Calculus Problems. Please help me

Question

There was initially 100mg of radioactive material. After 6 hours, its mass decreased by 3% . If at any instance its disintegration is proportional to the quantity of material present, determine how much is left after 24 hours.

Answer 1

d N / d t = k N

d N / N = k d t

∫ d N / N = ∫ k d t

ln | N | = k t + c

| N | = e ^{k t + c}

| N | = e ^{k t} e ^c

| N | = λ e ^{k t}

N = ± λ e ^{k t}

N = ξ e ^{k t} Now when t = 0 then N = 100

100 = ξ Then N = 100 e ^{k t}

Also when t = 6 then N = 97

That is 97 = 100 e ^{k 6} Then taking the natural log of both sides

ln ( 0.97 ) = ln ( e ^{6 k} ) ⇒ - 0.03046 = 6 k ⇒ k = -0.00508

Hence N = 100 e ^{-0.00508 t} and for t = 24 N = 100 e ^-(0.00508)24

N = 100 e ^{- 0.12183}

N = 100 ( .88529)

N = 88. 529

Answer 2

Let us do this;

Let N be the amount of mass at time t.

dN/dt α N

dN/dt = -λN

dN/N = -λdt

ln N = -λt + C

When t=0 N=100

ln N = -λt + ln 100

N = 100e^-λt... (I)

At t=6 N=6*97% = 97

97 = 100e^-λ*6

⁶√(97/100) = ⁶√(e^-λ6)

∴ N = 100(0.97)^t/6... (II)

At t = 24;

N = 100(0.97)^24/6

N = 88.53

After 24hrs, 88.53mg of Radioactive material is left.