Question

attempt 1: 10 attempts remaining. the number of visitors $n(t)$ on a website $t$ hours after launching a promotion is modeled by: $n(t)=120cdotln(180t + 240)$ (round your answers to the nearest whole number.) a) find the number of visitors 6 hours after the promotion begins: $n(6)=$ visitors b) find the rate at which the traffic is changing after 6 hours, $n(6)$. $n(6)=$

Explanation:

Step1: Calculate $N(6)$

Substitute $t = 6$ into $N(t)=120\cdot\ln(180t + 240)$.
$N(6)=120\cdot\ln(180\times6 + 240)=120\cdot\ln(1080 + 240)=120\cdot\ln(1320)$.
Using a calculator, $\ln(1320)\approx7.18$ and $120\times7.18 = 861.6\approx862$.

Step2: Differentiate $N(t)$

Use the chain - rule. If $N(t)=120\cdot\ln(180t + 240)$, let $u = 180t+240$, then $N(t)=120\cdot\ln(u)$.
The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$, and the derivative of $u = 180t + 240$ with respect to $t$ is $180$.
By the chain - rule $N'(t)=120\times\frac{180}{180t + 240}=\frac{21600}{180t + 240}$.

Step3: Calculate $N'(6)$

Substitute $t = 6$ into $N'(t)$.
$N'(6)=\frac{21600}{180\times6+240}=\frac{21600}{1080 + 240}=\frac{21600}{1320}\approx16.36\approx16$.

Answer:

a) 862
b) 16