Question

the derivative of the function $b$ is given by $b(t)=8e^{0.2cos t}$, 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 $b(t)=9$?
(a) 0.633
(b) 0.941
(c) 2.833
(d) 4.400

Explanation:

Step1: Recall linear - approximation formula

The linear - approximation formula is $L(t)=B(a)+B^{\prime}(a)(t - a)$, where $a = 2.2$, $B(a)=B(2.2)=4.5$, and $B^{\prime}(t)=8e^{0.2\cos t}$. First, find $B^{\prime}(2.2)$:
$B^{\prime}(2.2)=8e^{0.2\cos(2.2)}$. Calculate $\cos(2.2)\approx - 0.5298$, then $0.2\cos(2.2)\approx0.2\times(- 0.5298)=-0.10596$, and $e^{-0.10596}\approx0.8997$. So $B^{\prime}(2.2)=8\times0.8997 = 7.1976\approx7.2$.

Step2: Set up the linear - approximation equation

We want to find $t$ when $L(t)=9$, $a = 2.2$, $B(2.2)=4.5$, and $B^{\prime}(2.2)\approx7.2$. Substitute into the linear - approximation formula $L(t)=B(2.2)+B^{\prime}(2.2)(t - 2.2)$:
$9=4.5+7.2(t - 2.2)$.

Step3: Solve the equation for $t$

First, subtract 4.5 from both sides:
$9 - 4.5=7.2(t - 2.2)$.
$4.5=7.2(t - 2.2)$.
Then divide both sides by 7.2:
$\frac{4.5}{7.2}=t - 2.2$.
$0.625=t - 2.2$.
Finally, add 2.2 to both sides:
$t=0.625 + 2.2=2.825\approx2.833$.

Answer:

C. 2.833