Determine the percentage rate of change of F(t) = $\mathrm { e } ^ { - 0.09 \mathrm { t } ^ { 2 } }$ at t = 1 and t = 5.
A) $\frac { F ^ { \prime } ( 1 ) } { F ( 1 ) }$ = 18%, $\frac { F ^ { \prime } ( 5 ) } { F ( 5 ) }$ = 90%
B) $\frac { F ^ { \prime } ( 1 ) } { F ( 1 ) }$ = -18%, $\frac { F ^ { \prime } ( 5 ) } { F ( 5 ) }$ = -90%
C) $\frac { F ^ { \prime } ( 1 ) } { F ( 1 ) }$ = 19%, $\frac { F ^ { \prime } ( 5 ) } { F ( 5 ) }$ = 70%
D) $\frac { F ^ { \prime } ( 1 ) } { F ( 1 ) } = - 19 \% , \frac { F ^ { \prime } ( 5 ) } { F ( 5 ) } = - 70 \%$

Question

Quizplus · Accepted Answer

To find the percentage rate of change of $F(t) = e^{-0.09t^2}$ at specific values of $t$, we first calculate the derivative $F'(t)$. The derivative of $F(t)$ with respect to $t$ is $F'(t) = -0.18te^{-0.09t^2}$. Then, we evaluate $F'(t)$ and $F(t)$ at $t = 1$ and $t = 5$, and calculate the percentage rate of change as $\frac{F'(t)}{F(t)} \times 100\%$.At $t = 1$, $F(1) = e^{-0.09}$ and $F'(1) = -0.18e^{-0.09}$. The percentage rate of change is $\frac{-0.18e^{-0.09}}{e^{-0.09}} \times 100\% = -18\%$.At $t = 5$, $F(5) = e^{-2.25}$ and $F'(5) = -0.9e^{-2.25}$. The percentage rate of change is $\frac{-0.9e^{-2.25}}{e^{-2.25}} \times 100\% = -90\%$.Therefore, the correct answer is that the percentage rate of change of $F(t)$ at $t = 1$ is -18%, and at $t = 5$ is -90%.

Determine the Percentage Rate of Change of F(t) = e−0.09t2\mathrm { e } ^ { - 0.09 \mathrm { t } ^ { 2 } }e−0.09t2

Determine the Percentage Rate of Change of F(t) = $\mathrm { e } ^ { - 0.09 \mathrm { t } ^ { 2 } }$