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Partial Fractions and Integration

TMA 02 for MST125 Essential mathematics 2, covering Units 6, 7 and 8, to be submitted by 26 June 2019.

22 Pages4730 Words305 Views
   

Added on  2022-11-27

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This document explains the concept of partial fractions and integration. It provides step-by-step solutions for finding the partial fractions and evaluating the integral of a rational function. The domain and intercepts of the given function are also determined.

Partial Fractions and Integration

TMA 02 for MST125 Essential mathematics 2, covering Units 6, 7 and 8, to be submitted by 26 June 2019.

   Added on 2022-11-27

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Q 1.
Answer
a)
i)
g(x, y) = D[x, y] = (2x, 7y)
By simple inspection, the transformation matrix A is found as
D =
[2 0
0 7
]
This is verified as
g(x, y) =
[2 0
0 7
]
·
[x
y
]
=
[2x
7y
]
D is a scaling matrix.
ii)
h(x, y) = B · [x, y] = (x, 3x + y)
By simple inspection, the transformation matrix B is found as
B =
[1 0
3 1
]
This is verified as
h(x, y) =
[1 0
3 1
]
·
[x
y
]
=
[ x
3x + y
]
B is a shearing matrix with y shearing factor of 3.
iii)
k(x, y) = C · [x, y] = (y, x)
By simple inspection, the transformation matrix C is found as
C =
[0 1
1 0
]
1
Partial Fractions and Integration_1
This is verified as
k(x, y) =
[0 1
1 0
]
·
[x
y
]
=
[y
x
]
C is a reflection matrix through line y = x.
b)
The composition transformation required is
f = k h g
Applying the transformation to a vector x in R2. Therefore,
f (x) = (k h g)(x) = k(h(g(x)) = C(B(D · x))) = CBD(x)
Since, f (x) = A · x, we have
A = CBD
Matrix A is evaluated by matrix multiplication in the following way
A = CBD
=
[0 1
1 0
]
×
[1 0
3 1
]
×
[2 0
0 7
]
=
[0 1
1 0
]
×
[2 0
6 7
]
=
[6 7
2 0
]
c)
A transformation is invertible if the matrix of transformation is invertible,
that is the matrix has a non zero determinant (Clark and Mccune, 2013).
A is the matrix of transformation f . It’s determinant is
det(A) = 6 7
2 0 = (6 × 0) (2 × 14) = 14
2
Partial Fractions and Integration_2
Therefore, A is invertible.
Inverse of a 2d invertible matrix
M =
[a b
c d
]
is given by
M 1 = 1
det(M )
[ d b
c a
]
Therefore, inverse of A, which represents the transformation f 1 is
A1 = 1
14
[ 0 7
2 6
]
d)
Let C be a image in R2 to which the transformation f maps the unit circle C.
C = f (C)
Taking inverse on both sides
f 1(C) = C
Thus the inverse transformation maps the points on C back to the unit circle
C: x2 + y2 = 1.
Now, for a vector x in R2
f 1(x) = A1 · x = 1
14
[ 0 7
2 6
]
·
[x
y
]
= 1
14
[ 7y
2x + 6y
]
This transformation maps back to the unit circle, therefore
(
1
14(7y)
)2
+
(
1
14(2x + 6y)
)2
= 1
Simplifying
49y2 + 36y2 24xy + 4x2 = 142
or
4x2 24xy + 85y2 = 142
3
Partial Fractions and Integration_3
is the required equation of the image f (C).
e)
Area of the image f (C) is obtained using a simple property of transformation
Area of f (C) = ||det(A)|| × Area of C
where, A is the matrix of transformation of f .
For the given transformation f , determinant of transformation matrix was
calculated (refer c) as: det(A) = 14, which implies ||det(A)|| = 14. Since
the area of unit circle C is π, therefore
Area of f (C) = 14 × π = 14π
Q 2.
Answer
a)
i)
f maps points (0,0), (1,0) and (0,1) into points (3,3), (3,4) and (4,3).
Let, x’ represent the transformed coordinates of point x. Therefore,
x’ = f (x) = A · x + a
Let A be any arbitrary matrix
A =
[a1 a2
b1 b2
]
and a be a vector [c1, c2]T .
For the transformation (0, 0) (3, 3), we get
(3, 3)T = A · (0, 0)T + (c1, c2)T
Implies, (c1, c2)T = (3, 3)T .
For the transformation (1, 0) (3, 4), we get
(3, 4)T =
[a1 a2
b1 b2
]
· (1, 0)T + (3, 3)T
4
Partial Fractions and Integration_4
Implies, a1 + 3 = 3 = a1 = 0 and b1 + 3 = 4 = b1 = 1.
For the transformation (0, 1) (4, 3), we get
(4, 3)T =
[0 a2
1 b2
]
· (0, 1)T + (3, 3)T
Implies, a2 + 3 = 4 = a2 = 1 and b2 + 3 = 3 = b2 = 0.
Therefore, the transformation f is completely determined as
f =
[0 1
1 0
]
· (x, y)T + (3, 3)T
ii)
The are no fixed points visible in the transformation f . The transformation
is combination of translation, which happens 3 units along the line y = x,
as given by a and then reflection in the line y = x, as given by matrix A,
therefore, it is a case of glide-reflection.
b)
Reflection in the line y = x + 5.
To achieve this transformation, the idea is to translate everything to the ori-
gin, reflect about the line y = x and then move everything back.
Let us translate everything to origin using the transformation
x’ = x + h
where, h = (0, 5)T , is the translation vector.
Since we are at the origin, reflect in the line y = x. This is achieved using
transformation matrix
A =
[ 0 1
1 0
]
Therefore, the resulting transformation is
A(x + h) =
[ 0 1
1 0
] (
x +
[ 0
5
])
=
[ 0 1
1 0
]
· x +
[5
0
]
The final step is to move everything back to the original location which
is achieved by translating x by 5. This is achieved by the inverse vector
5
Partial Fractions and Integration_5
h1 = (0, 5)T . Therefore, the final transformation k, is
k = A(x + h) + h1 =
[ 0 1
1 0
]
· x +
[5
0
]
+
[0
5
]
=
[ 0 1
1 0
]
· x +
[5
5
]
︷︷
Ax+b
which is the required form of the transformation k.
Q 3.
2x3 3x2 18x + 17
x2 3x 4
Answer
a) Partial fractions
Since, for the given rational function, the degree of numerator is greater than
the degree of denominator, the function is first simplified by polynomial di-
vision in the following way
2x + 3
x2 3x 4
)
2x3 3x2 18x + 17
2x3 6x2 8x
3x2 10x + 17
3x2 9x 12
x + 29
Therefore, the rational function is simplified as
2x3 3x2 18x + 17
x2 3x 4 = 2x + 3 + 29 x
x2 3x 4
In this simplified, partial fractions of only the last term (rational) has to be
determined.
6
Partial Fractions and Integration_6

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