# A flu epidemic hits a college community, beginning with five cases on day t = 0. The rate of...

## Question:

A flu epidemic hits a college community, beginning with five cases on day t = 0. The rate of growth of the epidemic (new cases per day) is given by the following function r(t), where t is the number of days since the epidemic began. {eq}r(t) = 16e^{0.04t} {/eq}

(a) Find a formula F(t) for the total number of cases of flu in the first t days. F(t) = _____.

(b) Use your answer to part (a) to find the total number of cases in the first 16 days. F(16)=_____ cases.

## Exponential Growth Function:

1. The first integral of the rate of change of the function gives the function itself.
2. If initial values are given, use them as the lower limit of integration. By doing so, we omit one step where we introduce an integration constant and evaluate the same using the same boundary condition.
3. The value of the function can be found by putting the given value of the variable into the function.

Given Data

{eq}\begin{align} r(t) &= 16e^{0.04t} t &= 0, ~~~~ F(0) =5\\ \end{align} {/eq}

Solution

{eq}\begin{align} r(t) &= 16e^{0.04t}\\ t &= 0, ~~~~ F(0) =5\\ \Rightarrow F'(t) &=r(t)\\ \Rightarrow \int_5^{ F(t)} F'(t) dt &=\int_0^t r(t) dt& \left[ \text{ Use the initial conditions as the lower limit of integration} \right] \\ \Rightarrow\left[ F(t) \right]_5^{ F(t)} &=\int_0^t 16e^{0.04t} dt & \left[ \text{Integrate on both sides with respect to time }~t \right] \\ \Rightarrow F(t) -5 &=16 \left[ \frac{ e^{0.04t} }{0.04} \right]_0^t\\ \Rightarrow F(t) &=5+16 \left[ \frac{ e^{0.04t} }{0.04}- \frac{ e^{0} }{0.04} \right] \\ &=5+16 \left[ \frac{ e^{0.04t} }{0.04}- \frac{1 }{0.04} \right] \\ &=5+16 \times 25 \left( e^{0.04t}-1 \right) \\ &=5+400\left( e^{0.04t}-1 \right) \\ &=400 e^{0.04t} -395 \\ \end{align} {/eq}

Therefore, the formula F(t) for the total number of cases of flu in the first t days is{eq}\displaystyle \boxed{\color{blue} { F(t)=400 e^{0.04t} -395 }} {/eq}

(b)

{eq}\begin{align} F(t) &=400 e^{0.04t} -395 \\ \Rightarrow F(16) &=400 e^{0.04(16)} -395 \\ &= 363.59235 \\ \end{align} {/eq}

Therefore, the total number of cases in the first 16 days is{eq}\displaystyle \boxed{\color{blue} { F(16) \approx 364 }} {/eq}