Radium 226 has a decay rate that satisfies the differential equation D'(t)= -.00043D(t). Assume that D(0)=100.What is D(t)?How much is left after 100 years?(round to nearest hundredth.)To the nearest year, what is the life half life of Radium 226?

Answer:A) [tex]D(t)=100e^{-0.00043t}[/tex]B)[tex]{D(t=100)}=95.8[/tex]C) τ=2326 yearsStep-by-step explanation:A) Consider the differential equation:[tex]\frac{dD}{dt}=-0.00043D[/tex]Separate variables dD and dt:[tex]\frac{dD}{D}=-0.00043dt[/tex]Integrate: Remember that The integral of dD/D is ln(D).[tex]\int\limits^{D}_{D_0} {\frac{1}{D} } \, dD=\int\limits^{t}_{0} {-0.00043} \, dt[/tex][tex]ln(\frac{D}{D_0})=-0.00043t\\\frac{D}{D_0}=e^{-0.00043t}[/tex][tex]D=D_0*e^{-0.00043t}[/tex]And do [tex]D_0=100[/tex], so:[tex]D=100*e^{-0.00043t}[/tex]B)The function we have found is a function that let us know how many of Ra is after t years, so, evaluate the function at t=100:[tex]D(t=100)=100*e^{-0.00043*100}=95.791\\[/tex]Which rounded is 95.8.C) For this equation it is known that the life half life τ is equal to (1/λ), where λ is the exponential number that is multiplying to -t in the D(t) function; so for this case, λ=0.00043, and the life half life is [tex]\frac{1}{0.00043}=2325.58[/tex],which rounded is equal to 2326 years.