IDNLearn.com offers a comprehensive solution for finding accurate answers quickly. Get accurate and detailed answers to your questions from our dedicated community members who are always ready to help.
Sagot :
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].
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.
