Answer to Question #215349 in Differential Equations for Zerin

Question

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 .

Answer 1

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

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

$$A(24)=100e^{4\ln 0.97}$$

$$A(24)=100(0.97)^4$$

$$A=88.529281\ mg$$