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THE CHINESE UNIVERSITY OF HONG KONG
Department of Mathematics
MATH1510 Calculus for Engineers (Fall 2016)
Suggested solutions of coursework 1
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1a - i
2c
4a
1a - ii
2d
4b
1b
3a - i
2a
3a - ii
2b
3b
Total
/100 pts
Please attempt to solve all the problems. Your solutions of problems 1 - 4 are to be submitted,
a part of which will be graded. We strongly recommend that you study problems 5 - 6, though
you are not required to submit their solutions.
1. Answer the following questions:
(a) Consider the following functions :
√
f (x) = x + 1
and
g(x) = x2 − 2.
i. Find the formulas explicitly describing
f
(f + g)(x), (f − g)(x), (f g)(x) and
(x)
g
and state the domains of the functions.
ii. Find the domain of f ◦ g.
(b) Consider the two functions:
f (x) = |x|
Find the ranges of g ◦ f and f ◦ g.
and
g(x) = cos x.
(a)(i).
√
(f + g)(x) = f (x) + g(x) = x + 1 + (x2 − 2)
√
=
x + 1 + x2 − 2
Domain of (f + g)(x) : [−1, +∞).
√
Since the real-valued function x is only defined on [0, +∞).
√
(f − g)(x) = f (x) − g(x) = x + 1 − (x2 − 2)
√
=
x + 1 − x2 + 2
Domain of (f − g)(x) : [−1, +∞).
(f g)(x) = f (x)g(x) =
√
x + 1(x2 − 2)
Domain of (f g)(x) : [−1, +∞).
√
x+1
f (x)
= 2
(f /g)(x) =
g(x)
x −2
√
√
Domain of (f /g)(x) : [−1, 2) √
∪ ( 2, ∞).
√
√
2
That is because x − 2 = (x + 2)(x − 2) and only 2 ∈ [−1, +∞).
(a)(ii).
(f ◦ g)(x) = f (x2 − 2) =
p
(x2 − 2) + 1 =
√
x2 − 1
Dg = R and Df = [−1, ∞). So, domain of f ◦ g is given by
g(x) = x2 − 2 ≥ −1,
i.e., (−∞, −1] ∪ [1, ∞).
(b).
(g ◦ f )(x) = g(|x|) = cos |x|
Note that the range of f is [0, ∞), and the range of g on domain [0, ∞) is
[−1, 1] (see Figure 2i), so the range of g ◦ f is [−1, 1].
(f ◦ g)(x) = f (cos x) = | cos x|
Note that the range of g is [−1, 1], and the range of f on domain [−1, 1] is
[0, 1] (see Figure 2ii), so the range of f ◦ g is [0, 1].
f +g
f −g
10
4
2
8
0
−2
6
4
y
y
−4
−6
−8
2
−10
−12
0
−14
−2
−2
−1
0
1
x
2
3
−16
−2
4
−1
0
1
x
2
3
4
2
3
4
f
g
fg
40
10
35
8
30
6
25
4
20
y
y
2
15
10
0
−2
5
−4
0
−6
−5
−8
−10
−2
−10
−2
−1
0
1
x
2
3
4
−1
Figure 1: Graphs of Q1ai
0
1
x
y = cos x
w hen x ≥ 0
2
y = |x|
2
w hen − 1 ≤ x ≤ 1
1.5
1.5
1
1
y
0.5
y
0
0.5
−0.5
0
−1
−0.5
−1.5
−2
−2
−1
0
1
x
2
3
4
−1
−1.5
5
−1
−0.5
Figure 2i
0
4
4
3
3
2
2
1
1
0
0
−1
−1
−2
−2
−3
−3
−4
−4
−3
−2
−1
0
x
1.5
f ◦g
5
y
y
g◦f
−4
1
Figure 2ii
5
−5
−5
0.5
x
1
2
3
4
5
−5
−5
−4
Figure 2 : Graphs of Q1b
−3
−2
−1
0
x
1
2
3
4
5
2. Here are a few useful trigonometric identities:
sin A
;
cos A
• sin (A ± B) = sin A cos B ± cos A sin B;
• tan A =
• cos (A ± B) = cos A cos B ∓ sin A sin B
tan A ± tan B
• tan (A ± B) =
;
1 ∓ tan A tan B
• 2 sin(A) cos(B) = sin (A + B) + sin (A − B);
• 2 cos(A) cos(B) = cos (A + B) + cos (A − B);
• 2 sin(A) sin(B) = cos (A − B) − cos (A + B);
• sin2 A + cos2 A = 1;
• 1 + cot2 A = cosec2 A;
• tan2 A + 1 = sec2 A;
• cos(2A) = 1 − 2 sin2 A = 2 cos2 A − 1 = cos2 A − sin2 A;
• sin(2A) = 2 sin(A) cos(A);
1 − cos(2A)
;
2
1 + cos(2A)
;
cos2 A =
2
A+B
A−B
sin A + sin B = 2 sin
cos
;
2
2
A+B
A−B
cos
;
sin A − sin B = 2 sin
2
2
A+B
A−B
cos A + cos B = 2 cos
cos
;
2
2
A−B
A+B
sin
.
cos A − cos B = −2 sin
2
2
• sin2 A =
•
•
•
•
•
Note that sin2 A (which should be distinguished from sin A2 ) is the notation used
for (sin A)2 [which 6= sin (A2 ) in general]. Similarly, cos2 A means (cos A)2 .
Use the identities on the previous page to solve the following problems :
(a) Find the exact value of cos θ if sin θ = 0.2 and θ ∈ (π/2, π) (θ in radian).
(b) Given that cos(θ) = 2/3 (θ in radians), find sin(θ − π/2).
(c) It is well-known that
1
cos(60◦ ) = sin(30◦ ) = ;
2√
3
;
cos(30◦ ) = sin(60◦ ) =
√2
2
cos(45◦ ) = sin(45◦ ) =
.
2
Find the exact value of
(sec 10◦ )(sin 50◦ + sin 70◦ ).
(d) Show that
tan A + cot A
= csc 2A
2
for all A ∈ 0, π2 .
(a).
Note θ satisfies cos2 θ + sin2 θ = 1.
1
= 1
25
24
cos2 θ =
25√
cos2 θ +
cos θ = ±
Since θ ∈ (π/2, π), cos θ < 0, cos θ = −
(b).
24
5
√
24
.
5
Note that sin (A − B) = sin A cos B − cos A sin B, so
sin(θ − π/2) =
=
=
=
sin θ cos(π/2) − cos θ sin(π/2)
sin θ × 0 − cos θ × 1
− cos θ
−2/3
(c).
Since sin A + sin B = 2 sin
=
=
=
=
=
=
(d).
A+B
2
cos
A−B
2
(sec 10◦ )(sin 50◦ + sin 70◦ )
◦
◦
50 − 70◦
50 + 70◦
◦
cos
)
(sec 10 )(2 sin
2
2
(cos 10◦ )−1 (2 sin 60◦ cos(−10◦ ))
(cos 10◦ )−1 (2 sin 60◦ cos(10◦ ))
2 sin√
60◦
3
2×
√ 2
3
Note that by the identities
2 sin(A) cos(A) = sin(2A)
and
cos2 θ + sin2 θ = 1
1 sin A cos A
tan A + cot A
=
(
+
)
2
2 cos A sin A
1 sin2 A + cos2 A
=
(
)
2 sin A cos A
1
=
2 sin A cos A
1
=
sin 2A
= csc A
3. Answer the following questions:
(a) For each of the following sequences, fill the table and guess what lim an is.
n→∞
i. an = 3 + (−0.9)
n
an
for n ≥ 1.
1
n cos n o
,
n
ii. an =
n
n
an
2
3
10
100
3
10
100
for n ≥ 1.
1
2
(b) A sequence {an } is defined recursively by the following equations:
(
a1 = a2 = 1
√
an = 2an−1 − an−2 for n ≥ 3.
Find a5 .
(a)i.
n 1
2
3
10
100
an 2.1 3.81 2.271 3.3487(4 dp) 3.0000(4 dp)
lim (3 + (−0.9)n ) = lim 3 + lim (−0.9)n = 3 + 0 = 3
n→∞
n→∞
n→∞
because | − 0.9| < 1.
ii. (The answers in the table are corrected to 4dp)
n
1
2
3
10
100
an 0.5403 −0.2081 −0.3300 −0.0839 0.0086
Note that −1 ≤ cos n ≤ 1 for all n ≥ 1, so we have −
lim −
n→∞
1
1
= lim = 0,
n n→∞ n
by Sandwich Theorem, we have lim an = lim
n→∞
(b)
cos n
1
1
≤
≤ . Since
n
n
n
n→∞
cos n
= 0.
n
√
√
2a2 − a1 = 2 − 1
a3 =
√
√ √
√
a4 =
2a3 − a2 = 2( 2 − 1) − 1 = 1 − 2
√
√
√
√
a5 =
2a4 − a3 = 2(1 − 2) − ( 2 − 1) = −1
3(a)i.
a n = 3 + (−0.9) n ,
f or
n≥1
3(a)ii.
4
an =
1
cos n
,
n
f or
n≥1
3.8
0.8
3.6
0.6
3.4
0.4
0.2
an
3
2.8
0
−0.2
2.6
−0.4
2.4
−0.6
2.2
−0.8
2
0
5
10
15
20
25
30
n
35
40
3(b)
45
−1
50
an =
√
3
4
5
2a n −1 − a n −2 ,
10
f or
15
20
n≥3
2
1.5
1
0.5
an
an
3.2
0
−0.5
−1
−1.5
−2
1
2
5
n
6
7
8
Figure 2: Graphs of Q3
9
10
25
n
30
35
40
45
50
Definition 1 If v = a1 i + b1 j + c1 k and v = a2 i + b2 j + c2 k are two vectors, the dot
product v · w is defined as
v · w = a1 a2 + b1 b2 + c1 c2 .
(1)
Definition 2 If v = a1 i + b1 j + c1 k, the norm of a vector is given by
q
||v|| = a21 + b21 + c21 .
(2)
Theorem 1 If u and v are two nonzero vectors, the angle θ, 0 ≤ θ ≤ π, between u
and v is determined by the formula
cos θ =
u·v
.
||u||||v||
(3)
Definition 3 If v = a1 i + b1 j + c1k and w = a2 i + b2 j + c2 k are two vectors in space,
the cross product v × w is defined as the vector
v × w = (b1 c2 − b2 c1 )i − (a1 c2 − a2 c1 )j + (a1 b2 − a2 b1 )k.
(4)
Theorem 2 Let u and v be vectors in space.
u × v is orthogonal to both u and v.
(5)
||u × v|| = ||u||||v|| sin θ, where θ is the angle between u and v.
(6)
||u × v|| is the area of the parallelogram having u 6= 0 and v 6= 0
as adjacent sides.
u × v = 0 if and only if u and v are parallel.
(7)
(8)
4. Answer the following questions:
(a) Find the angle (in degree, 2 decimal places) between the vectors
u = j + 2k
and
v = −i + 3j.
(b) Suppose a plane P contains the points
A = (0, −1, 1), B = (0, 2, 3) and C = (−1, 1, 0).
Find a vector n which is perpendicular to P . Such a vector is called a normal
vector of the plane P .
(a). Let θ be the angle between u and v,
u·v
||u|| ||v||
(0, 1, 2) · (−1, 3, 0)
p
=√
2
1 + 02 + 22 (−1)2 + 32 + 02
3
=√ √
5 10
3
= √
5 2
3
−1
√
⇒ θ = cos
5 2
= 64.90◦
cos θ =
(b).
−→
u = AB = (0, 2, 3) − (0, −1, 1)
= (0, 3, 2)
−→
Let v = AC = (−1, 1, 0) − (0, −1, 1)
= (−1, 2, −1)
Clearly, u, v are parallel to P
n=u×v
= (−3 − 4)i − (0 + 2)j + (0 + 3)k
= −7i − 2j + 3k
Let
which is perpendicular to u and v, hence to the plane P .
Remarks:
1. Note that for any nonzero real number λ, λn is also perpendicular to P .
2. u · n = 0(a scalar) and v · n = 0
5. (Optional) A fourth-degree polynomial in x such as
3x4 + 5x3 + 4x2 + 3x + 1
contains all the powers of x from the first through the fourth. However, any polynomial can be written without powers of x. Evaluating a polynomial without powers of
x (Horner’s method) is somewhat easier than evaluating a polynomial with powers.
Answer the following questions :
(a) Show that
(((3x + 5)x + 4)x + 3)x + 1 = 3x4 + 5x3 + 4x2 + 3x + 1
is an identity.
(b) Rewrite the polynomial
P (x) = 6x5 − 3x4 + 9x3 + 6x2 − 8x + 12
without powers of x as in 5(a).
(c) For which form did you perform fewer arithmetic operations?
(a)
(((3x + 5)x + 4)x + 3)x + 1
= ((3x2 + 5x + 4)x + 3)x + 1
= (3x3 + 5x2 + 4x + 3)x + 1
= 3x4 + 5x3 + 4x2 + 3x + 1
(b)
(6x4 − 3x3 + 9x2 + 6x − 8)x + 12
= ((6x3 − 3x2 + 9x + 6)x − 8)x + 12
= (((6x2 − 3x + 9)x + 6)x − 8)x + 12
= ((((6x − 3)x + 9)x + 6)x − 8)x + 12
(c) Few arithmetic operations were done in finding P (x) using Horner’s
method. For example, using Horner’s method, we obtain
P (2) = ((((6 · 2 − 3)2 + 9)2 + 6)2 − 8)2 + 12 = 236.
By substitution, we get
P (2) = 6(2)5 − 3(2)4 + 9(2)3 + 6(2)2 − 8(2) + 12 = 236.
6. (Optional) If a football is kicked straight up with an initial velocity of 128 ft/sec
from a height of 5 ft, then its height above the earth is a function of time given by
h(t) = −16t2 + 128t + 5.
What is the maximum height reached by this ball?
Let us write
h(x) = ax2 + 2bx + c
as
2
ac − b2
b
,
+
h(x) = a x +
a
a
a 6= 0.
b
64
Since − = −
= 4, the maximum height is h(4) = 261 ft.
a
−16