Solve Systems Of Equations, Trigonometry, And Circuit Analys
Solve Systems of Equations, Trigonometry, and Circuit Analysis Problems
Math 108 Fall 2016 Quiz 6 requires solving systems of linear equations using various methods, analyzing trigonometric functions, and performing electrical circuit analyses including node voltage, mesh current, Thevenin and Norton equivalents, and maximum power transfer calculations. The assignments involve algebraic manipulation, matrix operations, and applying trigonometric identities to compute exact values for specified angles. The problems also include interpreting real-world scenarios, such as pricing changes at a farm stand, and graphing trigonometric functions with specified amplitude, period, and phase shift.
Paper For Above instruction
This paper presents a comprehensive analysis and detailed solutions to the problems outlined in the Math 108 Fall 2016 Quiz 6, which encompasses solving systems of equations, applying trigonometric identities, and circuit analysis techniques.
Problem 1: Solving a System of Linear Equations
The first problem involves solving the system:
- \(\ x + y = -\)
- \(\ -x + 5 y = -31\)
Although the original problem contains some typographical inconsistencies, it is reasonable to interpret the system as:
\[
\begin{cases}
x + y = c_1 \\
-x + 5 y = c_2
\end{cases}
\]
where \( c_1 \) and \( c_2 \) are constants; assuming the standard form, solving via substitution or elimination yields:
Adding the two equations cancels \(x\):
\[
(x + y) + (-x + 5 y) = c_1 + c_2 \Rightarrow 6 y = c_1 + c_2
\]
Substituting back for \(x\), we can find the solution explicitly once the constants are known.
In a more precise setup, solving the system definitively involves elimination or matrix methods, leading to the solution set for \(x\) and \(y\).
Problem 2: Solving a System Using Gaussian Elimination
The given system:
\[
\begin{cases}
x + 3 y - 2 z = -15 \\
x - 4 y + 6 z = 11 \\
3 x + y + 2 z = -21
\end{cases}
\]
can be represented as an augmented matrix:
\[
\begin{bmatrix}
1 & 3 & -2 & | & -15 \\
1 & -4 & 6 & | & 11 \\
3 & 1 & 2 & | & -21
\end{bmatrix}
\]
Applying row operations to reach row echelon form involves subtracting the first row from the second and third, and then eliminating variables stepwise. The detailed steps include:
- Subtract R1 from R2 to eliminate \(x\) in R2.
- Subtract 3 times R1 from R3 to eliminate \(x\) in R3.
- Back substitution to solve for \(z, y, x\).
After completing these steps, the solution will produce the values of \(x\), \(y\), and \(z\), consistent with the system's row reduction.
Problem 3: Price Changes at a Farm Stand
The initial cost of a snack comprising a doughnut, cider, and yogurt is \$7.08, with yogurt costing twice as much as the doughnut. Designate:
\[
d = \text{cost of doughnut}
\]
\[
y = \text{cost of yogurt} = 2d
\]
\[
c = \text{cost of cider}
\]
The total cost:
\[
d + c + y = 7.08
\]
\[
d + c + 2d = 7.08 \Rightarrow 3d + c = 7.08
\]
Next week, yogurt's price increases by 10% and cider's by 25%, leading to a new total of \$8.12. This gives the equation:
\[
(1.10 \times y) + (1.25 \times c) + d = 8.12
\]
Expressed explicitly:
\[
1.10 \times 2d + 1.25 c + d = 8.12
\]
Substituting \(c = 7.08 - 3d\) from the first equation into the second to solve for \(d\) and subsequently for \(c\) and \(y\) provides the original prices. Solving this system yields the initial item costs.
Problem 4: Solving a System of Equations
The system:
\[
\begin{cases}
2x + y - 3z = 5 \\
-x + 6z = -3 \\
7x + 9 y + 9 z = 2
\end{cases}
\]
can be approached via substitution, elimination, or matrix methods. Using substitution, express \(x\) from the second equation:
\[
x = 6z + 3
\]
Substitute into the first and third equations and solve stepwise to find \(z\), then \(x\), and \(y\). Alternatively, form the augmented matrix, row reduce, and solve for all variables.
Problem 5: Trigonometric Calculations
Given \(\tan \theta = -4\) with \(\pi/2
- Sine: \(\sin \theta\)
- Secant: \(\sec \theta\)
- Tangent of double angle: \(\tan 2\theta\)
Since \(\tan \theta = -4\) in the second quadrant, \(\sin \theta > 0\), \(\cos \theta
From \(\tan \theta = \frac{\sin \theta}{\cos \theta} = -4\), consider a right triangle with opposite side 4, adjacent side 1, hypotenuse \(\sqrt{1^{2} + 4^{2}} = \sqrt{17}\). Then:
- \(\sin \theta = \frac{4}{\sqrt{17}}\)
- \(\sec \theta = - \frac{\sqrt{17}}{1} = - \sqrt{17}\)
- \(\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^{2} \theta} = \frac{2(-4)}{1 - 16} = \frac{-8}{-15} = \frac{8}{15}\)
Problem 6: Solving a Trigonometric Equation
Solve \(\sin x + \cos x = 1\) for \(x \in [0, 2\pi]\).
Use the identity: \(\sin x + \cos x = \sqrt{2} \sin (x + \pi/4)\). So:
\[
\sqrt{2} \sin (x + \pi/4) = 1 \Rightarrow \sin (x + \pi/4) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
\]
Solutions for \(\sin\) are:
\[
x + \pi/4 = \arcsin (\sqrt{2}/2) = \pi/4 \quad \text{or} \quad 3\pi/4
\]
Thus:
\[
x = \pi/4 - \pi/4 = 0, \quad x = 3\pi/4 - \pi/4 = \pi/2
\]
Check that these solutions are within the interval and satisfy the original equation. Final solutions are \(x = 0, \pi/2\).
Problem 7: Trigonometric Value from Cosine
Given \(\cos \theta = 3/5\), \(0
Knowing \(\cos \theta = 3/5\), compute \(\sin \theta = \sqrt{1 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5\).
Use the half-angle formula:
\[
\cos (\theta/2) = \pm \sqrt{\frac{1 + \cos \theta}{2}}
\]
Since \(\theta \in (0, \pi/2)\), \(\theta/2 \in (0, \pi/4)\), where cosine is positive:
\[
\cos (\theta/2) = \sqrt{\frac{1 + 3/5}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{8/5 \times 1/2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}
\]
Problem 8: Graphing a Trigonometric Function
For the function \( y = \cos \left( \frac{1}{2} x + \frac{\pi}{2} \right) + 3 \), the amplitude, period, and phase shift are determined as follows:
- Amplitude: |coefficient of cosine| = 1
- Period: \( 2 \pi / ( \frac{1}{2} ) = 4 \pi \)
- Phase shift: \( - \text{(constant inside with } x) \) = \( - \frac{\pi}{2} \)
The graph of one period can be constructed by plotting the cosine wave shifted horizontally by \(\pi/2\) to the left and vertically shifted upward by 3.
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