# Growth problems: mold grows at a rate proportional to its present size

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Dutsky Kamdon
Growth problems: mold grows at a rate proportional to its present size

A mold grows at a rate proportional to its present size. Initially there is 2 oz of this mold,
and two days later there is 3 oz. Find (a) how much mold was present after one day and (b)
how much mold will be present in ten days.

Jhun Vert

$\dfrac{dP}{dt} = kP$

$\dfrac{dP}{P} = k \, dt$

$\displaystyle \int \dfrac{dP}{P} = k \int dt$

$\ln P = kt + C$

$\ln P = \ln e^{kt} + C$

$\ln P - \ln e^{kt} + C$

$C = \ln \dfrac{P}{e^{kt}}$

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

tyron

post lang ako uli sir

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