The relation between y and t is y(6) = 19.8 mm^2.
What is an area in math?
As the area covered increases according to a continous growth model, we start with these relation
[tex]y (t) = C e^{kt}[/tex]
We know that the initial area y(0) is 8.6. Then
[tex]y(0) = Ce^{0} = C = 8.6[/tex]
We also know that the area duplicates after 5 days.
y( t + 5 ) = 2y(t)
[tex]\frac{y(0+5)}{y(0)} = 2 = \frac{e^{0+ 5k} }{e^{0} } = e^{5k}[/tex]
5k = In(2)
k = In(2)/5 = 0.14
Then, the model becomes
[tex]y (t) = 8.6e^{In(2).t/5}[/tex]
b) We have to calculate y(6)
[tex]y(6) = 8.6e^{In(2). t /5}[/tex] [tex]= 8.6e^{0.832} = 8.6 * 2.297 = 19.758[/tex]
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