Question

the number of visitors ( n(t) ) on a website ( t ) hours after launching a promotion is modeled by: ( n(t) = 180 cdot ln(360t + 400) ) (round your answers to the nearest whole number.)
a) find the number of visitors 4 hours after the promotion begins: ( n(4) = ) (\boxed{quad}) visitors
b) find the rate at which the traffic is changing after 4 hours, ( n(4) ). ( n(4) = ) (\boxed{quad}) ?

Explanation:

Part (a)

Step1: Substitute \( t = 4 \) into \( N(t) \)

We have the function \( N(t)=180\cdot\ln(360t + 400) \). Substitute \( t = 4 \) into the function:
First, calculate the argument of the logarithm: \( 360\times4+400=1440 + 400=1840 \)
So \( N(4)=180\cdot\ln(1840) \)

Step2: Calculate the value

We know that \( \ln(1840)\approx7.517 \) (using a calculator to find the natural logarithm of 1840). Then:
\( N(4)=180\times7.517\approx1353.06 \)
Rounding to the nearest whole number, we get \( N(4)\approx1353 \)

Step1: Find the derivative of \( N(t) \)

We use the chain rule. The derivative of \( \ln(u) \) with respect to \( t \) is \( \frac{u'}{u} \), where \( u = 360t+400 \).
First, find \( u' \): the derivative of \( 360t + 400 \) with respect to \( t \) is \( u'=360 \)
Then, the derivative of \( N(t)=180\cdot\ln(360t + 400) \) is:
\( N'(t)=180\times\frac{360}{360t + 400}=\frac{180\times360}{360t + 400}=\frac{64800}{360t + 400} \)

Step2: Substitute \( t = 4 \) into \( N'(t) \)

Substitute \( t = 4 \) into the derivative:
First, calculate the denominator: \( 360\times4+400 = 1440+400=1840 \)
Then \( N'(4)=\frac{64800}{1840}\approx35.217 \)
Rounding to a reasonable decimal place (or whole number depending on context), if we round to the nearest whole number, \( N'(4)\approx35 \) (if we keep more decimals, it's approximately 35.22)

Answer:

\( 1353 \)

Part (b)