Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 1
To be solved during the tutorial session Thursday/Friday, 21/22 January 2016
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
where
A0 x02 + B 0 x0 y 0 + C 0 y 02 + D0 x0 + E 0 y 0 + F 0 = 0 ,
(3)
A0
B0
C0
D0
E0
F0
(4)
= 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.
If the angle φ satisfies
A−C
,
B
then the curve C is represented by the equation
(5)
cot 2φ =
A0 x02 + C 0 y 02 + D0 x0 + E 0 y 0 + F 0 = 0 ,
B0 = 0 .
(6)
1. 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) x2 − 2xy + y 2 − 4 2x − 4 2y = 0 .
Solution: One finds that φ = π4 , and in terms of x0 , y 0 variables the equation takes the form
y 02 − 4x0 = 0 .
(7)
Thus, the curve is a parabola, and its parametric representation is y 0 = t, x0 = t2 /4
or in terms of x, y
t
t2
t2
t
x = −√ + √ , y = √ + √ .
(8)
2 4 2
4 2
2
The parabola is shown below
1
y
25
20
15
10
5
5
10
15
20
25
x
√
√
(b) 31x2 − 10 3xy − 32x + 21y 2 − 32 3y − 80 = 0 .
Solution: One finds that φ = − π6 , and in terms of x0 , y 0 variables the equation takes the form
x02 (y 0 − 2)2
+
= 1.
4
9
(9)
Thus, the curve is an ellipse, and its parametric representation is
x0 = 2 cos t ,
y 0 = 2 + 3 sin t .
(10)
or in terms of x, y
x=
3 sin(t) √
+ 3 cos(t) + 1 ,
2
y=
1√
3(3 sin(t) + 2) − cos(t) .
2
The ellipse is shown below
y
4
3
2
1
1
-1
-1
2
2
3
x
(11)
√
√
(c) 32x2 − 7y 2 − 52xy − 144 5x + 72 5y + 900 = 0 .
Solution: One finds that cot(2φ) = − 34 , and in terms of x0 , y 0 variables the equation takes
the form
y 02 (x0 − 4)2
−
= 1.
(12)
9
4
Thus, the curve is a hyperbola, and its parametric representation (one branch of it)
is
y 0 = 3 cosh t , x0 = 4 + 2 sinh t .
(13)
or in terms of x, y
x=
4 sinh(t) + 3 cosh(t) + 8
√
,
5
y=
6 cosh(t) − 2(sinh(t) + 2)
√
.
5
(14)
The hyperbola is shown below
y
10
5
5
-5
10
15
x
-5
-10
-15
2. A curve C is the intersection of the cone
z 2 = x2 + y 2 ,
(15)
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 = 2.5 .
Solution: It is a circle of radius 2.5: x = 2.5 cos t , y = 2.5 sin t , z = 2.5
3. Consider the vector-valued function (with values in R3 )
r(t) = ln(−t) i − t j +
3
t2
k
4
(16)
(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.
2
Since D(ln(−t)) = (−∞, 0), D(t) = (−∞, ∞) and D( t4 ) = (−∞, ∞), one gets
D(r) = (−∞, 0)
(17)
that is the vector function r(t) is defined for t < 0.
(b) Find the derivative dr/dt.
Solution:
1
t
dr
= ( , −1 , ) .
dt
t
2
(18)
(c) Find the norm |dr/dt|.
Simplify the expressions obtained.
Solution: The magnitude or norm of this vector is
s 2 s
2 s 2
2
2
dr
1
1
t
1 t
t
1
t
1 t
2
|| || =
+
+
+ (−1) +
=
+2
=
=− − ,
dt
t
2
t
t 2
2
t 2
t 2
(19)
because t < 0.
(d) Find the unit tangent vector T for all values of t in D(r).
Solution: The unit tangent vector is
dr
2t
t2
2
dt
T = dr = −
,
,−
2 + t2 2 + t2
2 + t2
|| dt ||
(20)
(e) Find the vector equation of the line tangent to the graph of r(t) at the point
P0 (0, 1, 41 ) on the curve.
Solution: The point P0 (0, 1, 41 ) on the curve corresponds to t = −1. We find
r0 = r(−1) = j +
1
k,
4
4
v0 =
dr
1
(1) = −i − j − k .
dt
2
(21)
Thus the tangent line equation is
1
1
r = r0 + (t + 1) v0 = −(t + 1) i − t j − (t + ) k .
2
2
(22)
Note that the same line is also described by the following equation which is obtained
from the one above by the shift of the parameter t: t → t − 1
1
1
r = r0 + t v0 = −t i − (t − 1) j − (t − ) k .
2
2
5
(23)
MA1132, 2016, Tutorial Set 1
FS = FullSimplify;
Assumptions$ = {t ∈ Reals}
{t ∈ Reals}
Problem 1 of TS1
Clear[Ap, Bp, Cp, Dp, Ep, Fp]
Ap := Ac Cos[p]2 + Bc Cos[p] Sin[p] + Cc Sin[p]2
Bp := Bc Cos[2 p] -− Ac Sin[2 p] + Cc Sin[2 p]
Cp := Cc Cos[p]2 -− Bc Cos[p] Sin[p] + Ac Sin[p]2
Dp := Dc Cos[p] + Ec Sin[p]
Ep := Ec Cos[p] -− Dc Sin[p]
Fp := Fc
Part (a)
Clear[x, y, xp, yp, f, ff, p]
f = x ^ 2 -− 2 x y + y ^ 2 -− 4 × 2 ^ (1 /∕ 2) x -− 4 × 2 ^ (1 /∕ 2) y
-− 4
2 x + x2 -− 4
2 y -− 2 x y + y2
Ac = 1 /∕ 2 D[f, {x, 2}]
1
Bc = D[D[f, x, y]]
-− 2
Cc = 1 /∕ 2 D[f, {y, 2}]
1
Dc = D[f, x] /∕. {x -−> 0, y → 0}
-− 4
2
Ec = D[f, y] /∕. {x -−> 0, y → 0}
-− 4
2
Fc = f /∕. {x -−> 0, y → 0}
0
Ac -− Cc
Bc
0
2
Tutorial1.nb
p = ArcCot
Ac -− Cc
Bc
 2
π
4
Bp
0
ff = Collect[Ap xp ^ 2 + Bp xp yp + Cp yp ^ 2 + Dp xp + Ep yp + Fp,
{xp ^ 2, xp yp, yp ^ 2, xp, yp}, FS]
-− 8 xp + 2 yp2
Clear[xp, yp]
sub = Solve[ff ⩵ 0, xp][[1]]
xp →
yp2

4
yp = t
xp = xp /∕. sub
t
t2
4
x = xp Cos[p] -− yp Sin[p]
t
-−
t2
+
2
4
2
y = yp Cos[p] + xp Sin[p]
t
2
t2
+
4
2
Tutorial1.nb
ParametricPlot[{x, y}, {t, -− 10, 10}, AxesLabel → {"x", "y"}]
y
25
20
15
10
5
5
10
15
20
25
x
Part (b)
Clear[x, y, xp, yp, f, ff, p]
f = 31 x ^ 2 -− 10 × 3 ^ (1 /∕ 2) x y + 21 y ^ 2 -− 32 x -− 32 × 3 ^ (1 /∕ 2) y -− 80
-− 80 -− 32 x + 31 x2 -− 32
3 y -− 10
Ac = 1 /∕ 2 D[f, {x, 2}]
31
Bc = D[D[f, x, y]]
-− 10
3
Cc = 1 /∕ 2 D[f, {y, 2}]
21
Dc = D[f, x] /∕. {x -−> 0, y → 0}
-− 32
Ec = D[f, y] /∕. {x -−> 0, y → 0}
-− 32
3
Fc = f /∕. {x -−> 0, y → 0}
-− 80
3 x y + 21 y2
3
4
Tutorial1.nb
Ac -− Cc
Bc
1
-−
3
p = ArcCot
-−
Ac -− Cc
Bc
 2
π
6
Bp
0
ff = Collect[Ap xp ^ 2 + Bp xp yp + Cp yp ^ 2 + Dp xp + Ep yp + Fp,
{xp ^ 2, xp yp, yp ^ 2, xp, yp}, FS]
-− 80 + 36 xp2 -− 64 yp + 16 yp2
ff -− -− 144 + 36 xp2 + 16 (yp -− 2)2  /∕/∕ FS
0
-− 144 + 36 xxp2 + 16 (yyp -− 2)2   144 /∕/∕ Expand
-−
5
9
+
xxp2
-−
4
4 yyp
+
yyp2
9
9
xp = (144 /∕ 36) ^ (1 /∕ 2) Cos[t]
2 Cos[t]
yp = 2 + (144 /∕ 16) ^ (1 /∕ 2) Sin[t]
2 + 3 Sin[t]
ff /∕/∕ FS
0
x = xp Cos[p] -− yp Sin[p] /∕/∕ FS
1+
3 Cos[t] +
3 Sin[t]
2
y = yp Cos[p] + xp Sin[p] /∕/∕ FS
-− Cos[t] +
1
2
3 2 + 3 Sin[t]
Tutorial1.nb
ParametricPlot[{x, y}, {t, 0, 2 Pi}, AxesLabel → {"x", "y"}]
y
4
3
2
1
1
-−1
2
3
x
-−1
Part (c)
Clear[x, y, xp, yp, f, ff, p]
f = 32 x ^ 2 -− 52 x y -− 7 y ^ 2 + 72 × 5 ^ (1 /∕ 2) y -− 144 × 5 ^ (1 /∕ 2) x + 900
900 -− 144
5 x + 32 x2 + 72
5 y -− 52 x y -− 7 y2
Ac = 1 /∕ 2 D[f, {x, 2}]
32
Bc = D[D[f, x, y]]
-− 52
Cc = 1 /∕ 2 D[f, {y, 2}]
-− 7
Dc = D[f, x] /∕. {x -−> 0, y → 0}
-− 144
5
Ec = D[f, y] /∕. {x -−> 0, y → 0}
72
5
Fc = f /∕. {x -−> 0, y → 0}
900
5
6
Tutorial1.nb
Ac -− Cc
-−
Bc
3
4
p = ArcCot
-−
Ac -− Cc
Bc
  2 /∕/∕ FS
1
3
ArcCot 
2
4
Cot[2 p]
-−
3
4
Bp
0
ff = Collect[Ap xp ^ 2 + Bp xp yp + Cp yp ^ 2 + Dp xp + Ep yp + Fp,
{xp ^ 2, xp yp, yp ^ 2, xp, yp}, FS]
900 -− 360 xp + 45 xp2 -− 20 yp2
ff -− 180 -− 20 yp2 + 45 (xp -− 4)2  /∕/∕ FS
0
ff  (180) /∕/∕ Expand
5 -− 2 xp +
xp2
-−
yp2
4
9
ff  (180) -− 1 -−
(yp)2
+
(xp -− 4)2
9
4
0
yp = 3 /∕ 2 (t + 1 /∕ t)
3
1
2
t
+t
xp = 4 + (t -− 1 /∕ t)
4 -−
1
t
+t
ff /∕/∕ FS
0
yp = 3 Cosh[t];
xp = 4 + 2 Sinh[t];
ff /∕/∕ FS
0
x = xp Cos[p] -− yp Sin[p] /∕/∕ FS
8 + 3 Cosh[t] + 4 Sinh[t]
5
/∕/∕ FS
Tutorial1.nb
y = yp Cos[p] + xp Sin[p] /∕/∕ FS
6 Cosh[t] -− 2 2 + Sinh[t]
5
x1 = x /∕. {t → t + I Pi}; y1 = y /∕. {t → t + I Pi};
ParametricPlot[{{x, y}, {x1, y1}}, {t, -− 2, 2}, AxesLabel → {"x", "y"}]
y
10
5
5
-−5
-−5
-−10
-−15
Problem 2 of TS1
Part (a)
Clear[x, y, z, xp, yp, f, ff, p]
f = x ^ 2 + y ^ 2 -− z ^ 2
x2 + y2 -− z2
z = 2.5; x = z Cos[t]; y = z Sin[t];
f /∕/∕ FS
0.
10
15
x
7
8
Tutorial1.nb
Show[{ParametricPlot3D[{r Cos[t], r Sin[t], r}, {r, -− 3, 3},
{t, -− Pi, Pi}, AxesLabel → {"x", "y", "z"}, BoxRatios → Automatic,
PlotStyle → Directive[Opacity[0.2], Specularity[White, 5]]],
ParametricPlot3D[{x, y, z}, {t, 0, 2 Pi}, AxesLabel → {"x", "y", "z"},
PlotStyle → {Blue, Thick}]}, PlotRange → All]