Answer:
a) [tex]y(1) = 2.4[/tex] , [tex]y(2) = 13.2[/tex], b) [tex]y(x) = 2.4 + 5.2\cdot (x-1)[/tex], [tex]y(1) = 2.4[/tex], [tex]y(2) = 7.6[/tex]
Step-by-step explanation:
a) The values of the output for steady-state operation are:
[tex]y(1) = 1 + 1.4\cdot (1)^{3}[/tex]
[tex]y(1) = 2.4[/tex]
[tex]y(2) = 2 + 1.4\cdot (2)^{3}[/tex]
[tex]y(2) = 13.2[/tex]
b) The formula for linearization is:
[tex]y(x) = y(1) +\frac{dy}{dx}|_{x = 1} \cdot (x-1)[/tex]
The first derivative of the formula evaluated at x = 1 is:
[tex]\frac{dy}{dx}|_{x=1} = 1 + 4.2\cdot (1)^{2}[/tex]
[tex]\frac{dy}{dx}|_{x=1} = 5.2[/tex]
The linearized model is:
[tex]y(x) = 2.4 + 5.2\cdot (x-1)[/tex]
The output at x = 2 is presented below:
[tex]y(2) = 2.4 + 5.2\cdot (2-1)[/tex]
[tex]y(2) = 7.6[/tex]
Linearized model offers reasonable approximations for small intervals.
