AP Calculus ABhardmcq1 pt

The mass of a radioactive isotope decreases at a rate proportional to the mass present, and its half-life is 4040 days. Which differential equation models the mass mm, where tt is in days?

A.dmdt=ln240m\dfrac{dm}{dt} = -\dfrac{\ln 2}{40}\,m
B.dmdt=40m\dfrac{dm}{dt} = -40m
C.dmdt=ln240m\dfrac{dm}{dt} = \dfrac{\ln 2}{40}\,m
D.dmdt=140m\dfrac{dm}{dt} = -\dfrac{1}{40}\,m

Explanation

Core Concept

Solving dmdt=km\dfrac{dm}{dt} = km gives m=m0ektm = m_0e^{kt}; a half-life of 4040 days means e40k=12e^{40k} = \dfrac{1}{2}, so k=ln240k = -\dfrac{\ln 2}{40}.

Correct Answer

Admdt=ln240m\dfrac{dm}{dt} = -\dfrac{\ln 2}{40}\,m

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