A bacteria population starts with 400 bacteria and grows at a rate of r(t) = (450.268)e^{1.1256t}...

Question:

A bacteria population starts with {eq}\,400\, {/eq} bacteria and grows at a rate of {eq}\,r(t) = \left(450.268\right)e^{1.1256t}\; {/eq} bacteria per hour.

How many bacteria will there be after three hours?

Exponential growth:

Exponential growth can be measured by applying the formula,

{eq}A={{A}_{0}}{{e}^{kt}} {/eq}

where {eq}{{A}_{0}} {/eq} represents the original amount.

It is majorly used in biology to determine the micro-organism numbers.

It is too practiced in physics in the study of nuclear reactors.

Given:

{eq}r\left( t \right)=450.268{{e}^{1.1256t}} {/eq}

To determine:

Number of bacteria after 3 hours.

Let the number of bacteria be y

Therefore,

{eq}r\left( t \right)=\dfrac{dy}{dt} {/eq}

{eq}\begin{align*} \dfrac{dy}{dt}&=450.268{{e}^{1.1256t}} \\ dy&=450.268{{e}^{1.1256t}}dt \\ \end{align*} {/eq}

Integrate the above function,

{eq}\begin{align*} \int{dy}&=\int{450.268{{e}^{1.1256t}}dt} \\ y&=\dfrac{450.268}{1.1256}{{e}^{1.1256t}}+C \\ y&=400{{e}^{1.1256t}}+C \\ \end{align*} {/eq}

At initial level i.e, {eq}t=0 {/eq}, there are 400 bacteria.

Therefore,

{eq}\begin{align*} y\left( 0 \right)&=400 \\ 400&=400{{e}^{0}}+C \\ C&=0 \\ \end{align*} {/eq}

Therefore, the population function is given as,

{eq}y=400{{e}^{1.1256t}} {/eq}

Now,

After 3 hours, the number of bacteria will be given by substituting {eq}{/eq}.

{eq}\begin{align*} y\left( 3 \right)&=400{{e}^{1.1256\left( 3 \right)}} \\ &=400{{e}^{3.3768}} \\ &\approx 11711 \\ \end{align*} {/eq}