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Main points of this past exam are: Improved Euler Method, Double Integral, Fourth Quadrants, Polar Coordinates, Partial Derivatives, Arbitrary Function, Maximum and Minimum Values, Method of Undetermined Coefficients, Strut of Length
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Answer FIVE questions. All questions carry equal marks.
Examiners: Mr. T. Corcoran Prof. P. O’Donoghue Mr. T. O Leary
2 y 10 y(0) 3 dx
dy − = =
(ii) By using the Tree Term Taylor Method or the Improved Euler Method with
a step of 0.1 estimate the value of y at x=0.1. (12 marks)
(b) By evaluating a double integral locate the centroid of the region in the first and
fourth quadrants bounded by the lines y=2x, y=-2x and the circle x
2 +y
2 =20. (8 marks)
x=1,y=2 is given by
(y 2) ... 2
(x 1) 2(x 1)(y 2) 2
f(x, y) 3(x 1) (y 2)
2 2 = − − − − − + − − − − + (8 marks)
(b) Cartesian coordinates x and y are related to polar coordinates r and θ by the formulae
θ= (^)
x
y tan
1 r=
2 2 x +y
If stress T=g(r,θ ) is an arbitrary function in r and θ write down the relationships
between the partial derivatives of T with respect to x and y and those with r and θ.
Hence simplify as much as possible the expression
y
y x
x (7 marks)
(c) Find the maximum and minimum values of V=x
3 -12xy+6y
2 +48. (5 marks)
(i) Solve the differential equation
9x 90 x(0)= 12 x(0)= 0 dt
dx 6 dt
d x 2
2
(ii) Find the general solution of the differential equation
2x 2
2 8y 10e dx
dy 6 dx
d y − + = (14 marks)
(b) The deflection y at any point on a strut of length L is found by solving the
differential equation
wx ωy dx
d y 2 2
2
ω
Solve this differential equation where y is zero at x=0 and the slope of y is zero
at the midpoint of the strut. Show that the strut fails, that is, y becomes infinitely
large when P
2 = (^2)
π
. (6 marks)
(i) s 5 s 7 s 3
3 2
(ii) s(s 6s 8)
4s 24 2
(12 marks)
(b) By using Laplace Transforms solve the differential equations
25y 160e y(0) y(0) 0 dt
dy 6 dt
d y t 2
2
p(x)= 20
6x 3x
2
is an acceptable probability density function. Find P(0
For a function f(t) the Laplace Transform of f(t) is a function in s defined by
F(s) e f(t)dt
st
0
−
∞
f(t) F(s)
A=constant A
s
t
s
N + 1
e
at (^1)
s −a
sinhkt k
s k
2 2 −
coshkt s
s k
2 2 −
s ω
2 2
s
2 2
e f(t)
at (^) F(s-a)
f (t) ′ sF(s)-f(0)
f (t) ′′ (^) s F(s)^2 − sf(0) − f (o)′
f(u)du 0
t
F(s)
s
f(u)g(t u)du 0
t
F(s)G(s)
U(t-a) (^) e
s
-as
f(t-a)U(t-a) (^) e −asF(s)
δ ( t − a) e -as
Note: coshA
e e
2
sinhA
e e
2
A A A A
=
− −