Question

The spread of a fungal infection in an ant colony

Answer 1

Given the equation:

$\frac{dy}{dt}=\frac{y+1}{2\sqrt[\placeholder{⬚}]{t}}$

When t=1, y=3 (three percent of the ants are infected).

Replacing in the equation:

$\frac{dy}{dt}=\frac{3+1}{2\sqrt{1}}=2$

a) The equation of the line tangent to the graph is:

$\begin{gathered} y=3+2(t-1) \\ y=3+2t-2=2t+1 \end{gathered}$

In t= 1.2.

$\begin{gathered} y(t)=2t+1 \\ y(1.2)=2(1.2)+1=3.4 \end{gathered}$

The percentage of the colony infected in t=1.2 is 3.4.

b) y explicity:

$\begin{gathered} \frac{dy}{dt}=\frac{y+1}{2\sqrt{t}} \\ \\ \frac{dy}{y+1}=\frac{dt}{2\sqrt[\placeholder{⬚}]{t}} \end{gathered}$

integrating both sides:

$\int\frac{dy}{y+1}=\int\frac{dt}{2\sqrt[\placeholder{⬚}]{t}}$

The first one will be:

$\begin{gathered} \int\frac{dy}{y+1} \\ u=y+1 \\ du=dy \\ \int\frac{du}{u}=ln(u)=ln(y+1)+c \end{gathered}$

The second one, will be:

$\begin{gathered} \frac{1}{2}\int\frac{dt}{\sqrt[\placeholder{⬚}]{t}} \\ \frac{1}{2}\int t^{-\frac{1}{2}}dt=\frac{1}{2}*(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1})=\frac{1}{2}(\frac{t^{\frac{1}{2}}}{\frac{1}{2}})=t^{\frac{1}{2}}+k \\ \end{gathered}$

Substituing:

$ln(y+1)+c=t^{\frac{1}{2}}+k$

Using D=k-c

$\begin{gathered} ln(y+1)=t^{\frac{1}{2}}+D \\ y+1=e^{\sqrt[\placeholder{⬚}]{t}}*e^D \end{gathered}$

Finally:

$y=e^{\sqrt[\placeholder{⬚}]{t}}e^D-1$