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classify the critical point \\((0, 0)\\) of the given linear system by …

Question

classify the critical point \\((0, 0)\\) of the given linear system by computing the trace \\(\tau\\), determinant \\(\delta\\), and discriminant and using the figure.
\\
\

$$\begin{aligned} x &= -7x + 5y \\\\ y &= 3x - 9y \\end{aligned}$$

\\
trace \\(\tau =\\)
determinant \\(\delta =\\)
discriminant \\(\tau^2 - 4\delta =\\)
classify the critical point \\((0, 0)\\).

  • degenerate unstable node
  • unstable spiral
  • stable spiral
  • stable node
  • unstable node
  • center
  • saddle
  • degenerate stable node

Explanation:

Response

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<reasoning>

Construct the coefficient matrix

\[
A =

$$\begin{pmatrix} -7 & 5 \\ 3 & -9 \end{pmatrix}$$

\]

Compute trace, determinant, and discriminant

\[

$$\begin{aligned} \tau &= \operatorname{tr}(A) = -7 + (-9) = -16 \\ \Delta &= \det(A) = (-7)(-9) - (5)(3) = 63 - 15 = 48 \\ \tau^2 - 4\Delta &= (-16)^2 - 4(48) = 256 - 192 = 64 \end{aligned}$$

\]

Classify the critical point

\[

$$\begin{aligned} &\Delta = 48 > 0 \\ &\tau^2 - 4\Delta = 64 > 0 \\ &\tau = -16 < 0 \end{aligned}$$

\]
Since \(\Delta > 0\) and \(\tau^2 - 4\Delta > 0\), the eigenvalues are real, distinct, and have the same sign. Because \(\tau < 0\), both eigenvalues are negative, which classifies the critical point \((0,0)\) as a stable node.
</reasoning>

<answer>

Question 1

trace \(\tau =\) <blank>-16</blank>

Question 2

determinant \(\Delta =\) <blank>48</blank>

Question 3

discriminant \(\tau^2 - 4\Delta =\) <blank>64</blank>

Question 4

Classify the critical point \((0, 0)\).
<mcq-option>degenerate unstable node</mcq-option>
<mcq-option>unstable spiral</mcq-option>
<mcq-option>stable spiral</mcq-option>
<mcq-correct>stable node</mcq-correct>
<mcq-option>unstable node</mcq-option>
<mcq-option>center</mcq-option>
<mcq-option>saddle</mcq-option>
<mcq-option>degenerate stable node</mcq-option>
</answer>

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"Mathematics",
"Calculus",
"Phase Portrait Analysis"
]
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</post_analysis>

Answer:

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</pre_analysis>

<reasoning>

Construct the coefficient matrix

\[
A =

$$\begin{pmatrix} -7 & 5 \\ 3 & -9 \end{pmatrix}$$

\]

Compute trace, determinant, and discriminant

\[

$$\begin{aligned} \tau &= \operatorname{tr}(A) = -7 + (-9) = -16 \\ \Delta &= \det(A) = (-7)(-9) - (5)(3) = 63 - 15 = 48 \\ \tau^2 - 4\Delta &= (-16)^2 - 4(48) = 256 - 192 = 64 \end{aligned}$$

\]

Classify the critical point

\[

$$\begin{aligned} &\Delta = 48 > 0 \\ &\tau^2 - 4\Delta = 64 > 0 \\ &\tau = -16 < 0 \end{aligned}$$

\]
Since \(\Delta > 0\) and \(\tau^2 - 4\Delta > 0\), the eigenvalues are real, distinct, and have the same sign. Because \(\tau < 0\), both eigenvalues are negative, which classifies the critical point \((0,0)\) as a stable node.
</reasoning>

<answer>

Question 1

trace \(\tau =\) <blank>-16</blank>

Question 2

determinant \(\Delta =\) <blank>48</blank>

Question 3

discriminant \(\tau^2 - 4\Delta =\) <blank>64</blank>

Question 4

Classify the critical point \((0, 0)\).
<mcq-option>degenerate unstable node</mcq-option>
<mcq-option>unstable spiral</mcq-option>
<mcq-option>stable spiral</mcq-option>
<mcq-correct>stable node</mcq-correct>
<mcq-option>unstable node</mcq-option>
<mcq-option>center</mcq-option>
<mcq-option>saddle</mcq-option>
<mcq-option>degenerate stable node</mcq-option>
</answer>

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