi)
−eaπi
eaz
≈ (z−πi)
+ . . .. Thus, we get
around the pole: 1+e
z = 1−e(z−πi)
H
az
e
2πai
)I = 2πi [−eaπi ]. Solving for I gives
C dz 1+ez = (1 − e
aπi
π
2πie
I = − 1−e
2πai = sin πa .
H
2
log (z)
x
16. We want to evaluate I = 0∞ dx xlog
2 +b2 by considering the contour integral C dz (z 2 +b2 )
around the contour shown in Fig. 6.22 in the text. The contour has four contributions:
H
R
• A straight line just above the real axis, with phase θ = 0. Let z = x,
2
R
R∞
R
log2 (x)
log2 (x)
dz = dx. Then dz (zlog2 +b(z)2 ) = ρR dx (x
2 +b2 ) → 0 dx (x2 +b2 ) as ρ → 0 and
R → ∞.
• A large circle at radius R: Let z = Reiθ , dz = Reiθ idθ.
2
R
2
R
→ 0 as R → ∞.
Then dz (zlog2 +b(z)2 ) ∼ R log
R2
• A straight line just below the real axis, with phase θ = 2π. Let z = xe2πi ,
2
R
R
R
2
2
dz = dxe2πi . Then dz (zlog2 +b(z)2 ) = Rρ dx (log(x)+2πi)
→ − 0∞ dx (log(x)+2πi)
(x2 +b2 )
(x2 +b2 )
as ρ → 0 and R → ∞.
• A small circle at radius ρ: Let z = ρeiθ , dz = ρeiθ idθ.
2
R
2
ρ
→ 0 as ρ → 0.
Then dz (zlog2 +b(z)2 ) ∼ ρ log
b2
Combining these contributions, we obtain
R∞
R
R∞
H
2
log2 (x)−(log(x)+2πi)2
4πi log(x)
log2 (z)
= 0∞ dx (x(2π)
2 +b2 ) − 0 dx (x2 +b2 ) .
C dz (z 2 +b2 ) = 0 dx
(x2 +b2 )
Solving nfor the integral I that we want,
o we get
H
R∞
log2 (z)
1
2π 2
I = 4πi −∞ dx (x2 +b2 ) − C dz (z2 +b2 ) ,
where I’ve used the symmetry of the first integral to divide by 2 and change
the lower limit on the integration to −∞. The poles are at z = ±ib. The
first integral can be done by completing in the upper half-plane (hence only
picking up the residue at z = +ib), while the second integral uses the contour
of Fig. 6.22 and
½ picks up the residue at both poles. We get
¾
I=
1
(2πi)
4πi
n
2
Res
h
2π 2
(z 2 +b2 )
2
log2 (z)
−
2 +b2 )
(z
z=ib
z=ib
o
2
(log(b)+3iπ/2)
π
= 2b
log b.
−2ib
i
− Res
h
i
Res
h
log2 (z)
(z 2 +b2 ) z=−ib
i
− (log(b)+iπ/2)
−
= 12 2π
2ib
2ib
(A comment about phases: The choice of branch cut requires that θ range
between 0 and 2π. Thus, +ib = beπi/2 and −ib = be3πi/2 .)
∞
17. We want to evaluate the integral I = P −∞
dx cos(ax)−cos(bx)
by examining the
x2
H
eiaz −eibz
along the real axis and on a large semi-circle in the
contour integral C dz z2
upper half plane (radius R), with the origin excluded using a small semi-circle
(radius ρ). Assume that both a and b are positive.
R
• The contribution from the large semi-circle goes to zero as R → ∞.
• The contributionR of the segments along the real axis give the principle
iax
ibx
∞
as R → ∞ and ρ → 0.
dx e x−e
value integral P −∞
2
• The contribution of the small semi-circle gives −πiRes
0.
h
eiaz −eibz
z2
z=0
i
as ρ →
(I have left out details here, because the arguments are exactly as we did in
class.) Thus, we get
h iaz ibz i
R∞
H
iax
ibx
e −e
eiaz −eibz
dz
P −∞
dx e x−e
=
+
πiRes
.
2
C
z2
z2
z=0
The only singularities Hof the integrand are at z = 0, which is excluded from
iaz
ibz
= 0. The residue at zero can be found by
the contour C. Thus, C dz e z−e
2
1+iaz+...−(1+ibz+...)
eiaz −eibz
Taylor expansion:
=
= i(a−b)
+ . . .. Thus, we get
z2
z2
z
h
i
R∞
eiax −eibx
eiaz −eibz
P −∞ dx x2
= πiRes
= πi[i(a − b)] = π(b − a). Taking the
z2
z=0
real part of both sides of the equation gives I = π(b − a).
2a−1
18. We want to evaluate I = 0∞ dx bx2 +x2 with 0 < a < 1. Consider the contour
H
2a−1
integral C dz bz2 +z2 , where the contour C is the same as in problem 16. There
are four contributions:
R
• A straight line Rjust above the
real axis, with phase θ = 0. Let z = x,
RR
2a−1
z 2a−1
dz = dx. Then dz b2 +z2 = ρ dx bx2 +x2 → I as ρ → 0 and R → ∞.
• A large
circle at radius R: Let z = Reiθ , dz = Reiθ idθ.
R
2a−1
2a
Then dz bz2 +z2 ∼ RR2 → 0 as R → ∞.
• A straight line just below the real axis, with phase θ = 2π. Let z = xe2πi ,
R
R
2πi )2a−1
2a−1
→ −e2πi(2a−1) I as ρ → 0
dz = dxe2πi . Then dz bz2 +z2 = Rρ dx (xeb2 +x
2
and R → ∞.
• A small
circle at radius ρ: Let z = ρeiθ , dz = ρeiθ idθ.
R
2a
2a−1
Then dz bz2 +z2 ∼ ρb2 → 0 as ρ → 0.
Combining
these contributions, we obtain
H
z 2a−1
2πi(2a−1)
)I.
C dz b2 +z 2 = (1 − e
The poles of the complex function are at +ib = beπi/2 and −ib = be3πi/2 , where
we are careful to define
phase with 0 ≤
h πi/2the
i θ < 2π. So
(be
)2a−1
(be3πi/2 )2a−1
1
+
I = 1−e2πi(2a−1) 2πi
.
2ib
−2ib
1
After some algebra this can be reduced to I = π2 (b2 )a−1 sin (πa)
.
19. We are given χ̃I (ω) =
γ
.
(ω−ω0 )2 +γ 2
(a) We want to find χ̃R (ω). Since we cannot assume χ(t) is real, we cannot use
the Kramers-Kronig formula. We must use the Hilbert transform formula,
which makes Rno reality assumptions.
Thus, we have
R∞
χ̃I (z)
∞
γ
1
1
1
χ̃R (ω) = π P −∞ dz z−ω = π P −∞ dz z−ω
(z−ω0 )2 +γ 2
R∞
γ
1
dz z−ω
.
= π1 P −∞
(z−ω0 +iγ)(z−ω0 −iγ)
There are three poles. The one at z = ω is on the real axis and is treated
by the Principle-value prescription. The pole at z = ω0 + iγ is in the
upper half-plane. The pole at z = ω0 iγ is in the lower half-plane. Using
the method for finding Principle-value integrals from class we obtain
½
h
i
γ
1
1
2πiRes z−ω
π
(z−ω0 +iγ)(z−ω0 −iγ) z=ω0 +iγ
ω0 −ω
.
(ω−ω0 )2 +γ 2
χ̃R (ω) =
+ πiRes
h
=
Note that only the residue in the upper half plane is included in the first
term in brackets.
(b) We can combine χ̃I (ω) and χ̃R (ω) into a single complex function of a
complex variable,
ω0 −ω+iγ
−1
χ̃(ω) = χ̃R (ω) + iχ̃I (ω) = (ω−ω
2
2 = ω−(ω −iγ) .
0 ) +γ
0
This function has a pole in the lower half-plane. Thus, it is (1) analytic in
the upper half-plane, and (2) its magnitude goes to zero as ω → ∞. Thus,
the use of the Hilbert transform is justified.
(c) Include a sketch of χ̃I (ω) and χ̃R (ω) for γ = 1/10 and ω0 = 1 in some
units here.
20. Find the shortest distance between two points in polar coordinates. The differential line element is ds2 = dr2 + r2 dθ2 . One can choose r = r(θ) or θ = θ(r).
Either will work as easily as the other. I will choose r = r(θ). Then the
q distance
R θB
of a path between the two points is given by the integral L = θA dθ r2 + (r0 )2 .
q
0
∂f
√ 2r 0 2 . Since there
Letting f = r2 + (r0 )2 , we have ∂f
= √ 2 r 0 2 and ∂r
0 =
∂r
r +(r )
r +(r )
is no explicit dependence of f on θ, we can use the integrated Euler-Lagrange
∂f
equation r 0 ∂r
0 − f = c, where c is a constant. Plugging into this equation and
√
2
2
dr
0
= r rc −c . We can integrate this by quadrature:
solving for r gives r0 = dθ
R
Rr
√ 2dr̃ 2 = θ dθ̃ = θ − A, where A is an integration constant. The inter̃
r̃ /c −1
gration over r̃ can be done by change-of-variables. I will do this in two steps.
First, let r̃ R= c/x, dr̃ = −[c/x2 ]dx. This gives
dx
θ − A = − c/r √1−x
2.
Now let x =
sin φ, dx = cos φ dφ. This gives
R −1
θ − A = − sin c/r dφ = − sin−1 c/r.
We now have the equation r sin(θ − A) = −c, which is a straight line in polar
coordinates. This can be seen explicitly by rewriting it as
r sin θ cos A − r cos θ sin A = −c = y cos A − x sin A = −c.
√
R x2
R
1+(y 0 )2
=
dx
.
21. Fermat’s principle says to extremize I = ds
x1
u
u(y)
q
(a) We will use this to derive Snell’s law. The function f = 1 + (y 0 )2 /u(y)
does not depend explicitly on x, so we can use the integrated Euler∂f
Lagrange equation, y 0 ∂y
For this
0 − f = −c, where −c is a constant.
0
∂f
∂f
y
√
. Plugging f and ∂y0 into the integrated Eulerwe need ∂y0 =
0 2
u(y)
1+(y )
Lagrange equation, a little algebra gives
u(y)
√1
1+(y 0 )2
i
γ
1
z−ω (z−ω0 +iγ)(z−ω0 −iγ) z=ω
= c.
Now we just need to use geometry. Referring to Fig. 2.6 in the textbook,
¾
we see that sin φ =
dx
ds
=
dx
√dx
1+(y 0 )2
=√
1
.
1+(y 0 )2
Combining this with the result from the Euler-Lagrange equation gives
sin φ/u(y) = c = constant.
(b) Suppose the speed is proportional to y. Write u(y) = Ay, where A is some
√1 0 2 =
constant. The Euler-Lagrange equation from part (a) gives
u(y)
Ay
√1
1+(y 0 )2
1+(y )
= a constant. Let the constant be 1/(cA). We can solve to find
q
y 0 = c2 /y 2 − 1. We can integrate this by quadrature:
R
R y ỹdỹ
√
= x dx̃ = x − x0
c2 −ỹ 2
Use√change of variables, letting c2 − ỹ 2 = w2 , −2ỹdỹ = 2wdw. This gives
√
R c2 −y2
dw = − c2 − y 2 = x − x0 .
−
Squaring both sides gives (x − x0 )2 + y 2 = c2 , which is a circle centered at
(x0 , 0).
22. Consider the cycloid that connects the origin (0, 0) and the point (πa, −2a).
(a) Calculate the time required for a mass to slide from the origin to (πa, −2a).
The equations for the cycloid, from class are
x = a(θ − sin θ) and y = −a(1 − cos θ).
R (πa,−2a) ds
.
So θ goes from 0 to π. The time to the bottom is T = (0,0)
v
We can q
do the integral in terms of θ,qusing
√
ds = dθ (dx/dθ)2 + (dy/dθ)2 = dθ a 2(1 − cos θ) and v = −2gy.
Plugging these
for the total time gives
√ into the equation
q
Rπ
a 2(1−cos θ)
T = 0 dθ √
= π a/g.
2ga(1−cos θ)
(b) Suppose the mass is released from rest at an arbitrary point (xA , yA ) (other
than the origin) on this cycloid. We want to show that the time is unchanged from the result in part (a). Since it is released from rest at
a different point,
conservation of energy gives mv 2 /2 = mg(yA − y), so
q
we get v = 2g(yA − y). We can now write the integral for the total
time in terms of θ as before, except we start at an arbitrary initial θA
and the velocity formula is replaced
with the new one. Thus, we have
√
R (πa,−2a) ds
Rπ
a 2(1−cos θ)
T = (xA ,yA ) v = θA dθ √
√ 2 2g[a(1−cos θ)−a(1−cos θA )]
q
Rπ
sin (θ/2)
= a/g θA dθ √ 2
.
2
[cos (θA /2)−cos (θ/2)]
cos(θ/2)
, du = − 2sin(θ/2)dθ
cos(θA /2)
cos(θA /2)
q
2
gives T = a/g 01 du √1−u
Letting u =
2.
This integralqcan now be done q
by the substitution u = sin φ, du = cos φdφ,
R π/2
giving T = a/g 0 2dφ = π a/g, exactly as before.
R