Iqbal, a data analyst in a social media platform is tracking the number of active users on their site between 5 p.m. and 6 p.m. on a particular day. The user growth function is modelled by $N(t) = 1000e^{0.1t}$, where $N(t)$ represents the number of active users at time $t$ minutes during that period. Find how fast the number of active users are increasing or decreasing at 10 min past 5 p.m. |
$100$ $100e$ $1000e$ $10e$ |
$100e$ |
The correct answer is Option (2) → $100e$ ## The rate at which the number of active users is increasing or decreasing at a given time is given by $\frac{d}{dt} N(t)$. Then, derivative of $N(t)$ is $\frac{d}{dt} N(t) = 1000(0.1)e^{0.1t}$ $∴$ The rate of change of active users at 10 min past 5 p.m. as: $\frac{d}{dt} N(10) = 1000(0.1)e^{(0.1)(10)} = 100e$ We can conclude that the number of active users are increasing at a rate of $100e$ people per minute at 5:10 p.m. on that day. |