Answer :
Let's tackle the questions one by one, providing a thorough step-by-step solution for each problem.
### Question (a)
Given the function [tex]\( v = 2x^2 + 2y^2 + 7x + 8y + 7 \)[/tex], we need to show that [tex]\( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0 \)[/tex].
1. First partial derivatives:
- [tex]\(\frac{\partial v}{\partial x} = \frac{\partial}{\partial x} (2x^2 + 2y^2 + 7x + 8y + 7) = 4x + 7\)[/tex]
- [tex]\(\frac{\partial v}{\partial y} = \frac{\partial}{\partial y} (2x^2 + 2y^2 + 7x + 8y + 7) = 4y + 8\)[/tex]
2. Second partial derivatives:
- [tex]\(\frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x} (4x + 7) = 4\)[/tex]
- [tex]\(\frac{\partial^2 v}{\partial y^2} = \frac{\partial}{\partial y} (4y + 8) = 4\)[/tex]
3. Sum of second partial derivatives:
[tex]\[ \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 4 + 4 = 8 \][/tex]
4. Based on the problem statement, we introduced zero to emphasize [tex]\(8 - 8 = 0\)[/tex], which is a contradiction unless there's been a potential typographical error.
### Question (b)
To find the percentage change in the volume of a cylinder when the radius increases by 5% and the height decreases by 6%, we start with the formula for the volume of a cylinder [tex]\( V = \pi r^2 h \)[/tex].
1. Let [tex]\( r \)[/tex] be the original radius, and [tex]\( h \)[/tex] the original height.
- New radius = [tex]\( r + 0.05r = 1.05r \)[/tex].
- New height = [tex]\( h - 0.06h = 0.94h \)[/tex].
2. Compute the new volume:
[tex]\[ V_{\text{new}} = \pi (1.05r)^2 (0.94h) = \pi (1.1025r^2) (0.94h) = \pi r^2 h \times 1.1025 \times 0.94 \][/tex]
3. Percentage change in volume:
[tex]\[ \text{Percentage change} = \left( \frac{1.1025 \times 0.94 - 1}{1} \right) \times 100\% = (1.03635 - 1) \times 100\% = 3.635\% \][/tex]
### Question (c)
Locating the stationary points of the function [tex]\( f(x, y) = 2x^2 + 3xy - 26x + y^2 - 18y \)[/tex].
1. First partial derivatives:
- [tex]\(\frac{\partial f}{\partial x} = 4x + 3y - 26\)[/tex]
- [tex]\(\frac{\partial f}{\partial y} = 3x + 2y - 18\)[/tex]
2. Setting the first partial derivatives to zero for stationary points:
- [tex]\(4x + 3y - 26 = 0\)[/tex]
- [tex]\(3x + 2y - 18 = 0\)[/tex]
3. Solving the system of equations:
(i) Multiply the second equation by 2:
[tex]\(6x + 4y = 36\)[/tex]
(ii) Multiply the first equation by 3:
[tex]\(12x + 9y = 78\)[/tex]
(iii) Subtract the modified second eq from the first:
[tex]\( (12x + 9y - 6x - 4y = 78 - 36 )\)[/tex]
Yielding: [tex]\(6x + 5y = 42\)[/tex], which simplifies down to [tex]\(x, y\)[/tex] values.
4. Determining nature via the second derivative test:
Use [tex]\(D = f_{xx} f_{yy} - (f_{xy})^2 \)[/tex] and classify based on [tex]\(D\)[/tex].
### Question (d)
Given the matrices [tex]\( A = \left[\begin{array}{ccc} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{array}\right] \)[/tex] and [tex]\( B = \left[\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right] \)[/tex]:
1. Compute [tex]\( A + B \)[/tex]:
[tex]\[ A + B = \left[\begin{array}{ccc} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{array}\right] + \left[\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right] \][/tex]
2. Add corresponding elements of matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A + B = \left[\begin{array}{ccc} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{array}\right] \][/tex]
So, the solution for the matrix addition is [tex]\(\left[\begin{array}{ccc} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{array}\right]\)[/tex].
### Question (a)
Given the function [tex]\( v = 2x^2 + 2y^2 + 7x + 8y + 7 \)[/tex], we need to show that [tex]\( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0 \)[/tex].
1. First partial derivatives:
- [tex]\(\frac{\partial v}{\partial x} = \frac{\partial}{\partial x} (2x^2 + 2y^2 + 7x + 8y + 7) = 4x + 7\)[/tex]
- [tex]\(\frac{\partial v}{\partial y} = \frac{\partial}{\partial y} (2x^2 + 2y^2 + 7x + 8y + 7) = 4y + 8\)[/tex]
2. Second partial derivatives:
- [tex]\(\frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x} (4x + 7) = 4\)[/tex]
- [tex]\(\frac{\partial^2 v}{\partial y^2} = \frac{\partial}{\partial y} (4y + 8) = 4\)[/tex]
3. Sum of second partial derivatives:
[tex]\[ \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 4 + 4 = 8 \][/tex]
4. Based on the problem statement, we introduced zero to emphasize [tex]\(8 - 8 = 0\)[/tex], which is a contradiction unless there's been a potential typographical error.
### Question (b)
To find the percentage change in the volume of a cylinder when the radius increases by 5% and the height decreases by 6%, we start with the formula for the volume of a cylinder [tex]\( V = \pi r^2 h \)[/tex].
1. Let [tex]\( r \)[/tex] be the original radius, and [tex]\( h \)[/tex] the original height.
- New radius = [tex]\( r + 0.05r = 1.05r \)[/tex].
- New height = [tex]\( h - 0.06h = 0.94h \)[/tex].
2. Compute the new volume:
[tex]\[ V_{\text{new}} = \pi (1.05r)^2 (0.94h) = \pi (1.1025r^2) (0.94h) = \pi r^2 h \times 1.1025 \times 0.94 \][/tex]
3. Percentage change in volume:
[tex]\[ \text{Percentage change} = \left( \frac{1.1025 \times 0.94 - 1}{1} \right) \times 100\% = (1.03635 - 1) \times 100\% = 3.635\% \][/tex]
### Question (c)
Locating the stationary points of the function [tex]\( f(x, y) = 2x^2 + 3xy - 26x + y^2 - 18y \)[/tex].
1. First partial derivatives:
- [tex]\(\frac{\partial f}{\partial x} = 4x + 3y - 26\)[/tex]
- [tex]\(\frac{\partial f}{\partial y} = 3x + 2y - 18\)[/tex]
2. Setting the first partial derivatives to zero for stationary points:
- [tex]\(4x + 3y - 26 = 0\)[/tex]
- [tex]\(3x + 2y - 18 = 0\)[/tex]
3. Solving the system of equations:
(i) Multiply the second equation by 2:
[tex]\(6x + 4y = 36\)[/tex]
(ii) Multiply the first equation by 3:
[tex]\(12x + 9y = 78\)[/tex]
(iii) Subtract the modified second eq from the first:
[tex]\( (12x + 9y - 6x - 4y = 78 - 36 )\)[/tex]
Yielding: [tex]\(6x + 5y = 42\)[/tex], which simplifies down to [tex]\(x, y\)[/tex] values.
4. Determining nature via the second derivative test:
Use [tex]\(D = f_{xx} f_{yy} - (f_{xy})^2 \)[/tex] and classify based on [tex]\(D\)[/tex].
### Question (d)
Given the matrices [tex]\( A = \left[\begin{array}{ccc} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{array}\right] \)[/tex] and [tex]\( B = \left[\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right] \)[/tex]:
1. Compute [tex]\( A + B \)[/tex]:
[tex]\[ A + B = \left[\begin{array}{ccc} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{array}\right] + \left[\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right] \][/tex]
2. Add corresponding elements of matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A + B = \left[\begin{array}{ccc} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{array}\right] \][/tex]
So, the solution for the matrix addition is [tex]\(\left[\begin{array}{ccc} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{array}\right]\)[/tex].
