The University of Sydney
School of Mathematics and Statistics
Assignment 1: Solutions and Marking Scheme
MATH1001: Differential Calculus
Summer School, 2015
General feedback: The assignment was completed well by most students. The median
mark was 41 out of 50, and the mean was 37.72 with a standard deviation of 9.04.
Presentation was an issue for many students; the most common issues are
• folders which are difficult to open,
• illegible handwriting,
• pages stapled out of order or not stapled at all,
• important working not shown,
• questions not set out correctly,
• incorrect notation,
• incorrect cover sheet.
All of these are issues which can impact your final mark.
Question 1: Mostly well answered; no major recurring problems.
Question 2: Many students got confused by the Arg z term. The best way to approach
this question was to sketch the two regions seperately, then combine them. This way there
is no ambiguity as to how Arg z is measured. Many students also got the boundaries
incorrect; dashed boundaries are not included, full boundaries are.
Question 3: Many students got this correct, which was pleasing. Most students got
n = 2m reasonably easy, although many labelled this “equating real and imaginary parts”
when it is “equating modulus and argument”. Students also often had poor arguments
as to why the answer they found is the correct one; although this is to be expected, some
attempt should be made to justify your reasoning.
Question 4: Mostly well answered; no major recurring problems.
Question 5: Mostly well answered. Some students are still unsure how to write domain
and ranges for functions of two variables.
Question 6: Mostly well answered. Some students only checked a few paths; this does
not prove the limit exists! There may be another path to check that does NOT give the
same limit. A polar form argument, or an algebraic argument, were the best solutions
for this question.
Question 7: Badly answered by most students. Finding a global maximum or minimum is a very different procedure to finding a local maximum or minimum, and this
should be revised by many students to ensure they do not do a lot of incorrect working
in the future. Very few students correctly checked the boundaries and corners.
Question 8: A worryingly large number of students did not know what a differential
is. Part b) was well answered by many students, although most subtle points of the
c 2015 The University of Sydney
Copyright 1
question ignored by many (including the fact that the change in r and h can be positive
or negative).
Question 9: This was a very simple question, but some students took a very different
approach. For questions like this, the formulas from class can be applied immediately
and outright.
Question 10: This question was very well answered by most students, which was very
impressive. The main complaint would be that many figures were too small, but otherwise
the question was very well done. Students who got the question wrong mostly seemed not
to understand the premise, which is disappointing; students were given nearly a month
to prepare this assignment, which is more than enough time to ask a tutor or lecturer
how to understand the question.
1. Calculate the five fifth roots of i.
Solution: We wish to solve z 5 = i. Writing both sides in polar form,
π
(reiθ )5 = r5 e5iθ = e 2 i .
Thus r5 = 1 and 5θ =
π
2
+ 2kπ for k ∈ N.
θ=
So θ =
(1)
π π 9π −3π −7π
, , , 10 , 10 ,
10 2 10
π
2
(1 + 4k)π
+ kπ =
.
10 5
10
and
π
π
9π
z = e 10 , e 2 , e 10 , e
−3π
10
,e
−7π
10
.
(1 mark for writing i in exponential form, 1 mark for r = 1, 1 mark for any solution
for z, 1 mark for all solutions for z. 4 marks)
2. Sketch the region in the complex plane described by
π
3π
< Arg z <
and |z − 1 − i| ≤ 1 .
z ∈ C 8
8
Solution: See final page for sketch.
(1 mark for identifying the circle, 1 mark for identifying the segment, 1 mark for
combining correctly, 1 mark for correct boundaries. 4 marks total.)
3. Find the smallest
√ positive non-zero integers n,m such that
(1 − i)n = (1 + 3i)m . (Hint: Convert to polar exponential form.)
Solution: 1 − i =
√
√
π
π
2e− 4 i , 1 + 3i = 2e 3 i . So
√ n nπ
√
mπ
(1 − i)n = 2 e− 4 i , (1 + 3i)m = 2m e 3 i .
For these to be equal, their moduli should be equal, and the arguments equal up to a
√ n
√ 2
√ 2
multiple of 2π. So 2 = 2m = ( 2 )m = 2 m. Thus n = 2m.
2
For the arguments,
nπ
mπ
=−
+ 2πk for some k ∈ N.
3
4
But n = 2m, so substituting and rearranging,
2πk =
mπ mπ
5m
+
, so k =
.
2
2
12
But both k and m must be integers, so m must be a multiple of 12 (or k would not be
an integer). The √
smallest such value of m is m = 12, so n = 24, and we can confirm that
(1 − i)2 4 = (1 + 3i)1 2.
(1 mark for both polar forms correct, 1 mark for correctly applying index laws, 1
mark for n = 2m, 1 mark for correct relationship for arguments including stating k ∈ N,
2 marks for final correct answer. 6 marks total.)
4. Using sin θ =
1
(eiθ
2i
− e−iθ ) and the binomial theorem,
1
(a) Show that sin3 θ = (3 sin θ − sin 3θ).
4
Rπ
(b) Thus calculate 0 (1 + sin2 θ) sin θ dθ.
Solution: Using the
binomial theorem
3
3
3
3
(and noting that
= 1,
= 3,
= 3,
= 1),
0
1
2
3
3
1
eiθ − e−iθ
2i
−1 3iθ
e − e2iθ e−iθ + eiθ e−2iθ + e−3iθ
8i
1
1 3iθ
3 iθ
−3iθ
−iθ
− (e − e
) + (e − e )
4
2i
2i
1
(3 sin θ − sin 3θ)
4
3
sin (θ) =
=
=
=
(2)
(3)
(4)
(5)
So
Z
π
π
Z
2
sin θ + sin3 θdθ
(1 + sin θ) sin θ dθ =
0
(6)
0
Z
π
3
1
sin θ + sinθ − sin 3θdθ
4
4
0
π
7
1
= − cos θ +
cos 3θ
4
12
0
10
=
3
=
(7)
(8)
(9)
(1 mark for correct use of binomial theorem, 1 mark for correctly reversing the identity
for sin, 1 mark for correct integral in part b, 1 mark for fully correct solution. 4 marks
total)
3
5. Define a surface by z =
p
x2 + y 2 + 1.
(a) State the domain and range.
(b) Draw the level curves for z = 1, z = 2, and z = 3.
(c) Sketch the surface, showing any local maxima or minima.
(d) What is the shortest distance from the surface to the point (x, y, z) = (4, 6, 0)?
(Hint: follow example 7.6b from the notes.)
Solution: a) Domain is all (x, y) ∈ R2 , range is [1, ∞).
b), c) see pictures on final page. Note this is NOT a parabaloid; the cross sections
are not parabolas. In fact, it is a hyperboloid. Marks were not removed for making this
mistake.
d) Let the distance be s.
p
s = (x − x0 )2 + (y − y0 )2 + (z − z0 )2
where (x0 , y0 , z0 ) = (4, 6, 0) and (x, y, z) is some point on the surface. We want the
shortest distance from the point to the surface, so we want to minimise s. Define t = s2 .
If s is minimised, so is t, so we can work with t instead, which is simpler.
t = (x−x0 )2 +(y −y0 )2 +(z −z0 )2 = (x−4)2 +(y −6)2 +z 2 = (x−4)2 +(y −6)2 +x2 +y 2 +1.
p
(because (x, y, z) is on the surface, so z = x2 + y 2 + 1). We wish to minimise this.
∂t
∂t
= 2(x − 4) + 2x,
= 2(y − 6) + 2y.
∂x
∂y
Setting both of these to zero for critical points, we find that x = 2, y = 3 is a critical
point.
2
The determinant is D = fxx fyy − fxy
= 16 > 0, and fxx = 4 > 0, so this is a minimum
√
point.
√ Thus the closest point on the surface is (2, 3, 14) and the minimum distance is
s = 27 units by substitution.
(1 mark for domain and range, 1 mark for all three level curves, 1 mark for correct
shape for surface, 1 mark for correctly identifying and sketching the local minimum, 1
mark for correct formula for distance, 1 mark for correct working for maxima/minima, 1
mark for final correct distance. 7 marks total.)
6. Consider the limit
x4 − y 4
.
(x,y)→(0,0) x2 + y 2
lim
Does this limit exist? If it does, give the limit. You should justify your answer.
Solution: In polar coordinates x = r cos θ, y = r sin θ,
x4 − y 4
r4 cos4 θ − r4 sin4 θ
r4 (cos4 θ − sin4 θ)
=
lim
=
lim
= lim r2 (cos4 θ − sin4 θ).
r→0 r 2 cos2 θ + r 2 sin2 θ
r→0
r→0
(x,y)→(0,0) x2 + y 2
r2
lim
Now, by a squeeze law argument or similar, this is equal to zero as r → 0.
4
Alternatively, x4 − y 4 = (x2 − y 2 )(x2 + y 2 ). So
x4 − y 4
= lim x2 − y 2 = 02 − 02 = 0.
(x,y)→(0,0) x2 + y 2
(x,y)→(0,0)
lim
(1 mark for correct limit, 1 mark for using polar coordinates, 1 mark for correct simplification, 1 mark for squeeze law argument. If straight lines are used, 1 mark for correctly
substituting y = mx, 1 mark for correct conclusion. Note that this argument is not
complete, so does not warrant full marks. 4 marks total.)
7. Consider a square metal plate, which has been heated. Let
f (x, y) = 100(x − x2 )(y − y 2 ) be the temperature in degrees Celsius of the
plate at the point (x, y). The corners of the square metal plate are given by
the points (0, 0), (1, 0), (0, 1) and (1, 1). Which point on the plate has the
highest temperature? Fully justify your answer.
Solution: fx = 100(1 − 2x)(y − y 2 ), fy = 100(x − x2 )(1 − 2y). For interior critical
points, fx = fy = 0. fx = 0 if x = 12 or y = 0 or y = 1, and fy = 0 if y = 21 , x = 0 or
x = 1. So the critical points are ( 12 , 12 ), (0, 0), (0, 1), (1, 0) and (1, 1). Only ( 21 , 12 ) is not
on the boundary, so this is the only interior critical point.
The boundaries are x = 0, x = 1, y = 0, y = 1. On all of these boundaries, f (x, y) = 0
and fx = fy = 0. So all the boundaries are critical points.
The corners are also critical points; these are (0, 0), (0, 1), (1, 0) and (1, 1).
Now we need to check the value of f (x, y) at all our critical points.
1 1
f ( , ) = 25, f (x, 0) = f (x, 1) = f (0, y) = f (1, y) = 0,
2 2
f (0, 0) = f (0, 1) = f (1, 0) = f (1, 1) = 0.
Thus we can conclude that 25 degrees celcius is a global maximum of our function on
the domain occuring at ( 12 , 12 ), and is the highest temperature.
(1 mark for identifying ( 12 , 21 ) as a critical point, 1 mark for identifying the boundaries
correctly, 1 mark for checking boundaries, 1 mark for checking corner points, 1 mark for
correctly identifying that boundaries are constant, 1 mark for final correct conclusion. 6
marks total.)
8. Recall that the volume of a cylinder is V = πr2 h, where r is the radius and h
is the height.
(a) Write down the differential dV in terms of dr and dh.
(b) If the radius r changes by 5%, and the height h changes by 3%, find the
maximum possible change of the volume V as a percentage.
Solution:
a)
dV = 2πrhdr + πr2 dh.
b) If r changes by 5%, then |dr| = 0.05r (as the change can be positive or negative,
and
dV the change is a fraction of the original r). Similarly, |dh| = 0.03h. We want to find
, the percentage change of V .
V
5
dV
2πrhdr + πr2 dh
=
V
πr2 h
2dr dh
+
=
r
h
(10)
(11)
So
dV
V
≤ | 2dr | + | dh |
r
h
= 2 × 0.05 + 0.03
= 0.13
(12)
(13)
(14)
So the maximum percentage change in V is approximately 13%.
(1 mark for dV . 1 mark for identifying that we need dV
. 1 mark for correct formulas
V
dV
for dr and dh. 1 mark for correctly forming V . 1 mark for final correct answer. 5 marks
total.)
9. Suppose you are climbing a hill whose shape is given by the equation
z = 1000 − 5x2 − 10y 2 , where x, y, and z are measured in metres. You are
standing at the point x = 10, y = 5, z = 250.
(a) In what direction is the slope largest?
(b) What is the slope in that direction?
(c) In which directions is the slope 0?
Solution:
a) Slope is largest in the direction ∇f .
∇f = −10xi − 20yj.
So ∇f (10, 5) = −100i − 100j is the direction of largest magnitude.
b) The slope in
√ |.
√ this direction is |∇f
2
2
|∇f (10, 5)| = 100 + 100 = 100 2.
c) The slope is 0 in directions perpendicular to ∇f .
So any multiple of 100i − 100j.
In particular, i − j and j − i.
(1 mark for direction of largest magnitude, 1 mark for slope in that direction, 1 mark
for a direction with 0 slope, 1 mark for both directions with zero slope. 4 marks total.)
10. (a) Compute the equation of the tangent line to the curve y = f (x), where
f (x) = ex at (x, y) = (0, 1).
(b) Another way to think about a tangent line to the curve y = f (x) at a point
x = c is that it is a linear function g(x) = αx + β such that
• g(c) = f (c) (that is, the tangent line touches the curve at x = c), and
• g 0 (c) = f 0 (c) (that is, the tangent line has the same slope as the curve at
x = c).
6
Verify that these two statements are true for the tangent line you calculated
in part (a).
(c) Extend this idea by considering a quadratic equation h(x) = αx2 + βx + γ
such that
• h(c) = f (c), and
• h0 (c) = f 0 (c), and
• h00 (c) = f 00 (c) (that is, the second derivatives match at x = c).
Using the same function f (x) and c = 0 as above, calculate α, β, and γ so
that these three conditions are satisfied. Write down the function h(x).
(d) Sketch the curve y = f (x), the tangent line you found in part (b), and the
quadratic you found in part (c) on the same axis.
(Note: this is an example of a Taylor polynomial, which will be discussed in
the final lectures.)
Solution:
a) f 0 (x) = ex , f 0 (0) = 1. So the tangent line is y − 1 = 1(x − 0), or y = x + 1.
b) From the previous part, g(x) = x + 1 and c = 0.
f (0) = e0 = 1, g(0) = 0 + 1 = 1, so f (0) = g(0).
f 0 (0) = e0 = 1, g 0 (0) = 1, so f 0 (0) = g(0).
So the conditions are satisfied.
c) If h(x) = αx2 + βx + γ, h0 (x) = 2αx + β and h00 (x) = 2α. So h(0) = γ, h0 (0) = β
and h00 (0) = 2α.
We know that f (0) = f 0 (0) = f 00 (0) = 1. So the three conditions read
• h(c) = f (c) =⇒ γ = 1, and
• h0 (c) = f 0 (c) =⇒ β = 1, and
• h00 (c) = f 00 (c) =⇒ 2α = 1 =⇒ α =
1
2
.
Thus h(x) = 12 x2 + x + 1.
d) See final page
(1 mark for correct tangent line. 1 mark for correctly showing that f (0) = g(0),
f 0 (0) = g 0 (0). 1 mark for correctly calculating two of α, β, γ. 1 mark for correctly
calculating all three and writing down h(x). 1 mark for correct shape for plots, 1 mark
for showing that all three are tangent at (0, 1). 6 marks total.)
7
5c)
4.5
4
3.5
3
2.5
2
3
2
1.5
1
0
1
−1
3
2
1
−2
0
−1
−3
−2
−3
10d)
8
25
20
y=ex
y=x+1
y
15
y=0.5*x2+x+1
10
5
0
−5
−3
−2
−1
0
x
9
1
2
3