Ex Sheet
8
2A Multivariable Calculus 2013
.
Tutorial Exercises
T1
Let r = ( x, y, z), r = |r| and a be a constant vector. Prove the
following results
(a) div r = 3, (b) div (a r) = 0, (c) div (r n a) = nr n2 (a r).
Solution
(a) div r =
(b) a r = (c2 z c3
2
Ex Sheet
2A Multivariable Calculus 2013
.
Tutorial Exercises
T1
Find all second order partial derivatives of
(a) z = x log(1 + y),
(b) z = sin( xy),
(c) z =
x
y
2
.
Check in each case that z xy = zyx .
Solution
(a) z x = log(1 + y) and zy = x/(1 + y). T
Ex Sheet
2A Multivariable Calculus 2013
Tutorial Exercises
4
.
Lecture 7
Key Points:
By reversing the order of integration, evaluate
T1
1
(a)
0
1
dy
y
e
sinh( x2 ) dx,
(b)
1
1
dx
changing the order of integration
2
ey
log x
x
dy.
using double integrals
Ex Sheet
5
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 9
Key Points:
Evaluate the following using beta functions:
T1
/2
(a)
sin3 x cos2 x dx,
0
/2
(d)
0
/2
(b)
sin4 x cos2 x dx,
(e)
2
(g)
0
3
0
0
sin7 x cos3 x dx, (c)
sin5 x dx,
sin x cos
Ex Sheet
6
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 11
Key Points:
T1 Sketch the wedge shaped region W (in the rst octant) enclosed
by the ve planes x = 0, y = 0, z = 0, x = 1 and y + z = 1. Then
evaluate
xy dxdydz.
W
calculating trip
Ex Sheet
2A Multivariable Calculus 2013
4
Tutorial Exercises
T1
By reversing the order of integration, evaluate
1
(a)
0
1
dy
y
e
sinh( x2 ) dx,
(b)
1
2
1
dx
log x
ey
dy.
x
Solution
(a) Sketching the two formulations of the integral we get
So the integral
Tutorial Exercises
Evaluate the following using beta functions:
T1
/2
(a)
0
/2
(d)
0
/2
sin3 x cos2 x dx,
(b)
sin4 x cos2 x dx,
(e)
2
(g)
0
0
0
sin7 x cos3 x dx, (c)
sin5 x dx,
2
sin3 x cos3 x dx,
(h)
0
(f)
0
0
cos5 x dx,
sin2 x cos3 x dx,
sin4 x cos3 x d
Ex Sheet
2
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 3
Key Points:
T1
Find all second order partial derivatives of
(a) z = x log(1 + y),
(b) z = sin( xy),
implicit partial differentiation
(c) z =
x
y
2
Check in each case that z xy = zy
Chapter 0
Revision of dierentiation, integration
and vector algebra from level 1
0.1
Revision of dierentiation
(Stewart (Ed. 7): Chapter 2, p103.)
0.1.1
Three important rules for dierentiation
Product Rule This rule is for dierentiating the product of fun
Ex Sheet
3
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 5
Key Points:
T1 By making the change of variables indicated, nd the general
solution of each of the following partial differential equations.
f
f
a) x x + y y = 6xy. Change to u =
b)
Ex Sheet
3
2A Multivariable Calculus 2013
.
Tutorial Exercises
T1 By making the change of variables indicated, nd the general
solution of each of the following partial differential equations.
f
f
a) x x + y y = 6xy. Change to u =
y
x
and v = x
f
f
b) 2x x
Ex Sheet
6
2A Multivariable Calculus 2013
Tutorial Exercises
T1 Sketch the wedge shaped region W (in the rst octant) enclosed
by the ve planes x = 0, y = 0, z = 0, x = 1 and y + z = 1. Then
evaluate
xy dxdydz.
W
Solution
Hence the integral is
1
I=
0
1 y
1
7
Ex Sheet
2A Multivariable Calculus 2013
.
Tutorial Exercises
T1
Find the directional derivative of xyz2 at the point (1, 5, 1) in
the direction of the vector (1, 1, 2).
Solution
The unit vector in the direction of (1, 1, 2) is n = (1, 1, 2)/ 6 and
(5, 1
Ex Sheet
8
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 15
Key Points:
T1
Let r = ( x, y, z), r = |r| and a be a constant vector. Prove the
following results
(a) div r = 3, (b) div (a r) = 0, (c) div (r n a) = nr n2 (a r).
T2
Let r = ( x,
Ex Sheet
9
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 17
Key Points:
T1
Evaluate
P
xy2 dx + x4 y dy,
where P is the arc of the parabola y = 2x2 from A(0, 0) to B(1, 2). (a)
by parametrising the curve, (b) using x, y coordinates.
T2 The c
Tutorial Exercises
10
Ex Sheet
2A Multivariable Calculus 2013
.
Lecture 19
T1
In R3 let S be the part of the plane 4x + 2y z = 37 enclosed
within the innite cylinder with rectangular section dened by 0
x 5, 0 y 2. Evaluate
S
2y dS.
Key Points:
calculat
1
Ex Sheet
2A Multivariable Calculus 2013
.
Tutorial Exercises
T1 State the type of surface given by each of the following equations
in three dimensional space.
(a) 4x + 5y 2z = 20,
(d) x2 +
y2
z2
+
= 1,
4
9
(b) x2 + y2 = 1,
(c) x2 + y2 + z2 2x = 10,
(e)
Ex Sheet
10
2A Multivariable Calculus 2013
.
Tutorial Exercises
T1
In R3 let S be the part of the plane 4x + 2y z = 37 enclosed
within the innite cylinder with rectangular section dened by 0
x 5, 0 y 2. Evaluate
S
2y dS.
Solution
1+
We need to calculate
Ex Sheet
1
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 1
Key Points:
T1 State the type of surface given by each of the following equations
in three dimensional space.
(a) 4x + 5y 2z = 20,
(d) x2 +
y2
z2
+
= 1,
4
9
(b) x2 + y2 = 1,
(c) x2
Ex Sheet
7
2A Multivariable Calculus 2013
Tutorial Exercises
.
Lecture 13
T1
Find the directional derivative of xyz2 at the point (1, 5, 1) in
the direction of the vector (1, 1, 2).
Let f be a scalar eld, u a unit vector and let be the angle
T2
between u
Ex Sheet
9
2A Multivariable Calculus 2013
Tutorial Exercises
T1
Evaluate
P
xy2 dx + x4 y dy,
where P is the arc of the parabola y = 2x2 from A(0, 0) to B(1, 2). (a)
by parametrising the curve, (b) using x, y coordinates.
Solution
(a) Parameterise P
y = 2t