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The University of Sydney
MATH1111 Introduction to Calculus
Semester 1
1.
Longer Solutions to Week 4 Exercises (Thurs/Fri)
(i) If a is the length of the adjacent side to θ then
1
3
= cos θ = a6 , so that a =
2017
6
3
= 2.
(ii) If tan θ = 3 then we can choose a√right-angled triangle with adjacent side 1 and
opposite side 3, with hypotenuse 10 (by Pythagoras), so that
3
sin θ = √
10
1
sin θ = √ .
10
and
(iii) If cos θ = 23 then we can choose a right-angled triangle with adjacent side 2 and
√
hypotenuse 3, with opposite side 5 (by Pythagoras), so that
√
5
5
and tan θ =
.
sin θ =
3
2
√
6
10
3
3
θ
θ
2.
√
5
θ
a
1
2
Assuming the string is approximately straight and forms the hypotenuse of a rightangled triangle, the height of the kite above the ground is
50 × sin 50◦ ∼
= 38 m (to the nearest metre).
3.
We have a right-angled triangle with base 1, 000 metres and angle 24◦. The opposite
side length represents the height of the Empire State Building, which we estimate to
be
1, 000 × tan 24◦ ∼
= 450 m (to the nearest 10 metres).
4.
The graphs for parts (i) to (viii) are as follows:
y
f (x) = x2 + 1
domain = R, range = [1, ∞)
1
x
1
y
f (x) = x2 − 2x + 3 = x2 − 2x + 1 + 2 = (x − 1)2 + 2
domain = R, range = [2, ∞)
2
1
x
y
f (x) =
1
x
domain = range = R\{0}
1
−1
1
−1
x
y
f (x) =
1
x−1
domain = R\{1}, range = R\{0}
1
1
x
−1
y
f (x) =
1
x+1
domain = R\{−1}, range = R\{0}
1
−1
x
y
f (x) =
√
x
domain = range = [0, ∞)
1
1
x
2
y
f (x) =
√
x−1
x
1
y
f (x) =
√
x+1
domain = [−1, ∞), range = [0, ∞)
1
−1
5.
domain = [1, ∞), range = [0, ∞)
x
To get the number of apples, we multiply the density of trees by the number of apples
per tree, and use the trick of completing the square:
A(n) = n(3000 − 10n) = −10(n2 − 300n) = −10(n − 150)2 + 10(150)2 ,
which is maximised when n = 150 and takes maximimum value 10(150)2 = 225, 000.
Thus the model predicts that 150 trees should be planted per hectare, to produce a
mamimum yield of 225, 000 apples per hectare.
6.
The graphs for parts (i) to (iv) are as follows:
y
f (x) = sin x
1
− 3π
2
π
2
−1
x
2π
The domain of f (x) = sin x is R and the range is [−1, 1].
y
f (x) = cos x
−π
−1
1
π
2
3
2π
x
The domain of f (x) = cos x is R and the range is [−1, 1].
y
f (x) = tan x
−7π
2
−5π
2
−3π
2
π
2
−π
2
3π
2
5π
2
7π
2
x
sin x
is only defined when cos x 6= 0, and this occurs precisely when x
The value tan x = cos
x
avoids odd multiples of π/2. Thus the domain of f (x) = tan x is R\{± π2 , ± 3π
, ± 5π
, . . .}.
2
2
All real values are achieved by tan x, so the range is R.
y
f (x) =
1π
−π
2
2
3π
2
5π
2
−1
1
cos x
7π
2
x
The value cos1 x is only defined when cos x 6= 0, and this occurs precisely when x avoids
odd multiples of π/2. Thus the domain of f (x) = cos1 x is R\{± π2 , ± 3π
, ± 5π
, . . .}.
2
2
Reciprocals of cos x vary over all reals whose magnitude is at least 1, so the range is
(−∞, −1] ∪ [1, ∞).
7.
(i) Observe that
3
5
+
= 2 holds if and only if
x x+2
2 =
3x + 6 + 5x
8x + 6
3(x + 2) + 5x
=
=
,
x(x + 2)
x(x + 2)
x(x + 2)
which is equivalent to
8x + 6 = 2x(x + 2) = 2x2 + 4x and x 6= 0, −2 ,
yielding
0 = 2x2 + 4x − 8x − 6 = 2x2 − 4x − 6 = 2(x2 − 2x − 3) = 2(x − 3)(x + 1)
with solutions x = 3 and x = −1.
(ii) Observe that the equation
0 = x4 − 5x2 + 4 = (x2 − 4)(x2 − 1) = (x + 2)(x − 2)(x + 1)(x − 1)
holds if and only if one of the factors on the right is zero, so that either x + 2 = 0,
x − 2 = 0, x + 1 = 0 or x − 1 = 0, with roots ±2 and ±1.
4
√
(iii) Suppose that 2x − 1 = − 2 − x holds. Squaring both sides yields
(2x − 1)2 = 2 − x ,
which becomes
4x2 − 4x + 1 = 2 − x ,
so that
0 = 4x2 − 3x + 1 = (4x + 1)(x − 1) .
The roots of this quadratic are x = − 14 and x = 1. But only the first can satisfy
the original equation, because 2x − 1 has to be negative or zero. Hence x = − 14
is the only solution.
8.
(i) Observe that
x2 + y 2 − 2x − 4y − 11 = (x2 − 2x) + (y 2 − 4y) − 11
= (x2 − 2x + 1) − 1 + (y 2 − 4y + 4) − 4 − 11
= (x − 1)2 + (y − 2)2 − 16
so that the equation becomes
(x − 1)2 + (y − 2)2 = 16 = 42
with centre (1, 2) and radius 4.
(ii) Observe that
4x2 + 4y 2 − 16x − 24y + 51 = 4(x2 − 4x) + 4(y 2 − 6y) + 51
= 4(x2 − 4x + 4) − 16 + 4(y 2 − 6y + 9) − 36 + 51
= 4(x − 2)2 + 4(y − 3)2 − 1
so that the equation becomes
4(x − 2)2 + 4(y − 3)2 = 1 ,
which in turn becomes
1
=
(x − 2) + (y − 3) =
4
2
2
2
1
,
2
with centre (2, 3) and radius 21 .
(iii) Observe that
x2 + y 2 + 10y + 24 = x2 + (y 2 + 10y + 25) − 25 + 24
= x2 + (y + 5)2 − 1
so that the equation becomes
x2 + (y + 5)2 = 1 = 12
with centre (0, −5) and radius 1.
5
(iv) Observe that
x2 + y 2 + 2x + 2y + 2 = (x2 + 2x) + (y 2 + 2y) + 2
= (x2 + 2x + 1) − 1 + (y 2 + 2y + 1) − 1 + 2
= (x + 1)2 + (y + 1)2
so that the equation becomes (x + 1)2 + (y + 1)2 = 0 = 02 with centre (−1, −1)
and radius 0, so that the circle is degenerate (a single point coinciding with its
centre).
9.
The trigonometric identities tell us
cos 2θ = cos2 θ − sin2 θ = cos2 θ − (1 − cos2 θ) = 2 cos2 θ − 1 ,
so that
cos2 θ =
cos 2θ + 1
.
2
Hence
cos 15◦ =
and so
r
vq
sp
u
p√
u 1− 1 +1
2
◦
◦
t
cos 30 + 1
1 − sin 30 + 1
3+2
4
=
=
=
,
2
2
2
2
sin 15 =
◦
√
1 − cos2 15◦ =
s
1−
√
3+2
=
4
p
2−
2
√
3
.
Hence
tan 15◦ =
sin 15
=
cos 15◦
◦
√
√
2− 3
√ 2√
2+ 3
2
=
s
6
s
√
√
√
2− 3
(2 − 3)2
√ =
= 2− 3.
4−3
2+ 3