Vera studies the extinction of the bear population of Siberia over time. The following function gives the number of bears $t$ years since Vera started tracking it: $\newline$ $B(t)=2190 \cdot e^{-0.3 t}$ $\newline$ What is the instantaneous rate of change of the number of bears after $2$ years? $\newline$ Choose $1$ answer: $\newline$ (A) $1202$ years per bear $\newline$ (B) $1202$ bears per year $\newline$ (C) $-361$ years per bear $\newline$ (D) $-361$ bears per year

Full solution

Q. Vera studies the extinction of the bear population of Siberia over time. The following function gives the number of bears

t

years since Vera started tracking it:

\newline

B(t)=2190 \cdot e^{-0.3 t}

\newline

What is the instantaneous rate of change of the number of bears after

2

years?

\newline

Choose

1

answer:

\newline

(A)

1202

years per bear

\newline

(B)

1202

bears per year

\newline

(C)

-361

years per bear

\newline

(D)

-361

bears per year

Differentiate function $B(t)$ : To find the instantaneous rate of change, we need to differentiate the function $B(t)$ with respect to $t$ .
Apply chain rule: Differentiate $B(t) = 2190 \cdot e^{-0.3t}$ using the chain rule. $\frac{dB}{dt} = 2190 \cdot (-0.3) \cdot e^{-0.3t}$
Calculate derivative at $t=2$ : Now, plug in $t = 2$ years into the derivative to find the rate of change at that specific time. $\frac{dB}{dt}$ at $t = 2 = 2190 \times (-0.3) \times e^{(-0.3\times2)}$
Find $e^{-0.6}$ : Calculate the value. $\newline$ $dB/dt$ at $t = 2 = 2190 \times (-0.3) \times e^{-0.6}$
Multiply values: Use a calculator to find the value of $e^{-0.6}$ . $e^{-0.6} \approx 0.5488$
Perform final multiplication: Now multiply the values together. $\frac{dB}{dt}$ at $t = 2$ $\approx 2190 \times (-0.3) \times 0.5488$
Perform final multiplication: Now multiply the values together. $\newline$ dB/dt at $t = 2$ $\approx 2190 \times (-0.3) \times 0.5488$ Perform the multiplication. $\newline$ dB/dt at $t = 2$ $\approx -2190 \times 0.16464$