We begin with 20g of a certain radioactive isotope. As it decays, the amount remaining at time t (hrs) is A = 20e-0.08t. What is its rate of decay at time t = 8 ? solution

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Leaving the answer to two decimal places

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Answer 1 · 2018-05-07T11:52:05

#-0.84# grams per hour, or loss of #0.84# grams per hour.

Well, the decay function here is:

#A=20e^(-0.08t)#

where:

#A# is the amount remaining

#t# is the time taken in hours

So, the rate of decay will be the derivative of the amount remaining, as it will the rate of change of the amount left.

Differentiating respect to time, we get:

#A'=d/dt(20e^(-0.08t))#

#=20d/dt(e^(-0.08t))#

Let's compute #d/dt(e^(-0.08t))# by the chain rule, which states that,

#dy/dx=dy/(du)*(du)/dx#

Let #u=-0.08t,:.(du)/dt=-0.08#.

Then #y=e^u,:.dy/(du)=e^u#.

Combining, we get:

#dy/dx=e^u*-0.08#

#=-0.08e^u#

Substituting back #u=-0.08t#, we get:

#=-0.08e^(-0.08t)#

So, the derivative of #A# becomes,

#A'=20*-0.08e^(-0.08t)#

#=-1.6e^(-0.08t)#

At #t=8#, its rate of decay is:

#A'(8)=-1.6e^(-0.08*8)#

#=-1.6e^-0.64#

#~~-0.84# to two decimal places