UNIVERSITY OF DUBLIN TRINITY COLLEGE

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UNIVERSITY OF DUBLIN
XMA1S121
TRINITY COLLEGE
Faculty of Engineering, Mathematics
and Science
school of mathematics
JF Natural Science
JF Human Genetics
JF Medicinal Chemistry
JF Physics and Chemistry of Advanced
Materials
JF Chemistry with Molecular Modelling
Trinity Term 2010
Maths 1S12
Dr. D. Kitson and Dr. C. Ó Dúnlaing
Attempt 6 questions: 3 from each section.
Log tables and statistical tables are available from the invigilators, if required.
Non-programmable calculators are permitted for this examination,—please indicate the make
and model of your calculator on each answer book used.
Page 2 of 10
XMA1S121
SECTION I
1. (a) Use the adjoint matrix method to invert


3 4


5 6
(b) Use Cramer’s Rule to solve
3x + 4y + z = 6
x + 2y − z = 2
2x + y + 2z = 1
(c) Let P = (2, 3, 1), Q = (3, 5, 4), and R = (5, 5, 2). Compute P~Q × P~R and hence
give an equation for the plane through P, Q, and R.
2. Let W3 = (3, 4, 0), and let H be the plane through O perpendicular to OW3 . In this
question, you will compute the matrix for perpendicular reflection in H.
(a) Give the matrix A0 for (perpendicular) reflection in the xy-plane.
(b) Compute an orthonormal basis X1 , X2 , X3 where X3 is on the axis OW3 .
(c) Hence compute the matrix for perpendicular reflection in H, the plane through O
perpendicular to OW3 .
3. (a) Compute the least-squared-error quadratic curve fitting the data
(0, 1), (1, 5), (2, 8), (3, 16)
(b) Compute the least-squared-error straight line fitting the same data.
4. (a) State the formula for the Binomial distribution B(n, p).
(b) A jar contains 30 white balls and 20 black. The following trial is repeated several
times: a ball is selected at random from the jar, its colour is noted, and the ball is
returned to the jar.
Page 3 of 10
XMA1S121
Compute the probability that a white ball is extracted 7 times and a black ball 3
times in a series of 10 trials.
(c) Compute the sample mean and sample standard deviation for the following set of
data
5.03, 5.42, 4.51, 6.30, 5.35, 5.75, 4.50, 6.78, 6.11, 4.18
(d) Given that the above data gives 10 measurements from a normal distribution with
mean µ, use Student’s t-distribution to give a symmetric 95% confidence interval
for µ.
Page 4 of 10
XMA1S121
SECTION II
1. Compute the following integrals
(a)
Z
x2 sin 2x dx
Solution: Use integration by parts twice or use tabular integration. We use tabular
integration with
p(x) = x2 ,
g(x) = sin 2x
Derivatives of p(x) Integrals of g(x)
x2
sin 2x
2x
− 12 cos 2x
2
− 14 sin 2x
1
8
0
cos 2x
Forming products and adding (with alternating + and − signs) we get
Z
1
1
1
x2 sin 2x dx = −x2 cos 2x + 2x sin 2x + 2 cos 2x + c
2
4
8
(b)
Z
√
x2
dx
9 − x2
Solution: Use the trigonometric substitution x = 3 sin θ.
√
9 − x2 = 3 cos θ
dx = 3 cos θ dθ
Z
x2
√
dx =
9 − x2
9 sin2 θ 3 cos θ
dθ
3 cos θ
Z
= 9
Z
sin2 θ dθ
1
1
= 9 [θ − sin 2θ] + c
2
2
√
9
x 9 − x2
−1 x
=
sin ( ) −
+c
2
3
3
3
Page 5 of 10
XMA1S121
(c)
∞
Z
2
2x e−x dx
0
Solution:
∞
Z
−x2
2x e
dx =
0
=
lim
Z
lim
Z
b→∞
b→∞
b
2
2x e−x dx
0
b
−eu du
0
lim [−eu ]b0
i
h
2 b
= lim −e−x
b→∞
0
h
i
2
= lim −e−b + e0
=
b→∞
b→∞
= 1
(d)
x2 + 2x + 1
dx
(x2 + 1)2
Z
Solution: Write
x2 + 2x + 1
A1 x + B1 A2 x + B2
=
+ 2
2
2
(x + 1)
(x2 + 1)
(x + 1)2
Then
x2 + 2x + 1 = (A1 x + B1 )(x2 + 1) + (A2 x + B2 )
= A1 x3 + B1 x2 + (A1 + A2 )x + (B1 + B2 )
=⇒ A1 = 0,
Z
B1 = 1,
x2 + 2x + 1
dx =
(x2 + 1)2
A2 = 2,
Z
B2 = 0
Z
1
2x
dx +
dx
2
2
(x + 1)
(x + 1)2
1
= tan−1 x − 2
+c
(x + 1)
2. (a) Solve the initial value problem
dy
= 3x2 e−y ,
dx
y(1) = 0
Solution:
Z
y
e dy =
Z
3x2 dx
Page 6 of 10
XMA1S121
=⇒ ey = x3 + c
From the initial condition, c = 0.
y = ln |x3 |
(b) Find a general solution for the linear differential equation
x
dy
cos x
= 2 − 3y,
dx
x
x>0
Solution: We need to find an integrating factor.
dy 3
cos x
+ y= 3
dx x
x
Let
p(x) =
3
x
q(x) =
cos x
x3
Then
Z
p(x) dx = 3 ln x + c
Let
µ(x) = e3 ln x = x3
Then
Z
1
y(x) =
µ(x) q(x) dx
µ(x)
Z
1
= 3
cos x dx
x
sin x
c
=
+ 3
3
x
x
(c) Using Euler’s method with increment 4x = 0.2 find the first three approximations
for the solution to the initial value problem
y 0 = 3xy + 2y,
y(0) = 2
Solution: The x-values are
x0 = 0,
x1 = 0.2,
x2 = 0.4,
x3 = 0.6
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XMA1S121
The approximate y-values are
y0 = 2,
yn+1 = yn + f (xn , yn )4x
where f (x, y) = 3xy + 2y.
y1 = 2 + f (0, 2)(0.2)
= 2 + 4(0.2) = 2.8
y2 = 2.8 + f (0.2, 2.8)(0.2)
= 2.8 + (7.28)(0.2)
= 4.256
y3 = 4.256 + f (0.4, 4.256)(0.2)
= 4.256 + 2.72384
= 6.97
3. (a) Determine if the following sequences converge or diverge
)∞
(r
∞
3n
sin n
n+1
n
n=1
n=1
Solution: The first sequence converges.
lim
3n
3
= lim
n+1
1+
=⇒ lim
r
1
n
=3
√
3n
= 3
n+1
The second sequence also converges by the “Sandwich Theorem”.
−
1
sin n
1
≤
≤
n
n
n
=⇒ lim
sin n
=0
n
Page 8 of 10
XMA1S121
(b) Test the following series for convergence
∞
X
n=1
∞ n
X
1
n
n=1
n
2n + 1
Solution: The first series diverges since
lim
1
n
= lim
2n + 1
2+
1
n
=
1
6= 0
2
The second series converges by the nth-root test
lim
1
=0<1
n
(c) Find the radius of convergence for the power series
∞
X
(3x)n
n!
n=1
Solution: Applying the ratio test
lim
(3x)n+1 n!
3x
an+1
= lim
= lim
=0
n
an
(n + 1)! (3x)
n+1
and so the radius of convergence is ∞.
(d) Find the first four terms of the Maclaurin series for
1
x+1
Solution: Writing f (x) =
1
,
x+1
f (0) = 1
1
=⇒ f 0 (0) = −1
(x + 1)2
2
f 00 (x) =
=⇒ f 00 (0) = 2
(x + 1)3
6
f 000 (x) = −
=⇒ f 000 (0) = −6
4
(x + 1)
f 0 (x) = −
The first four terms of the Maclaurin series are
f (0) + f 0 (0)x +
f 00 (0) 2 f 000 (0) 3
x +
x
2!
3!
= 1 − x + x2 − x3
Page 9 of 10
XMA1S121
4. (a) Find the slope of the tangent line to the ellipse
x = 3 cos(4t)
y = 2 sin(4t)
π
.
16
at the point where t =
Solution:
dy
8 cos(4t)
=−
dx
12 sin(4t)
At t =
π
16
the slope is − 23 .
(b) Compute the length of the cardioid
r = 1 + cos θ
Solution:
Length =
Z
=
Z
2π
s
r2 +
0
2π
dr 2
dθ
dθ
√
2 + 2 cos θ dθ
Z 2π
θ
2
| cos
| dθ
2
0
Z π
θ
4
cos
dθ
2
0
π
θ
8 sin
2 0
8
0
=
=
=
=
(c) Show that the function
u(x, t) = cos (x + ct)
is a solution to the wave equation
2
∂2u
2 ∂ u
=c
∂t2
∂x2
Solution:
∂u
= −c sin(x + ct)
∂t
∂2u
= −c2 cos(x + ct)
∂t2
Page 10 of 10
∂u
= − sin(x + ct)
∂x
∂2u
= − cos(x + ct)
∂x2
c UNIVERSITY OF DUBLIN 2011
XMA1S121
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