Q. 72 - Calculus

Question

 A colony of bacteria is growing in a large petri dish in such away that the shape of the colony is a

disc whose area A(t) after t days grows at rate proportional to the diameter d of the disk. Suppose the colony has an area of 2 cm² on the first day and 5 cm² on the third day.

 (a) Set up a differential equation that describes this situation, and solve it to get an equation for the area A(t) of the colony after t days. Your answer will involve a proportionality constant k.

 (b) Use the information in the problem to determine k.

 (c) Given that the petri dish has a diameter of 6 inches and that the colony started in the exact center of the dish, how long will it take for the colony to fill the entire petri dish? 

Step-by-Step Solution

Answer

(a) $\mathrm{A}\left(\mathrm{t}\right)=2{\mathrm{e}}^{\mathrm{kt}}$

(b) $k=0.4581$

(c)$\mathrm{t}=5.782$

Step 1. Given

A colony of bacteria is growing in a large petri dish in such away that the shape of the colony is a

disc whose area A(t) after t days grows at rate proportional to the diameter d of the disk. Suppose the colony has an area of 2 cm² on the first day and 5 cm² on the third day.

Part(a) Step 2. Finding the differential equation

Note that the problem resembles exponential growth model. Note that the rate of growth of the area of the colony of bacteria is proportional to the diameter of the disk. This means that the area of colony of bacteria is proportional to the area of the disk (since area of the disk is equal to

$$\begin{gathered} \frac{1}{4}{\mathrm{\pi d}}^2 \\ \frac{\mathrm{dA}}{\mathrm{dt}}\mathrm{\alpha A} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{kA} \\ \mathrm{Now}, \mathrm{proceed} \mathrm{to} \mathrm{solve} \mathrm{the} \mathrm{differential} \mathrm{equation}. \\ \mathrm{Observe} \mathrm{that} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{does} \mathrm{not} \mathrm{contain} \mathrm{the} \mathrm{independent} \mathrm{variable} \mathrm{at} \mathrm{all}, \mathrm{so} \mathrm{solve} \mathrm{it} \mathrm{by} \mathrm{using} \mathrm{antidifferentiating} \mathrm{method}.\backslash \\ \int \frac{\mathrm{dA}}{\mathrm{A}}=\mathrm{k}\int \mathrm{dt} \\ \ln \left|\mathrm{A}\right|=\mathrm{kt}+\mathrm{C} \\ \mathrm{A}={\mathrm{e}}^{\mathrm{kt}+{\mathrm{C}}_1} \\ ={\mathrm{Ce}}^{\mathrm{kt}} \\ \mathrm{Use} \mathrm{the} \mathrm{initial} \mathrm{condition}, \mathrm{namely} \mathrm{A} = 2 \mathrm{for} \mathrm{t} = 0 \mathrm{corresponding} \mathrm{to} \mathrm{the} \mathrm{first} \mathrm{day}) \mathrm{and} \mathrm{solve} \mathrm{to} \mathrm{evaluate} \mathrm{the} \mathrm{constant} \mathrm{C} \\ 2=\mathrm{C} \\ \mathrm{Substitute} \mathrm{this} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{constant} \mathrm{and} \mathrm{write} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{problem} \mathrm{as} \\ \mathrm{A}(\mathrm{t})=2{\mathrm{e}}^{\mathrm{kt}} \end{gathered}$$

Part(b) Step 3. Calculating proportionality constant

It is given that the area of the colony of bacteria has grown to 5 cm² on the third day. This means that after 2 days the area of the colony of bacteria becomes 5 cm². Hence, take A = 5 and t = 2 (corresponding to the third day) in equation (2) and solve for k

$$\begin{gathered} 5=2e^{2k} \\ {e}^{2k}=2.5 \\ k=\frac{1}{2}\ln (2.5) \\ k=0.4581 \end{gathered}$$

Part(c) Step 4. Calculating the total time

Substitute the value of k obtained in the part(b) in equation (1) to find the solution  of theh model as $A\left(t\right)=2e^{0.4581t}$

Now, the diameter of the dish is given to be 6 inches, so its radius is 3 inches implying that the area of the dish to be filled by the colony of bacteria is  A(1) = $\mathrm{\pi }$(3)³= 9$\mathrm{\pi }$. In order to find the time taken by the colony of bacteria to fill this area take A(1)=9$\mathrm{\pi }$ in equation (2) and solve for t

$$\begin{gathered} e^{0.4581 }=4.5\mathrm{\pi } \\ \mathrm{t}=\frac{1}{0.4581}\ln (4.5\mathrm{\pi }) \\ \mathrm{t}=5.782 \end{gathered}$$