It's amazing how you can have this question whose answer hardly involves any numbers and uses niche terms like e and ln and c and still manages to get 180. God maths is beautiful!
@9:28 Shouldn't the left side of the equation integrate to (ln |p| / 2)? Don't you factor out the 1/2 as a constant instead of using the reverse chain rule?
The last two are wrong actually. The volume one should be V=e^((-10ln(50)+t)/-10)+150, and hence the volume is ~91ml The population one is P= e^(2t+20√10)
Idk about the volume one but I know the population one is correct. You don't have to divide by 2P, you can just divide by P giving us 1/P dP = 2dt ∫1/P dP = ∫2dt lnP = 2t + C P sub 0 = 1000 therefore ln1000 = 2(0) + C P = e^(2t + ln1000) = 1000e^2t
