1. In a laboratory experiment you measure the amount of some radioactive substan

Question

1. In a laboratory experiment you measure the amount of some radioactive substance to be 20
grams. You know that exactly five days ago you still had 25 grams. Find the specic decay rate;
determine and graph the amount of material as a function of time; and compute the following
quantities:

(1) the amount that will remain ten days from now;

(2) the amount that must have been present ten days ago; and

(3) the period of time that it takes for 50% of the initial amount
to decay.
Hint: Let x(t) denote the amount (in grams) present at time t (in days). Then x(0) = 20 and
x(-5) = 25. Use these data to determine the constants c and k in the formula x(t) = ce^-kt (the
general solution of the exponential-decay equation dx/dt = -kx).

Please show all work :)

Explanation / Answer

In a laboratory experiment you measure the amount of some radioactive substance to be 20
grams. You know that exactly five days ago you still had 25 grams. Find the specic decay rate;
determine and graph the amount of material as a function of time; and compute the following
quantities:

Let x(t) denote the amount (in grams) present at time t (in days).

Then x(0) = 20 and x(-5) = 25.

dx/dt = -kx

dx/x = -k dt

take integration

lnx = -kt+C

x(t) = C e^-kt

x(0) = 20 thus

x(0) = C = 20

x(t) = 20 e^-kt

x(-5) = 25

x(-5) = 20 e^5k = 25

5k = ln(25/20) = ln(1.25) = 0.2231435513142098

k = 0.04462871026284196

decay constant = 0.04462871026284196


(1) the amount that will remain ten days from now;

x(10) = 20 e^(-0.04462871026284196*10) = 12.799999999999998

= 12.8 grams.

(2) the amount that must have been present ten days ago; and

x(-10) = 20 e^(-0.04462871026284196*-10) = 31.25 grams.

(3) the period of time that it takes for 50% of the initial amount
to decay.

x(t) = x(t)/2

x(0) = 20

x(0)/2 = 10

10 = 20 e^(-0.04462871026284196*t)

1/2 = e^(-0.04462871026284196t)

ln(0.5) = -0.04462871026284196t

t = 15.53141859752694609021

after 15.53 dats

it reduces to half of its inital amount.

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