Step 1: Gradient of function.
\[
\nabla f = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2} \right)
\]
\[
\frac{\partial f}{\partial x_1} = 2x_1 + x_2 + 3
\]
\[
\frac{\partial f}{\partial x_2} = 4x_2 + x_1 + 3
\]
Step 2: Evaluate at (1,1).
\[
\frac{\partial f}{\partial x_1} = 2(1) + 1 + 3 = 6
\]
\[
\frac{\partial f}{\partial x_2} = 4(1) + 1 + 3 = 8
\]
So gradient = (6, 8).
Step 3: Maximum rate of change.
Magnitude of gradient =
\[
|\nabla f| = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10
\]
Correction: Exactly 10, not 9.
Final Answer:
\[
\boxed{10}
\]
Given an open-loop transfer function \(GH = \frac{100}{s}(s+100)\) for a unity feedback system with a unit step input \(r(t)=u(t)\), determine the rise time \(t_r\).
Consider a linear time-invariant system represented by the state-space equation: \[ \dot{x} = \begin{bmatrix} a & b -a & 0 \end{bmatrix} x + \begin{bmatrix} 1 0 \end{bmatrix} u \] The closed-loop poles of the system are located at \(-2 \pm j3\). The value of the parameter \(b\) is: