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Re: Growth problems
$\dfrac{dP}{dt} = kP$
$\dfrac{dP}{P} = k \, dt$
$\displaystyle \int \dfrac{dP}{P} = k \int dt$
$\ln P = kt + C$
$\ln P = \ln e^{kt} + C$
$\ln P - \ln e^{kt} + C$
$C = \ln \dfrac{P}{e^{kt}}$
When t = 0, P = 2
$C = \ln \dfrac{2}{e^{0}}$
$C = \ln 2$
Hence,
$\ln 2 = \ln \dfrac{P}{e^{kt}}$
$2 = \dfrac{P}{e^{kt}}$
$P = 2e^{kt}$
When t = 2, P = 3
$3 = 2e^{2k}$
$\dfrac{3}{2} = e^{2k}$
$e^k = \left( \dfrac{3}{2} \right)^{1/2}$
Thus,
$P = 2\left( \dfrac{3}{2} \right)^{t/2}$
(a) for t = 1
$P = 2\left( \dfrac{3}{2} \right)^{1/2} = 2.4495 ~ \text{oz}$ answer
(b) for t = 10
$P = 2\left( \dfrac{3}{2} \right)^{5} = 15.1875 ~ \text{oz}$ answer
Re: Growth problems: mold grows at a rate proportional to its...
post lang ako uli sir