Advanced Calculus (MAU11202),
Homework Assignment 1 Solutions
Straightforward
1. [3 points] Find the derivative of the vector-valued function
2
t −1
, sin(t) ln(t), cos(sin(t)) .
r(t) = 2
t +2
What is r0 (1) equal to?
Solution:
To differentiate r(t) = (x(t), y(t), z(t)), we differentiate the three component functions separately.
Using the Quotient Rule,
x0 (t) =
2t3 + 4t − 2t3 + 2t
6t
2t(t2 + 2) − 2t(t2 − 1)
=
= 2
.
2
2
2
2
(t + 2)
(t + 2)
(t + 2)2
Using the Product Rule,
y 0 (t) = cos t ln(t) + sin(t) ·
sin(t)
1
= cos(t) ln(t) +
.
t
t
Using the Chain Rule, z 0 (t) = cos(t)(− sin(sin(t))) = − cos(t) sin(sin(t)).
Hence
sin(t)
6t
, cos(t) ln(t) +
, − cos(t) sin(sin(t)) .
r (t) =
(t2 + 2)2
t
Letting t = 1 in this expression, we obtain
6(1)
sin(1)
0
r (1) =
, cos(1) ln(1) +
, − cos(1) sin(sin(1)
(12 + 2)2
1
2
=
, sin(1), − cos(1) sin(sin(1)) .
3
0
2. [3 points] Find the derivative of the vector-valued function
2
t −1
3
2
r(t) = (2t − 3t + 5t + 3) 2
, sin(t) ln(t), cos(sin(t)) .
t +2
What is the domain of this function?
Solution:
Here we use Theorem 1.2.11(c) (Chapter 1 of course notes by Anthony Brown), the solution
to Question 1, and the fact that the derivative of f (t) = 2t3 −3t2 +5t+3 is Fb(t) = 6t2 −6t+5,
to obtain
6t
sin(t)
0
3
2
r (t) =(2t − 3t + 5t + 3)
, cos(t) ln(t) +
, cos(t) cos(sin(t))
(t2 + 2)2
t
2
t −1
2
+ (6t − 6t + 5) 2
, sin(t) ln(t), cos(sin(t)) .
t +2
The domain of r is R ∩ (0, ∞) ∩ R = (0, ∞).
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3. [3+1 points] Find an arc length parameterization of the curve
h πi
r(t) = cos3 (t)i + sin3 (t)j t ∈ 0,
,
2
in the same direction as the original curve. Try to write your result without sin and cos.
Solution:
Here we are given our reference point t0 = 0, so we don’t have to calculate it.
Next, let us calculate the arc length of the curve from t = 0 to t.
Using the Chain Rule, we obtain
r0 (t) = −3 sin(t) cos2 (t)i + 3 cos(t) sin2 (t)j.
Hence
q
kr (t)k = 9 sin2 (t) cos4 (t) + 9 cos2 (t) sin4 (t)
q
= 9 sin2 (t) cos2 (t)(cos2 (t) + sin2 (t))
q
= 9 sin2 (t) cos2 (t)
0
= 3 sin(t) cos(t).
Thus the required arc length is
t
Z
0
Z
kr (v)k dv =
s(t) =
0
0
t
t
3 2
3
sin (v) = sin2 (t),
3 sin(v) cos(v) dv =
2
2
0
where we have integrated ‘by inspection’ If you can’t spot the integral, then you can use the
substitution u = sin(v).
So we have s = 32 sin2 (t), and solving this for t we obtain
3
s = sin2 (t)
2
=⇒
r
2s
sin2 (t) =
3
=⇒
sin t =
2s
3
r
t = sin−1
=⇒
2s
3
!
.
Substituting this into r(t), we obtain
r
r(s) = cos3 sin−1
2s
3
!!
r
i + sin3 sin−1
2s
3
!!
j
Now, when t = 0, s = 0 and when t = π2 , s = 32 , so we obtain the arc length parametrization
r !!
r !!
3
2s
2s
3
−1
3
−1
r(s) = cos sin
i + sin sin
j s ∈ 0,
.
3
3
2
The additional (+1) point goes to students trying to simplify this expression:
Although this is a perfectly good answer, it doesn’t look particularly ‘nice’. We can get
a ‘nicer’ answer (although of course this is a matter of opinion) by instead of solving s =
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r
3
2
2
sin (t) for t, stopping when we got an expression for sin(t), that is sin t =
32
2s
Then sin3 (t) =
and
3
2s
=
3
2s
3
12
.
q
3
3
32
2s 2
2
2
cos (t) =
1 − sin (t) = 1 − sin (t) = 1 −
.
3
3
So we obtain the parametrization
r(s) =
2s
1−
3
32
i+
2s
3
32
3
.
j s ∈ 0,
2
t2 − 9 2
4. [3 points] Let r(t) = 2t + 1,
, t . Find limt→3 r(t). Is r(t) continuous t = 3.
t−3
Solution:
From the lecture, we know that
t2 − 9
2
lim r(t) = lim 2t + 1, lim
, lim t = (7, 6, 9)
t→3
t→3
t→3 t − 3 t→3
(1)
It is not continuous at t = 3 because it is note defined there.
√
R
5. [3 points] Find r(t) = r0 (t)dt, where r0 (t) = e2t , t, sin t , and r(π) = (0, 0, 0).
Solution:
From the lecture, we know that
Z
Z
Z
Z √
1 2t 2 3
0
2t
tdt, sin tdt =
r (t)dt =
e dt,
e , t 2 , − cos t + c
2
3
(2)
Evaluating at π yields the equation
1 2π 2 3
1 2π 2 3
e , π 2 , − cos π + c =
e , π 2 , 1 + c = (0, 0, 0)
2
3
2
3
1 2π 2 3
which implies that c = −
e , π2,1 .
2
3
(3)
Intermediate
6. [3 points] Prove the Serret-Frenet formulae
(a)
dT
ds
= κ(s)N(s)
Solution:
By definition
N(s) =
dT
ds
dT
| ds |
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=
dT
ds
κ(s)
.
(4)
(b)
dB
ds
= −τ (s)N(s), where τ is a scalar called the torsion of r(s).
Solution:
By definition
dB
dT
dN
dN
=
×N+T×
=T×
.
(5)
ds
ds
ds
ds
Thus dB
is normal to T. Since B is normal, dB
is also normal to B. Thus, it is
ds
ds
proportional to N.
(c)
dN
ds
= −κ(s)T(s) + τ (s)B
Solution:
We know that
N = B × T,
B = T × N,
T = N × B.
(6)
Thus
dN
dB
dT
=
×T+B×
= −τ (s)N × T + κ(s)B × N = −κ(s)T(s) + τ (s)B . (7)
ds
ds
ds
7. [6 points] Consider the vector-valued function
r(t) = (4 cos(t), 4 sin(t), 3t),
t ∈ R.
(a) Find an arc-length parameterization for the function with reference point t0 = 0.
Solution:
Here we are given our reference point t0 = 0, so we don’t have to calculate it.
Next, let us calculate the arc length of the curve from t = 0 to t.
We already did this in lectures, finding that s = 5t.
s
Thus t = and substituting this into r(t), we obtain the arc length parametrization
5
s
s 3s r(s) = 4 cos
, 4 sin
,
s ∈ R.
5
5
5
(b) Calculate the curvature κ of r, using the arc-length parameterization you found in Part
(a) and the formula κ(s) = kT0 (s)k.
Solution:
Since we now have a unit speed parametrization, using Remark 1.4.2(b),
s 4
s 3
4
0
T(s) = r (s) = − sin
, cos
,
.
5
5 5
5 5
Hence
0
T (s) =
s
s 4
4
− cos
, − sin
,0 .
25
5
25
5
Thus
r
0
κ(s) = kT (s)k =
s
42
42
2 s
2
cos
+ 2 sin
=
252
5
25
5
r
42
4
= .
2
25
25
Note that the curvature is independent of the parametrization, as we would want for
any reasonable measure of ‘curvature’.
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(c) Calculate the torsion τ of r, using the arc-length parameterization you found in Part
(a) and the formula τ (s) = −N(s) · B0 (s).
Solution:
We need to calculate N(s), B(s) and hence B0 (s). Firstly
T0 (s)
kT0 (s)k
s
s 1
4
4
=
− cos
, − sin
4/25
25
5
25
5
s
s = − cos
, − sin
,0 .
5
5
N(s) =
Next, using B(s) = T(s) × N(s),
i
j
k
s 4
s 3
4
cos
B(s) = − 5 sin 5
s 5
5s 5
− cos
− sin
0
5
5
s 3
s 4
s
4
4
4
3
− sin
− sin
cos
cos
5 5 +k
5 5 5
5s 5 − j
5s =i 5
s
s
− sin
0
− cos
0
− cos
− sin
5
5
5
5
s
s
s
4 2 s 4
3
3
= sin
i − cos
j+
sin
+ cos2
k
5
5
5
5
5
5
5
5
s 3
s 4
3
=
sin
, − cos
,
.
5
5
5
5 5
Then
0
B (s) =
s 3
s 3
cos
,
sin
,0 .
25
5 25
5
Finally,
τ (s) = − N(s) · B0 (s)
s 3
s 3
s s
, − sin
,0 ·
cos
,
sin
,0
= − − cos
5
5
25
5 25
5
3
3
2 s
2 s
= − − cos
−
sin
25
5
25
5
3
= .
25
Again, as we would want, the torsion does not depend on the parametrization.
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8. [4 points] A curve C is the intersection of the cone z 2 = x2 + y 2 , with a plane defined
through (a), (b), (c) or (d). Identify the curve, find a parametric representation.
(a) z = 2 .
Solution:
It is a circle of radius 2: x = 2 cos t , y = 2 sin t , z = 2
(b) y = 0 .
Solution:
It is two intersecting lines: y = 0 , x = t , z = ±t
(c) x + y = 1 .
Solution:
It is a hyperbola: x =
1
2
+ 41 (t − 1t ) , y =
1
2
− 14 (t − 1t ) , z =
(d) x + y + z = 1 .
Solution:
It is a hyperbola: x = 1 + t , y = 1 +
1
2t
, z = −1 − t −
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1
2t
1
(t
23/2
+ 1t )
9. [3 points] Consider the motion of a particle moving along a curve, described at time t by
ds
r(t). The velocity of the particle at time t is then v(t) = r0 (t) = T(t).
dt
(a) How do we find the acceleration a(t) of the particle at time t?
Solution:
First derivative is velocity v(t) = r0 (t), and the derivative of velocity is acceleration;
so, a(t) = v0 (t) = r00 (t).
(b) Using v(t), write an expression for calculating the distance traveled between times t1
and t2 .
Solution:
We know that distance traveled is the arc length between t1 and t2 which can be
expressed as
Z t2
Z t2
0
||r (t)||dt =
||v(t)||dt
(8)
t1
t1
(c) Using v(t), write an expression to calculate the length of the displacement of the particle
occurring from time t1 to time t2 , r(t2 ) − r(t1 ).
Solution:
Using
R t the fundamental theorem of calculus for vector-valued functions, we know that
|| t12 v(t)dt|| = ||r(t2 ) − r(t1 )||
Challenge
10. [9 points] The parameterization
x = α + a cosh t ,
y = β + b sinh t ,
−∞ < t < ∞
(9)
of the hyperbola
(x − α)2 (y − β)2
−
= 1,
a2
b2
represents only one branch of the hyperbola.
a > 0,
b > 0,
Find a parameterization t = γ(p) which represents both branches of the hyperbola.
Solution:
7 of 9
(10)
Recall that using the definitions for sinh t and cosh t, we can rewrite the given parameterization as
et − e−t
et + e−t
, y(r) = β + b
, −∞ < t < ∞.
(11)
x(r) = α + a
2
2
Since 0 < et < ∞, we can reparameterize using r = et with 0 < r < ∞ yielding (with some
algebraic manipulation)
a
1
b
1
x=α+
r+
, y=β+
r−
, 0 < r < ∞.
(12)
2
r
2
r
1
a
r+
, and it is not hard
Lastly, observe that since −r < 0, we have that x(−r) = α −
2
r
to show that for all negative values −r, y(−r) delivers the correct corresponding y-value, so
that the full parameterization of both branches is
1
a
x = α + (r + ) ,
2
r
b
1
y = β + (r − ) ,
2
r
−∞ < r < ∞ ,
r 6= 0 .
(13)
11. [9 points] A curve C in the xy-plane is represented by the equation
Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 .
(14)
In the uv-plane obtained by rotating the xy-plane through an angle φ
u = x cos φ + y sin φ ,
v = −x sin φ + y cos φ ,
(15)
the curve C is represented by a similar equation
b 2 + Buv
b + Cv
b 2 + Du
b + Ev
b + Fb = 0 .
Au
(16)
b B,
b C,
b D,
b E,
b Fb in terms of A, B, C, D, E, F and φ.
(a) Express A,
Solution:
Expressing x, y in terms of u, v, one gets
x = u cos φ − v sin φ ,
y = u sin φ + v cos φ .
(17)
Substituting these expressions in (14), one finds
b = A cos2 (φ) + B sin(φ) cos(φ) + C sin2 (φ) ,
A
b = −A sin(2φ) + B cos(2φ) + C sin(2φ) ,
B
b = A sin2 (φ) − B sin(φ) cos(φ) + C cos2 (φ) ,
C
b = D cos(φ) + E sin(φ) ,
D
b = E cos(φ) − D sin(φ) ,
E
(18)
Fb = F .
(b) Prove that if the angle φ satisfies
cot 2φ =
A−C
,
B
(19)
then the curve C is represented by the equation
b 2 + Cv
b 2 + Du
b + Ev
b + Fb = 0 ,
Au
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(20)
b = 0.
i.e. B
Solution:
Solving the equation
b = −A sin(2φ) + B cos(2φ) + C sin(2φ) = 0
B
(21)
for φ, one obviously finds (19).
!
b2 E
b2
D
+
. Give a parameterization
b
b
A
C
1
(c) Suppose now that (19) holds and that Fb <
4
of this curve.
Solution:
The key observation here is that if we complete the square correctly, we might have the
b
equation describing a circle. First, let us rewrite the equation (with B=0)
as
!
!
b
b
E
D
b v 2 + v + Fb = 0.
b u2 + u + C
(22)
A
b
b
A
C
We simply must complete the square in both parentheses, i.e.,

b
b u2 + D u+ 1
A
b
4
A
b
D
b
A
!2 

b
b v 2 + E v+ 1
+C
b
4
C
b
E
b
C
!2 
 + Fb− 1
4
b2
D
b
A
!
1
−
4
b2
E
b
C
!
= 0.
(23)
We can factor both polynomials now as perfect squares and move the compensation
terms to the right, yielding
b
b u+ 1D
A
b
2A
!2
b
b v+ 1E
+C
b
2C
!2
1
=
4
b2
b2 E
D
+
b
b
A
C
!
− Fb.
(24)
This equation describes a circle in the new coordinate system, as long !
as the right2
2
b
b
1 D
E
hand side is positive. This leads to the constraint that Fb <
+
. Thus, the
b
b
4 A
C
parameterization is
x(t) = −
y(t) = −
b
1D
+
b
2A
1
4
b
1E
+
b
2C
1
4
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b
A
!
cos t
b2 E
b2
D
+
− Fb
b
b
A
C
b
C
!
sin t
b2 E
b2
D
+
− Fb
b
b
A
C