Answer:
t=2.83
Step-by-step explanation:
Linear Approximation Of Functions
The equation of a line is given by
[tex]y=y_o+m(x-x_o)[/tex]
Where m is the slope of the line and [tex](x_o,y_o)[/tex] are the coordinates of a point through which the line goes.
Given a function B(t), we can build an approximate line to model the function near one point. The value of m is the derivative of B in a specific point [tex](t_o, B_o)[/tex]. The equation becomes
[tex]B(t)=B_o+B'(t_o)(t-t_o)[/tex]
Let's collect our data.
[tex]B'(t)=8e^{0.2cost},\ B_o=B(2.2)=4.5[/tex]
Let's find the required values to build the approximate function near [tex]t_0=2.2[/tex]. We evaluate the derivative in 2.2
[tex]B'(2.2)=8e^{0.2cos2.2}=7.11[/tex]
The function can be approximated by
[tex]B(t)=4.5+7.11(t-2.2)[/tex]
Once we have B(t), we are required to find the value of t, such that
[tex]B(t)=9[/tex]
Or equivalently:
[tex]4.5+7.11(t-2.2)=9[/tex]
Rearranging
[tex]\displaystyle t-2.2=\frac{9-4.5}{7.11}[/tex]
Solving for t
[tex]\displaystyle t=\frac{9-4.5}{7.11}+2.2[/tex]
[tex]\boxed{t=2.83}[/tex]
