Answer to Question #215338 in Differential Equations for Zerin

Question

Initially milligrams of a radioactive . After hours the mass had decayed by the rate of decay is proportional to the amount of substance present at time amount remaining after hours .

Answer 1

Initially 100 milligrams of a radioactive . After 6 hours the mass had decayed by 3% the rate of decay is proportional to the amount of substance present at time t amount remaining after 24 hours.

The rate of decay is proportional to the amount of substance present at time $t$

$$\dfrac{dA}{dt}=kA$$ $$\dfrac{dA}{A}=kdt$$ $$\int\dfrac{dA}{A}=\int kdt$$ $$\ln |A|=kt+\ln C$$ $$A=Ce^{kt }$$

Initially 100 milligrams of a radioactive

$$100=Ce^{k(0)}=>C=100$$ $$A(t)=100e^{kt}$$

After 6 hours the mass had decayed by 3%

$$97=100e^{6k}$$ $$6k=\ln0.97$$ $$k=\dfrac{1}{6}\ln 0.97$$

After 24 hours

$$A(24)=100e^{{\ln 0.97 \over 6}(24)}$$ $$A(24)=100e^{4\ln 0.97}$$ $$A(24)=100(0.97)^4$$ $$A=88.529281\ mg$$