Which way spiral
in Assignments
(5 pts)
Show the origin for the system
$$
\begin{align}
x' &= -9x -10y \\
y' &= 8x + 8y
\end{align}
$$
is a spiral and determine whether it's attractive or repulsive.
Comments
The spiral is attractive because the trace is negative.
It is attractive because the trace is negative.
The spiral is attractive because the trace is negative
The Spiral is attractive because the trace is negative.
The spiral is attractive because the trace is negative.
The trace is negative so the spiral is attractive
The trace is negative therefore it is an attractive spiral
The trace is negative so the spiral is attractive
The Spiral is attractive because the trace is negative.
The trace is -1 so the spiral is attractive.
Let $A = \begin{matrix} -9 & -10 \ 8 & 8 \end{matrix}$, so the whole system can be rewritten $\mathbf{x\prime} = A * \mathbf{x}$.
To classify the system, we only need to calculate the determinant and trace of $A$:
$det(A) = (-9 * 8) - (-10 * 8) = (-9 - -10) * 8 = (-9 + 10) * 8 = 1 * 8 = 8$
$tr(A) = -9 + 8 = -1$
Since $det(A) > 0$, the system is a spiral; since $tr(A) < 0$, this spiral is attractive.
The trace is negative and so the spiral is attractive.
The trace is negative and because of that the spiral is attractive.
The trace is -1 and the determinate is 8 so the spiral is attractive.
$$
\begin{align}
x' &= -9x - 10y \\
y' &= 8x + 8y
\end{align}
$$
determinant = $8$
trace = $-1$
$tr^2 < 4det$ so it's a spiral, and it's attractive since the trace is negative.
According to thm. 4.41, since the $det$, 8, is greater than 0, and the $tr$, -1, is less than 0, and $4det > tr => 32 > -1$, this is an attractive spiral.
Since our $tr=-1$, then we can state that this spiral is attractive since the trace is negative.
Since $tr=-1$ the spiral is attractive
since tr = -1, the spiral is attractive