Question

The derivative of the function B is given by B′(t)=8e0.2cost, and B(2.2)=4.5. If the linear approximation to B(t) at t=2.2 is used to estimate B(t), at what value of t does the linear approximation estimate that B(t)=9 ?

Answer 1

Answer:

t=2.83

Step-by-step explanation:

Linear Approximation Of Functions

The equation of a line is given by

$y=y_o+m(x-x_o)$

Where m is the slope of the line and $(x_o,y_o)$ 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 $(t_o, B_o)$. The equation becomes

$B(t)=B_o+B'(t_o)(t-t_o)$

Let's collect our data.

$B'(t)=8e^{0.2cost},\ B_o=B(2.2)=4.5$

Let's find the required values to build the approximate function near $t_0=2.2$. We evaluate the derivative in 2.2

$B'(2.2)=8e^{0.2cos2.2}=7.11$

The function can be approximated by

$B(t)=4.5+7.11(t-2.2)$

Once we have B(t), we are required to find the value of t, such that

$B(t)=9$

Or equivalently:

$4.5+7.11(t-2.2)=9$

Rearranging

$\displaystyle t-2.2=\frac{9-4.5}{7.11}$

Solving for t

$\displaystyle t=\frac{9-4.5}{7.11}+2.2$

$\boxed{t=2.83}$