#Exponential Growth Calculus
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rate = 3000e^(3t/5)
let A = initial number of bacteria, then the number of bacteria at time t is:
B(t) = A(1 + 3000e^(3t/5))
t = 0 -> B(t) = 7500
7500 = A(1 + 1)
7500 = 2A
A = 3750
B(t) = 3750(1 + 3000e^(3t/5))
B(5) = 3750(1 + 3000³)
B(5) = 3750(1 + 27 × 10⁹) = ...
@pseudo harbor
im pretty sure we are supposed to sue derivatiaves
so like
so using smth like this
@pseudo harbor
oh so like
dy/dt = 3000e^(3t/5)
then solve for y
?
dy = 3000e^(3t/5) dt
u = 3t/5
du = (3/5) dt
dt = (5/3) du
dy = 3000e^u * (5/3) du
dy = 5000e^u du
integrate both sides
y = 5000e^u + C
y = 5000e^(3t/5) + C
t = 0 -> y = 7500
7500 = 5000 + C
C = 2500
y = 5000e^(3t/5) + 2500
t = 5:
y = 5000e³ + 2500 = ...
wdym
it's already taken into consideration in the working I gave
yes
ight thx a lot