Target Exam

CUET

Subject

Section B1

Chapter

Applications of Derivatives

Question:

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.

Options:

$100$

$100e$

$1000e$

$10e$

Correct Answer:

$100e$

Explanation:

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.