Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 1
Each set of homework questions is worth 100 marks
Due: at the beginning of the tutorial session Thursday/Friday, 28/29 January 2016
Name:
1. A curve C in the xy-plane is represented by the equation
Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 .
(1)
In the x0 y 0 -plane obtained by rotating the xy-plane through an angle φ
x0 = x cos φ + y sin φ ,
y 0 = −x sin φ + y cos φ ,
(2)
the curve C is represented by a similar equation
A0 x02 + B 0 x0 y 0 + C 0 y 02 + D0 x0 + E 0 y 0 + F 0 = 0 .
(3)
(a) Express A0 , B 0 , C 0 , D0 , E 0 , F 0 in terms of A, B, C, D, E, F and φ.
Solution: Expressing x, y in terms of x0 , y 0 , one gets
x = x0 cos φ − y 0 sin φ ,
y = x0 sin φ + y 0 cos φ .
(4)
Substituting these expressions in (1), one finds
A0
B0
C0
D0
E0
F0
= A cos2 (φ) + B sin(φ) cos(φ) + C sin2 (φ) ,
= −A sin(2φ) + B cos(2φ) + C sin(2φ) ,
= A sin2 (φ) − B sin(φ) cos(φ) + C cos2 (φ) ,
= D cos(φ) + E sin(φ) ,
= E cos(φ) − D sin(φ) ,
=F.
(5)
(b) Prove that if the angle φ satisfies
cot 2φ =
A−C
,
B
(6)
then the curve C is represented by the equation
A0 x02 + C 0 y 02 + D0 x0 + E 0 y 0 + F 0 = 0 ,
(7)
B 0 = −A sin(2φ) + B cos(2φ) + C sin(2φ) = 0
(8)
i.e. B 0 = 0.
Solution: Solving the equation
for φ, one obviously finds (6).
1
2. Use Mathematica, and the result of the previous question to identify the curve. Find a
parametric representation and plot the curve in the xy-plane. The Mathematica function
ParametricPlot can be used to plot parametric curves in the xy-plane.
√
√
(a) 3x2 + y 2 − 2 3xy − 8x − 8 3y = 0 .
Solution: One finds that φ = − π6 , and in terms of x0 , y 0 variables the equation takes the form
x02 − 4y 0 = 0 .
(9)
Thus, the curve is a parabola, and its parametric representation is x0 = t, y 0 = t2 /4
or in terms of x, y
√
√ 2
3t
3t
t2
t
x= +
, y=
− .
(10)
8
2
8
2
The parabola is shown below
y
25
20
15
10
5
5
10
15
20
x
√
√
(b) 57x2 + 14 3xy + 36x + 43y 2 − 36 3y − 540 = 0 .
Solution: One finds that φ = π6 , and in terms of x0 , y 0 variables the equation takes the form
x02 (y 0 − 1)2
+
= 1.
9
16
(11)
Thus, the curve is an ellipse, and its parametric representation is
x0 = 3 sin t ,
or in terms of x, y
1 √
3 3 sin(t) − 4 cos(t) − 1 ,
x=
2
The ellipse is shown below
2
y 0 = 1 + 4 cos t .
y=
√
1
3 sin(t) + 3(4 cos(t) + 1) .
2
(12)
(13)
y
4
2
-4
-3
-2
1
-1
2
3
x
-2
√
√
(c) 2x2 + 5xy + 9 2x + 2y 2 + 9 2y + 36 = 0 .
Solution: One finds that φ = π4 , and in terms of x0 , y 0 variables the equation takes the form
−
(x0 + 2)2 y 02
+
= 1.
4
36
(14)
Thus, the curve is a hyperbola, and its parametric representation is
x0 = −2 + t −
1
y 0 = 3(t + ) .
t
1
,
t
(15)
or in terms of x, y
√ 2
2 (t + t + 2)
x=−
,
t
√
2(t(2t − 1) + 1)
y=
.
t
The hyperbola is shown below
y
30
20
10
-30
-20
10
-10
-10
-20
-30
3
20
30
x
(16)
3. A curve C is the intersection of the cone
z 2 = x2 + y 2 ,
(17)
with a plane.
Identify the curve, find a parametric representation and plot the curve in the xyz-space
for the planes below. The Mathematica function ParametricPlot3D can be used to plot
parametric curves in the xyz-space.
(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 = 1 .
Solution: It is a hyperbola: x = 1 , y = 12 (t − 1t ) , z = 12 (t + 1t )
4
(d) x + y = 1 .
Solution: It is a hyperbola: x =
1
2
(e) x + z = 1 .
Solution: It is a parabola: x =
−
1
2
+ 14 (t − 1t ) , y =
t2
2
1
2
− 14 (t − 1t ) , z =
, y = t, z = 1 − t
(f) x + y + z = 1 .
Solution: It is a hyperbola: x = 1 + t , y = 1 +
5
1
2t
, z = −1 − t −
1
2t
1
(t
23/2
+ 1t )
4. Consider the vector-valued function (with values in R3 )
r(t) = ln(3 −
√
t) i + (1 +
√
√
(3 − t)2
k
t) j +
4
(18)
(a) Find the domain D(r) of the vector-valued function r(t).
Solution: The domain D(r) of r(t) is the intersection of domains of its
component functions.
√
√
√
(3− t)2
Since D(ln(3 − t)) = [0 , 9 ), D(1 + t) = [0, ∞) and D( 4 ) = [0, ∞), one gets
D(r) = [0 , 9 )
(19)
that is the vector function r(t) is defined for 0 ≤ t < 9.
(b) Find the derivative dr/dt.
Solution:
√
dr
1
1 3− t
1
1
√ i+ √ j− √
=− √
k.
dt
2 t 3− t
2 t
2 t 2
(20)
(c) Find the norm |dr/dt|.
Simplify the expressions obtained.
Solution: The magnitude or norm of this vector is
s
s
√ 2
√ 2
2 2 2
dr
1
1
1 3− t
3− t
1
1
1
√
√
√
| |=
+
+ − √
+1+
− √
= √
dt
2
2 t 3− t
2 t
2 t 2
2 t
3− t
s
√ 2
√
2
3− t
1
1 3− t
1
√
√
+
+2
= √
2
2 t
3− t
3− t 2
s
√ 2
√ √
1
3− t
1
3− t
t − 6 t + 11
1
1
√ +
√ +
√ ,
= √
= √
=−
2
2
2 t
3− t
2 t 3− t
4 t−3 t
(21)
√
because t < 3.
(d) Find the unit tangent vector T for all values of t in D(r).
Solution: The unit tangent vector is
√
√
dr
2
t
−
3
2
−t + 6 t − 9
dt
√
√
√
T = dr = −
i−
j+
k.
| dt |
t − 6 t + 11
t − 6 t + 11
t − 6 t + 11
(22)
(e) Find the vector equation of the line tangent to the graph of r(t) at the point
P0 (0, 3, 41 ) on the curve.
Solution: The point P0 (0, 3, 14 ) on the curve corresponds to t = 4. We find
dr
1
1
1
(4) = − i + j − k .
dt
4
4
8
(23)
t
t
3 t
r = r0 + (t − 4) v0 = (1 − ) i + (2 + ) j + ( − ) k .
4
4
4 8
(24)
r0 = r(4) = 3j +
1
k,
4
v0 =
Thus the tangent line equation is
6
Note that the same line is also described by the following equation which is obtained
from the one above by rescaling and shifting the parameter t: t → −4t + 4
1
r = r0 + t v0 = t i + (3 − t) j + (2t + 1) k .
4
7
(25)