MAT2615/103/3/2012
Tutorial Letter 103/3/2012
Calculus in Higher Dimensions
MAT2615
Department of Mathematical Sciences
This tutorial letter contains two trial papers, with their memorandums
2
MAT2615/103/3
Dear Student,
The January 2010 and October 2010 exams and memos follow below. This is for self study; we recommend
that you answer the exam questions for each exam within two hours and only then compare your answers
to the memo.
Best of luck for the exams!
January 2010 Exam Questions
MAT2615
CALCULUS IN HIGHER DIMENSIONS
Time: 2 Hours
Examiners:
First: Mr. R.J. de Beer
Second: Prof. W.M. Lesame
January/February 2010
100 Marks
Instructions to the candidate:
• The use of programmable calculators is allowed.
• This paper consists of 3 pages.
• Show ALL calculations.
• Try to answer ALL the questions.
This examination paper remains the property of the University of South Africa and may NOT be removed
from the examination venue.
1. L1 and L2 are lines in R3 . A parametric equation for L1 is
(x, y, z) = (1, 0, 2) + t(2, 4, −1); t ∈ R.
L2 is parallel to L1 and contains the point (3, 4, 3). V is the plane that contains both the lines L1 and L2 . Find
an equation for V .
[7]
2. Prove from first principles that
lim
(x − y) = 0.
(x,y)→(2,2)
[8]
3. Prove that
x2 + y
does not exist.
(x,y)→(0,0) x − y
[8]
lim
4. Consider the R2 − R function f defined by
f (x, y) = x2 − 2xy + y 2 .
Let V be the plane that is tangent to the graph of f at the point (x, y, z) = (1, −1, 4) .
(a) Find a vector that is perpendicular to the plane V .
(4)
(b) Find an equation for the plane V .
(4)
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MAT2615/103/3
(c) Find an equation for the line L that contains the point (1, −1, 4) and is perpendicular to the plane V .
(3)
[11]
5. Consider the R2 − R function f defined by
f (x, y) =
√
xy
2
and the R − R function r defined by
r (t) = e2t , cos t .
0
Use the General Chain Rule to determine the value of (f ◦ r) (0).
[7]
6. Consider the R2 − R function f defined by
f (x, y) = 2x + 3y 2 − x2 + y 3 .
Determine how many saddle points the function f has. (Make use of the Second Order Partial Derivatives
Test for Local Extrema.)
[9]
7. Use the Method of Lagrange to determine the largest possible volume that a rectangular block with a square
base can have if the height of the block plus the perimeter of its base equals 30cm.
[9]
8. Evaluate the line integral
Z
(xy + z) ds
C
3
where C is the line segment in R with initial point (1, 0 − 1) and endpoint (2, 1, 1).
[8]
9. Let D be the region in R3 that lies inside the sphere x2 + y 2 + z 2 = 2 and above the plane z = 1.
(a) Sketch the region D in R3 .
(3)
(b) Express the volume of D in terms of a triple integral, using spherical coordinates.
EVALUATE THE INTEGRAL.
10. Let S be that part of the surface z =
p
x2 + y 2 that lies between the plane z = 1 and the plane z = 3.
(a) Sketch the surface S, together with its XY-projection.
(b) Use a surface integral to determine the area of S.
11.
DO NOT
(7)
[10]
(3)
(8)
[11]
(a) State Stokes’ Theorem for converting a flux integral over a bounded surface to a line integral over the
curve that bounds the surface.
(4)
4
MAT2615/103/3
(b) Consider the surface
S = (x, y, z) | z = x2 + y 2 ; z ≤ 9 .
Sketch the surface S in R3 and show the curve that bounds S on your sketch. Then use Stokes’ Theorem
to evaluate the flux integral
ZZ
(curl F ) · n dS
S
where
F (x, y, z) = (y, −x, z) .
(8)
[12]
TOTAL: [100]
c
UNISA 2010/2012
5
MAT2615/103/3
October 2010 Exam Questions
MAT2615
CALCULUS IN HIGHER DIMENSIONS
Time: 2 Hours
Examiners:
First: Mr. R.J. de Beer
Second: Dr. J. Singleton
October/November 2010
100 Marks
Instructions to the candidate:
• The use of non-programmable calculators is allowed.
• This paper consists of 3 pages.
• Show ALL calculations.
• Try to answer ALL the questions.
This examination paper remains the property of the University of South Africa and may NOT be removed
from the examination venue.
1. L1 and L2 are lines in R3 that both contain the point (1, 0, 1). L1 is parallel to the vector (2, 4, 1) and L2 is
parallel to the vector (3, 1, −1). Find an equation for the plane that contains both the lines L1 and L2 .
[6]
2. Show that
y3
does not exist.
(x,y)→(0,0) x2
lim
[5]
3. Write down the integral below as a sum of double integrals in which the order of integration has been reversed.
Z
4
Z
4|x|
(2x + y) dydx.
−4
x2
Do not evaluate the integral.
[8]
6
MAT2615/103/3
4. Let D be the region in R3 below the plane z = 2 and above the surface z = x2 + y 2 − 2.
4.1 Make a 3-dimensional sketch of D.
(3)
4.2 Express the volume of D as a triple integral, using cylindrical co-ordinates.
(7)
Do not evaluate the integral.
[10]
5. Let V be the plane perpendicular to the vector (2, 2, 1) and containing the point (0, 0, 2). Let S be that part of V
in the first octant (i.e. where x ≥ 0, y ≥ 0 and z ≥ 0).
5.1 Find an equation for V .
(3)
5.2 Sketch S and its XY-projection.
(3)
5.3 Determine the downward flux of F through S, where
F = (x + z, y − z, z).
(11)
[17]
6. Consider the R2 − R function f defined by
f (x, y) = x3 − 3x + y 2 .
Find the critical points of f and determine their nature.
[8]
7. Use the method of Lagrange to determine the radius and height of a cylindrical tank that will have the maximum volume subject to the constraint that the area of its outer surface (top and bottom included) must be
600π m2 .
[10]
8. Let F (x, y) = (6x2 − y 2 , 2y − 2xy + 3) and C be the curve in the sketch below
Y
3
X
y = 2x − x2
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MAT2615/103/3
Use the Fundamental theorem of Line Integrals to evaluate the line integral
Z
F · r.
C
[11]
9. Let S be the sphere x2 + y 2 + z 2 = 1 and let
F (x, y, z) = (5x3 , 5y 3 , xy).
9.1 State Gauss’s Theorem.
(4)
9.2 Use this theorem to determine the outward flux of F through S.
(9)
[13]
10. Consider the surface S in R3 defined by
S = {(x, y, z) ∈ R3 : z = 4 − x2 − y 2 ; z ≥ 3}.
Let C be the boundary of S.
10.1 Sketch S, showing C clearly.
(2)
10.2 Give a parametrisation of C.
(2)
10.3 Assuming that S is oriented upward, use Stokes’s Theorem to evaluate
Z Z
curl F · n dS,
S
where F (x, y, z) = (y, −x, z).
(8)
[12]
TOTAL: [100]
c
UNISA 2010/2012
8
MAT2615/103/3
January 2010 Memorandum
1. (2, 4, −1) is parallel to the plane, as is
(3, 4, 3) − (1, 0, 2) = (2, 4, 1).
So the normal to the plane is given by
(2, 4, −1) × (2, 4, 1)
i
2
2
=
=
j
4
4
k
−1
1
8i − 4j + 0k = (8, −4, 0).
We know that (1, 0, 2) lies in the plane, so its equation is given by
x − (1, 0, 2) · (8, −4, 0) =
0
(x − 1, y, z − 2) · (8, −4, 0)
=
0
8x − 8 − 4y
=
0
2x − y
=
2.
2. Let > 0 be given.
|x − y| = |(x − 2) − (y − 2)| ≤ |x − 2| + |y − 2|
p
by the triangle inequality. Also, |x − 2| ≤ |x − 2|2 ≤ k(x − 2, y − 2)k, and similarly, |y − 2| ≤ k(x − 2, y − 2)k.
Therefore,
|x − y| ≤ 2k(x − 2, y − 2)k.
Set δ = /2. Then if 0 < k(x − 2, y − 2)k < δ, we have
|x − y| ≤ 2k(x − 2, y − 2)k < 2δ = .
3. On the curve C1 defined by x = 0, we compute the limit:
x2 + y
y
lim
=
lim
= −1.
(x,y)−→(0,0) x − y
(x,y)−→(0,0) −y
C1
C1
On the curve C2 defined by y = 0, we compute the limit:
x2
x2 + y
lim
=
lim
= 0.
(x,y)−→(0,0) x − y
(x,y)−→(0,0) x
C2
As the limits are not equal,
4.
C2
x2 + y
cannot have a limit at (0, 0).
x−y
(a) Let g(x, y, z) = x2 − 2xy − y 2 − z.
gradg is normal to the graph of f .
gradg(x, y, z) = (2x − 2y, 2y − 2x, −1). Therefore gradg(1, −1, 4) = (4, −4, 1) will be normal to the tangent
plane V .
(b)
(x, y, z) · (4, −4, −1)
=
(1, −1, 4) · (4, −4, −1)
4x − 4y − z
=
4.
(c) Such a line is parallel to (4, −4, −1), so its equation is:
(x, y, z) = (1, −1, 4) + t(4, −4, −1),
t∈R
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MAT2615/103/3
or
x =
4t + 1
y
=
−4t − 1
z
=
4−t
for t ∈ R.
5.
(f ◦ r)0 (0) = gradf (r(0)) · r0 (0).
We compute:
r0 (t)
=
(2e2t , − sin t)
r(0)
=
(1, 1)
0
=
r (0)
(2, 0)
1 −1/2 1/2 1 1/2 −1/2 x
y , x y
gradg(x, y) =
2
2
1 1
gradg(r(0)) =
, .
2 2
Therefore
(f ◦ r)0 (0) = (1/2, 1/2) · (2, 0) = 1.
6. For critical points, we require gradf (x, y) = (0, 0).
gradf (x, y)
(2 − 2x, 6y + 3y 2 ), so
=
2 − 2x =
6y + 3y
2
0
=
0.
Therefore, x = 1 and y = 0 or y = −2. The two critical points are (1, 0) and (1, −2).
For the second derivatives test, we calculate:
fxx (x, y)
= −2
fyy (x, y)
=
6y + 6
fxy (x, y)
=
0
∆(x, y)
= fxx fyy − (fxy )2 = −12y − 12.
Hence ∆(1, 0) = −12 < 0, so (1, 0) is a saddle point. ∆(1, −2) = 12 > 0, and so is not a saddle point.
Therefore the function f has one saddle point.
7. Let x be the base length, z the height. We must maximise g(x, z) = x2 z subject to the constraint
f (x, z) = z + 4x = 30.
We compute:
gradg(x, z)
gradf (x, z)
=
=
(2xz, x2 )
(4, 1)
Setting up the Lagrange multiplier equation,
gradg
2
(2xz, x )
= λgradf
= λ(4, 1)
x2 = λ.
2xz = 4λ
From the first equation, λ = xz/2, so
xz/2
z
= x2
=
2x.
But z + 4x = 30, so 6x = 30. Consequently, x = 5 and z = 10 and the largest possible volume is 250cm3 .
10
MAT2615/103/3
8. Parametrise the line as follows:
r(t)
=
(1, 0, −1) + t (2, 1, 1) − (1, 0, −1) ,
=
(1, 0, −1) + t(1, 1, 2)
=
(1 + t, t, 2t − 1).
t ∈ [0, 1]
To calculate the relationship between ds and dt:
r0 (t)
=
(1, 1, 2)
√
√
1+1+4= 6
kr (t)k =
√
ds = kr0 (t)kdt = 6dt.
0
Setting f (x, y, z) = xy + z, we get f (r(t)) = f (1 + t, t, 2t − 1) = t2 + 3t − 1. Now we can calculate the integral.
Z
Z
(xy + z) ds =
C
1
f (r(t))kr0 (t)kdt
0
=
√ Z
6
0
1
t2 + 3t − 1dt
√ 1 3 3 2
1
6 t + t −t 0
=
3
2
√ 5
=
6
6
5
= √ .
6
11
9.
MAT2615/103/3
(a) The sketch:
Z
√
2
1
√
Y
2
Z
(b) Calculating the bounds of φ with the XZ-projection:
Z
z=1
1 φ √2
X
Therefore the maximum value of φ, which is shown above, is
cos φ =
φ
=
1
√
2
π
.
4
Calculating the lower bound of ρ for a given angle φ:
Z
z=1
1
φ
ρ
X
cos φ
=
ρ
=
1
ρ
sec φ
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MAT2615/103/3
So:
0≤
θ
0≤ φ
0≤
therefore the volume is given by the integral
Z Z Z
Z
dV =
d
10.
ρ
2π
Z
2π
π
4
sec φ.
π/4
√
Z
0
0
2
ρ2 sin φ dρdφdθ.
sec φ
(a) The sketch:
Z
3
1
3
X
The XY -projection:
Y
3
1
R
X
(b)
f (x, y) =
∂f
=
∂x
∂f
=
∂y
p
x2 + y 2
1 2
(x + y 2 )−1/2 .2x
2
y(x2 + y 2 )−1/2 .
Hence
s
∂f 2
∂f 2
+
+1
∂x
∂y
s
=
=
x2
√
2.
x2
y2
+ 2
+1
2
+y
x + y2
Y
13
Z Z
Therefore
dS =
Z Z √
S
MAT2615/103/3
2 dR. Now in polar co-ordinates, R can be described as
R
R = {(r, θ) : 1 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}
so the integral is
Z Z √
Z
2 dR
Z
3
=
√
2.r dr dθ
1
√
Z 2π
2 23
=
r 1 dθ
2
0
Z 2π √
4 2 dθ
=
0
√
= 8 2π.
R
11.
2π
0
(a)
I
Z Z
curlF · n dS =
F · dr
C
S
where F is a C 1 -vector field defined on an open region in R3 containing S and S and C are piecewise
smooth, and n is the outward normal of the surface and C is positively oriented (anti-clockwise).
(b) The sketch:
Z
9
C
3
Y
X
C = {(x, y, z) : z = 9, x2 + y 2 = 9}.
To parametrise C, we use
r(t) = (3 cos t, 3 sin t, 9), where 0 ≤ t ≤ 2π.
0
Therefore r (t) = (−3 sin t, 3 cos t, 0), and
F (r(t)) = F (3 cos t, 3 sin t, 9) = (3 sin t, −3 cos t, 9).
Using Stokes’s Theorem,
Z Z
curlF · n dS
I
F · dr
=
S
C
Z 2π
=
F (r(t)) · r0 (t) dt
0
Z
2π
(3 sin t, −3 cos t, 9) · (−3 sin t, 3 cos t, 0) dt
=
0
Z
=
2π
−9 sin2 t − 9 cos2 t dt
0
= −18π.
14
MAT2615/103/3
October 2010 Memorandum
1. The normal to the plane is given by
(2, 4, 1) × (3, 1, −1)
i
2
3
=
j
4
1
k
1
−1
(−5, 5, −10).
=
Thus the equation of the plane is
(x, y, z) − (1, 0, 1) · (−5, 5, −10)
=
0
(x − 1, y, z − 1) · (−1, 1, −2)
=
0
x − y + 2z
=
3.
2. Consider the curves C1 given by y = x and C2 given by y = x2/3 , x ≥ 0.
y3
C1 →(0,0) x2
y3
lim
C2 →(0,0) x2
lim
=
lim x = 0
x→0
=
lim
C2 →(0,0)
1 = 1.
As these limits are unequal, the limit doesn’t exist.
3. The two curves y = x2 and y = 4|x| are drawn below.
Y
X
They intersect where
x2
=
4|x|
4
=
16x2
x2 (x2 − 16)
=
0.
x
Thus they intersect where x = 0 or ±4. So R is split into R1 and R2 where
Therefore
Z
4
Z
= {(x, y) ∈ R2 : 0 ≤ y ≤ 16, −
R2
=
{(x, y) ∈ R2
4|x|
Z
2x + y dydx =
−4
4.
4.1
x2
0
16
√
1
y ≤ x ≤ − y}
4
1
√
: 0 ≤ y ≤ 16, y ≤ x ≤ y}.
4
R1
Z
−y/4
√
− y
Z
16
√
Z
y
2x + y dxdy +
2x + y dxdy.
0
y/4
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MAT2615/103/3
Z
2
Y
2
X
2
4.2 D = {(r, θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2, r2 − 2 ≤ z ≤ 2}. Therefore the volume is
Z Z Z
Z
2π
Z
2
Z
2
r dzdrdθ.
1 dV =
D
5.
0
r 2 −2
0
5.1
(x, y, z) − (0, 0, 2) · (2, 2, 1)
=
0
(x, y, z − 2) · (2, 2, 1)
=
0
2x + 2y + z − 2
=
0.
5.2 S is the shaded triangle given below.
Z
2
Y
1
1
X
The XY -projection of this triangle is the following triangle:
Y
1
1
X
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MAT2615/103/3
5.3 The normal to S is the unit vector in the direction (−2, −, 2 − 1). The length of this vector is
p
(−2)2 + (−2)2 + (−1)2 = 3.
Therefore n =
1
(−2, −, 2 − 1).
3
1
(x + z, y − z, z) · (−2, −2, −1)
3
1
(−2x − 2z − 2y + 2z − z)
3
1
(−2x − 2y − z).
3
F ·n =
=
=
s
dS
∂z 2
∂z 2
+
+ 1 dA
∂x
∂y
=
√
=
=
4 + 4 + 1 dA
3dA.
Now we can calculate the flux by converting the surface integral to an ordinary double integral over R, the
XY -projection drawn in 5b). It is described in cartesian co-oridnates as
R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x}.
We compute:
Z Z
Z Z
F · n dS
−2x − 2y − z dA
=
S
R
Z
1Z
1−x
− 2 dydx
=
0
Z
0
1
−2y
=
0
Z
1−x
0
dx
1
−2(1 − x) dx
=
0
=
−2x + x2
1
0
= −1.
6. To find the critical points, we solve grad f = (3x2 − 3, 2y) = (0, 0). Therefore
x = ±1, y = 0.
Thus the critical points are (1, 0) and (−1, 0). For the second derivative test we compute
2
∆ = fxx fyy − fxy
= 6x.2 − 0 = 12x.
∆(1, 0) = 12 > 0, so (1, 0) is an extreme point. As fxx (1, 0) = 6 > 0, it is a local minimum.
∆(−1, 0) = −12 < 0, so (−1, 0) is a saddle point.
7. We minimise the function g(r, h) = πr2 h subject to the constraint f (h, r) = 2πr2 + 2πrh = 2πr(r + h) = 600π.
grad g(r, h)
=
(2πrh, πr2 )
grad f (r, h)
=
(4πr + 2πh, 2πr).
Now Lagrange’s method requires that we solve
grad g = λgrad f.
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MAT2615/103/3
Therefore
2πrh = λ(4πr + 2πh)
πr2
= λ(2πr).
Hence
rh = λ(2r + h)
r
=
(1)
2λ
and consequently
2λh =
2
λ(4λ + h)
4λ − 2λh + λh =
0
4λ2 − λh =
0
λ(4λ − h)
=
0.
Thus λ = 0 or λ = h/4. As λ = 0 is not possible, we see that λ = h4 . Therefore
h
r
=
4
2
λ=
where we used equation (1). So
h = 2r.
2
Using the fact that f (h, r) = 2πr + 2πrh = 2πr(r + h) = 600π,
2πr(r + 2r)
=
600π
2
=
300
r
=
10 m
h =
20 m.
3r
8. We seek an R2 − R function, call it f , such that grad f = F .
fx
=
6x2 − y 2
fy
=
2y − 2xy + 3.
Therefore f (x, y) = 2x3 − xy 2 + g(y) where g is some function of y that we now compute.
fy = −2xy + g 0 (y) = 2y − 2xy + 3
and so g 0 (y) = 2y + 3 and g(y) = y 2 + 3y + c. Therefore
f (x, y) = 2x3 − xy 2 + y 2 + 3y.
[Note that the constant term c is unnecessary].
The Fundamental Theorem says that
Z
F · dr = f (b) − f (a)
C
where b is the endpoint and a is the starting point of the curve C.
For the end point, if x = 3 then y = −3, so b = (3, −3) and a = (0, 0). Therefore
Z
F · dr = f (3, −3) − f (0, 0)
C
=
27 − 0
=
27.
18
9.
MAT2615/103/3
9.1 Let D be a closed, bounded region in R3 with a piecewise smooth outer surface S, oriented outward. If
F = (M, N, P ) is a smooth 3-dimensional vector field defined on S, then
Z Z Z
Z Z
divF dV =
F · n dS.
D
S
9.2 We will give two calculations to answer this question. Either are acceptable.
• Using cylindrical co-ordinates:
divf = 15x2 + 15y 2 .
Z Z Z
Z Z Z
Z Z
divf dV =
15(x2 + y 2 ) dV.
F · n dS =
V
S
V
We write this in cylindrical co-ordinates.
V = {(r, θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, −
p
p
1 − r2 ≤ z ≤ 1 − r2 }.
Therefore
Z Z
Z
F · n dS
2π
1
Z
√
Z
=
S
0
Z
−
0
2π
0
2π
Z
drdθ
30r3 (1 − r2 )1/2 drdθ
0
2π
1
Z
−10r2 − 3r(1 − r2 )1/2 drdθ
=
0
Z
√
1−r 2
√
− 1−r 2
1
Z
=
0
15r3 dzdrdθ
1−r 2
15r3 z
=
Z
√
1
Z
0
1−r 2
0
2π
=
− 10r2 (1 − r2 )3/2
0
Z
=
2π
−4(1 −
0+
0
Z
1
r2 )5/2 0
1
0
Z
1
−20r(1 − r2 )3/2 dr dθ
−
0
dθ
2π
=
4 dθ
0
=
8π.
• Using spherical co-ordinates:
divf = 15(x2 + y 2 ) = 15ρ2 sin2 φ.
Z Z
Z Z Z
Z Z Z
F · n dS =
divf dV =
15ρ2 sin2 φ dV.
S
V
V
We write this in spherical co-ordinates.
V = {(θ, φ, ρ) : 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π, 0 ≤ ρ ≤ 1}.
19
MAT2615/103/3
Therefore
Z Z
2π
Z
F · n dS
π
Z
1
Z
15ρ4 sin3 φ dρdφdθ
=
S
0
0
0
2π
Z
π
Z
3ρ5 sin3 φ
=
0
0
3 sin3 φ dφdθ
=
0
0
2π
Z
dφdθ
π
2π Z
Z
1
0
π
Z
3(1 − cos2 φ) sin φ dφdθ
=
0
0
2π
Z
3 sin φ − 3 cos2 φ sin φ dφdθ
=
0
2π
Z
−3 cos φ + cos3 φ
=
0
2π
Z
2 − (−2) dθ
=
0
=
8π.
10. 10.1
Z
4
3
C
1
Y
1
X
10.2 If z = 3 then 1 = x2 + y 2 , so the parametrisation of this circle is
r(t) = (cos t, sin t, 3).
10.3
Z Z
Z
curl F · n dS =
F · dr.
C
S
Now
F (r(t))
0
r (t)
=
(− sin t, cos t, 3)
=
(− sin t, cos t, 0).
Therefore
F (r(t)) · r0 (t) = sin2 t + cos2 t = 1
and consequently
Z
Z
F · dr =
C
2π
1 dt = 2π.
0
π
0
dθ