Answer to Question #273170 in Differential Equations for Adnan

Question

Initial 100 milligram of a radio active substance was present.After 6 hour the mass has been decreased by 3%. If the rate of decay is proportional to the amount of the substance present at time t,find the amount remaining after 24hours.

Answer 1

Let $A(t)=$ the amount of radio active substance that re,aims after $t$ hours in

milligram.

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

$$\dfrac{dA}{dt}=kA$$

$$\dfrac{dA}{A}=ktdt$$

Integrate

$$\int \dfrac{dA}{A}=\int ktdt$$

$$A(t)=Ce^{kt}$$

Initial 100 milligram of a radio active substance was present

$$A(0)=C=100$$

Then

$$A(t)=100e^{kt}$$

After 6 hour the mass has been decreased by 3%

$$e^{6t}=1-0.3$$

$$t=\dfrac{\ln 0.97}{6}$$

Then

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

$$A(t)=100(0.97)^{t/6}$$

$$A(24)=100(0.97)^{24/6}=100(0.97)^4$$

$$=88.529281\approx88.53 \ (mg)$$

88.53 milligram of a radio active substance remained after 24 hours.