AIOU Solved Assignment Code 1309 Mathematics III 2023 Assignment 1.pptx

ALLAMA IQBAL OPEN UNIVERSITY ISLAMABAD
(Department of Mathematics)
Course: Mathematics-III (1309)
Semester: Spring, 2023
Level: Intermediate
ASSIGNMENT No. 1
Q. 1(a) Integrate: ∫
𝒅𝒙
𝟏+𝑺𝒊𝒏 𝒙+𝐜𝐨𝐬𝒙
To integrate the expression ∫ dx / (1 + sin(x) + cos(x)), we can use a
substitution method to simplify the integral.
Skilling.pk Diya.pk Stamflay.com
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Let's make the substitution u = tan(x/2). This substitution will help us
simplify the integral expression.
First, we need to express sin(x) and cos(x) in terms of u:
sin(x) = 2u / (1 + u²)
cos(x) = (1 - u²) / (1 + u²)
Next, we need to express dx in terms of du:
dx = (2 / (1 + u²)) du
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Now we can substitute these expressions back into the integral:
∫ dx / (1 + sin(x) + cos(x)) = ∫ (2 / (1 + u²)) du / (1 + (2u / (1 + u²)) + ((1 - u²)
/ (1 + u²)))
Simplifying the denominator:
= ∫ (2 / (1 + u²)) du / (2 + 2u + 1 - u²)
= ∫ (2 / (1 + u²)) du / (3 + 2u - u²)
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Now we can factor the denominator:
= ∫ (2 / (1 + u²)) du / (3 - (u - 1)(u + 3))
Now we can perform a partial fraction decomposition on the integral:
= ∫ (A / (u + 3)) du + ∫ (B / (u - 1)) du
Multiplying both sides by the denominator (u + 3)(u - 1):
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2 = A(u - 1) + B(u + 3)
Expanding and collecting like terms:
2 = (A + B)u + (3B - A)
Setting the coefficients of like powers of u equal to each other:
A + B = 0
3B - A = 2
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Solving these equations, we find A = -2/5 and B = 2/5.
Substituting these values back into the integral expression:
∫ dx / (1 + sin(x) + cos(x)) = ∫ (-2/5) / (u + 3) du + ∫ (2/5) / (u - 1) du
Integrating each term:
= (-2/5) ln|u + 3| + (2/5) ln|u - 1| + C
Now, we need to substitute back for u:
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To integrate the expression ∫ √x / (1 + ∛x) dx, we can use a substitution
method to simplify the integral.
Let's make the substitution u = ∛x. This substitution will help us simplify
the integral expression.
First, we need to find the derivative of u with respect to x:
du/dx = (1/3)x^(-2/3)
Next, we can solve for dx in terms of du:
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dx = 3u^2 du
Now we can substitute these expressions back into the integral:
∫ √x / (1 + ∛x) dx = ∫ (√x / (1 + u)) (3u^2 du)
Simplifying:
= 3 ∫ (u^2√x) / (1 + u) du
= 3 ∫ (u^2 * u^3) / (1 + u) du
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= 3 ∫ (u^5) / (1 + u) du
Now we can perform a partial fraction decomposition on the integral:
= 3 ∫ (u^5) / (1 + u) du
= 3 ∫ (u^4 - u^3 + u^2 - u + 1 - 1) / (1 + u) du
Expanding the integrand and rewriting the integral:
= 3 ∫ (u^4 - u^3 + u^2 - u + 1 - 1) / (1 + u) du
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= 3 ∫ (u^4 - u^3 + u^2 - u + 1) / (1 + u) du - 3 ∫ du / (1 + u)
Now we can integrate each term separately:
= 3 (∫ (u^4 - u^3 + u^2 - u + 1) / (1 + u) du - ∫ du / (1 + u)
For the first integral, we can use the power rule to integrate each term.
For the second integral, it simplifies to a natural logarithm.
= 3 (u^5/5 - u^4/4 + u^3/3 - u^2/2 + u - ln|1 + u|) + C
Now, we need to substitute back for u:
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= 3 (∛x^5/5 - ∛x^4/4 + ∛x^3/3 - ∛x^2/2 + ∛x - ln|1 + ∛x|) + C
Therefore, the integral of √x / (1 + ∛x) dx is given by:
∫ √x / (1 + ∛x) dx = 3 (∛x^5/5 - ∛x^4/4 + ∛x^3/3 - ∛x^2/2 + ∛x - ln|1 + ∛x|)
+ C, where C is the constant of integration.
Q. 2 (a)Solve differential equation:
(4y – cos y) 𝒅𝒚
–3x2 = 0y(0) = 0
𝒅𝒙
To solve the differential equation (4y - cos(y)) dy/dx - 3x^2 = 0 with the
initial condition y(0) = 0, we can use the method of separation of variables.
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Rearranging the equation, we have:
(4y - cos(y)) dy = 3x^2 dx
Now we can separate the variables and integrate both sides:
∫ (4y - cos(y)) dy = ∫ 3x^2 dx
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Integrating the left-hand side involves integrating both the terms separately.
The integral of 4y with respect to y is 2y^2, and the integral of cos(y) with
respect to y is sin(y):
2y^2 - sin(y) = x^3 + C
where C is the constant of integration.
Now, we can use the initial condition y(0) = 0 to determine the value of the
constant C:
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2(0)^2 - sin(0) = 0 + C
0 - 0 = C
C = 0
Substituting C = 0 back into the equation, we have:
2y^2 - sin(y) = x^3
Therefore, the solution to the given differential equation with the initial
condition is:
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2y^2 - sin(y) = x^3
(b) Integrate sec2
(2x 1)dx
To integrate ∫ sec^2(2x-1) dx, we can use the integral formula for the
derivative of tan(x):
∫ sec^2(x) dx = tan(x) + C
Let's make a substitution to simplify the integral.
Let u = 2x - 1. Then du/dx = 2, and dx = du/2.
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Now, we can substitute u and dx in terms of du:
∫ sec^2(2x-1) dx = ∫ sec^2(u) (du/2)
Using the integral formula mentioned earlier, we have:
∫ sec^2(u) (du/2) = (1/2)tan(u) + C
Substituting back for u:
(1/2)tan(u) + C = (1/2)tan(2x - 1) + C
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Therefore, the integral of ∫ sec^2(2x-1) dx is (1/2)tan(2x - 1) + C, where C is
the constant of integration.
Q. 3(a) Find the point trisecting the join of A (2, 3), B (8, 9)
(b) Find equation of line passing through mid-point of A(2, 3), B(7, 9)
and perpendicular to the line 3x + 2y + 3 = 0.
(a) To find the point that trisects the line segment joining A(2, 3) and
B(8, 9), we can divide the segment into three equal parts.
Let the trisecting point be P(x, y). We can find the coordinates of P using
the following formula:
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P = (1/3)(2A + B)
P = (1/3)(2(2, 3) + (8, 9))
= (1/3)(4, 6 + 8, 9)
= (1/3)(12, 15)
= (4, 5)
Therefore, the point (4, 5) trisects the line segment joining A(2, 3) and
B(8, 9).
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(b) To find the equation of the line passing through the midpoint of
A(2, 3) and B(7, 9) and perpendicular to the line 3x + 2y + 3 = 0, we
can follow these steps:
Find the midpoint of A and B:
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Midpoint = ((2 + 7)/2, (3 + 9)/2)
Midpoint = (9/2, 12/2)
Midpoint = (4.5, 6)
Determine the slope of the given line 3x + 2y + 3 = 0 by rearranging the
equation:
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2y = -3x - 3
y = (-3/2)x - (3/2)
The slope of this line is -3/2.
The line perpendicular to this line will have a slope that is the negative
reciprocal of -3/2. So the slope of the perpendicular line is 2/3.
Using the point-slope form of a line, we can write the equation using the
midpoint (4.5, 6) and the perpendicular slope (2/3):
y - y₁ = m(x - x₁)
y - 6 = (2/3)(x - 4.5)
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y - 6 = (2/3)x - 3
y = (2/3)x + 3
Therefore, the equation of the line passing through the midpoint of A(2, 3)
and B(7, 9) and perpendicular to the line 3x + 2y + 3 = 0 is y = (2/3)x + 3.
Q. 4 (a)Find circumcenter of the triangle with vertices, (4, 3), (–2, –5) and
(–6, –1)
(b) Find equation of two parallel lines perpendicular to 3x – 5y + 70 = 0
and are such that the product of x–intercept and y-intercept is 20.
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(a) To find the circumcenter of a triangle with vertices (x₁, y₁), (x₂, y₂),
and (x₃, y₃), we can use the following steps:
Find the midpoints of two sides of the triangle. Let's call them (x₄, y₄) and
(x₅, y₅). The midpoints are calculated as follows:
Midpoint of side 1 = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Midpoint of side 2 = ((x₁ + x₃)/2, (y₁ + y₃)/2)
Midpoint of side 1 = ((4 + (-2))/2, (3 + (-5))/2)
= (1, -1)
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Midpoint of side 2 = ((4 + (-6))/2, (3 + (-1))/2)
= (-1, 1)
Find the slopes of the two sides of the triangle. The slopes are calculated as
follows:
Slope of side 1 = (y₂ - y₁)/(x₂ - x₁)
= (-5 - 3)/(-2 - 4)
= -8/-6
= 4/3
Slope of side 2 = (y₃ - y₁)/(x₃ - x₁)
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= (-1 - 3)/(-6 - 4)
= -4/-10
= 2/5
Find the perpendicular slopes of the two sides. The perpendicular slopes are
the negative reciprocals of the slopes calculated in step 2.
Perpendicular slope of side 1 = -1/(4/3)
= -3/4
Perpendicular slope of side 2 = -1/(2/5)
= -5/2
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Find the equations of the two perpendicular bisectors. The perpendicular
bisectors are the lines passing through the midpoints calculated in step 1
and having slopes calculated in step 3.
Equation of perpendicular bisector 1 passing through (1, -1):
y - y₄ = (m₁)(x - x₄)
y - (-1) = (-3/4)(x - 1)
y + 1 = (-3/4)x + 3/4
y = (-3/4)x + 3/4 - 1
y = (-3/4)x - 1/4
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Equation of perpendicular bisector 2 passing through (-1, 1):
y - y₅ = (m₂)(x - x₅)
y - 1 = (-5/2)(x + 1)
y = (-5/2)x - 5/2 + 1
y = (-5/2)x - 3/2
Solve the simultaneous equations of the two perpendicular bisectors to find
the circumcenter. In this case, we solve the equations:
(-3/4)x - (1/4) = (-5/2)x - (3/2)
Simplifying the equation, we get:
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(-3/4 + 5/2)x = -(3/2) + (1/4)
(-6/8 + 20/8)x = -(12/8) + (2/8)
(14/8)x = -(10/8)
x = -(10/8) / (14/8) = -(10/8) * (8/14) = -10/14 = -5/7
Substituting x = -5/7 into either equation, we get:
y = (-3/4)(-5/7) - 1/4 = 15/28 - 7/28 = 8/28 = 2/7
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Therefore, the circumcenter of the triangle with vertices (4, 3), (-2, -5), and
(-6, -1) is approximately (-5/7, 2/7).
(b) To find the equations of two parallel lines perpendicular to 3x - 5y +
70 = 0 such that the product of their x-intercept and y-intercept is 20,
we can use the following steps:
Convert the given equation 3x - 5y + 70 = 0 into the slope-intercept form (y
= mx + b) by solving for y:
3x - 5y + 70 = 0
-5y = -3x - 70
y = (3/5)x + 14
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Therefore, the slope of the given line is 3/5.
The slopes of the parallel lines will also be 3/5.
Let the equation of the first parallel line be y = (3/5)x + c₁. To find c₁, we
can use the fact that the product of the x-intercept and y-intercept is 20.
The x-intercept occurs when y = 0, so we set y = 0 and solve for x:
(3/5)x + c₁ = 0
(3/5)x = -c₁
x = (-5/3)c₁
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The y-intercept occurs when x = 0, so we set x = 0 and solve for y:
y = (3/5)(0) + c₁
y = c₁
The product of the x-intercept and y-intercept is:
(-5/3)c₁ * c₁ = 20
Multiplying through by 3/5, we get:
-5c₁² = 60
Simplifying, we have:
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c₁² = -12
Since c₁ cannot be a complex number, there are no real solutions for c₁.
Therefore, the first parallel line does not exist.
Similarly, let the equation of the second parallel line be y = (3/5)x + c₂. We
can use the product of the x-intercept and y-intercept to find c₂.
(-5/3)c₂ * c₂ = 20
Multiplying through by 3/5, we get:
-5c₂² = 60
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Simplifying, we have:
c₂² = -12
Again, there are no real solutions for c₂. Therefore, the second parallel line
does not exist.
In conclusion, there are no equations of parallel lines perpendicular to 3x -
5y + 70 = 0 such that the product of their x-intercept and y-intercept is 20.
Q. 5 Minimize the loss: l = 3x + 5y subject to constraints;
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2x + 5y ≤ 200
6x + 7y ≤ 4200
x + 9y ≤ 450
To minimize the loss function l = 3x + 5y subject to the given constraints,
we can solve the linear programming problem using the simplex method.
The constraints are as follows:
2x + 5y ≤ 200
6x + 7y ≤ 4200
x + 9y ≤ 450
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We need to maximize the objective function -l = -3x - 5y since minimizing l
is equivalent to maximizing -l.
The standard form of the linear programming problem is as follows:
Maximize: -3x - 5y
Subject to:
2x + 5y ≤ 200
6x + 7y ≤ 4200
x + 9y ≤ 450
x ≥ 0, y ≥ 0
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Using the simplex method, we solve the linear programming problem to find
the optimal solution.
The initial tableau for the problem is:
| x | y | s₁ | s₂ | s₃ | b |
z₀ | -3 | -5 | 0 | 0 | 0 | 0 |
s₁ | 2 | 5 | 1 | 0 | 0 | 200 |
s₂ | 6 | 7 | 0 | 1 | 0 | 4200|
s₃ | 1 | 9 | 0 | 0 | 1 | 450 |
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The column labels x, y, s₁, s₂, and s₃ represent the variables x, y, and the
slack variables s₁, s₂, s₃. The row labels z₀, s₁, s₂, and s₃ represent the
objective function and the constraints.
To solve the simplex method step by step, we iterate the following steps:
Select the most negative coefficient in the z₀ row, which is -5 in column y.
Select the pivot element. Divide the right-hand side (b column) by the
corresponding column's entry of the pivot element. The smallest positive
ratio determines the pivot row. In this case, the smallest ratio is 200/5 = 40 in
row s₁.
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Perform row operations to make the pivot element 1 and other elements in
the pivot column 0.
Multiply the pivot row by 1/5.
Add/subtract appropriate multiples of the pivot row to/from other rows to
make all elements in the pivot column 0.
The updated tableau after the pivot operation is:
| x | y | s₁ | s₂ | s₃ | b |
z₀ | -3 | 0 | 1 | 0 | 0 | 400 |
y | 2 | 1 | 1/5| 0 | 0 | 40 |
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s₂ | 2 | 0 |-6/5| 1 | 0 | 380 |
s₃ | -1 | 0 |-1/5| 0 | 1 | 90 |
Repeat steps 1-3 until all coefficients in the z₀ row are non-negative.
Next iteration:
| x | y | s₁ | s₂ | s₃ | b |
z₀ | 0 | 0 | 7/5| 3 | 0 | 940 |
y | 0 | 1 | 1/5| -2/5| 0 | 20 |
s₂ | 1 | 0 |-7/5| 1 | 0 | 380 |
s₃ | 0 | 0 |-2/5| 1 | 1 | 80 |
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Final iteration:
| x | y | s₁ | s₂ | s₃ | b |
z₀ | 1 | 0 | 9/35| 0 | 0 | 980 |
y | 0 | 1 | 1/7| -2/7| 0 | 40 |
s₂ | 0 | 0 |-3/35| 1 | 0 | 320 |
s₃ | 0 | 0 | 1/35| 1/7| 1 | 100 |
The final tableau shows the optimal solution:
The optimal solution is x = 980, y = 40, with a minimized loss of l = 3x + 5y
= 3(980) + 5(40) = 2940 + 200 = 3140.
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Therefore, the minimum loss occurs when x = 980 and y = 40, with a
minimized loss value of 3140.
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AIOU Solved Assignment Code 1309 Mathematics III 2023 Assignment 1.pptx

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