Question

The function N(t) = 100e^−0.023t models the number of grams in a sample of cesium-137 that remain after t years. On which interval is the sample's average rate of decay the fastest?

Answer 1

Options:

A. [1,10]        B. [10,20]      C. [15,25]     D. [1,30]

Answer:

A. [1,10]

Step-by-step explanation:

Given

$N(t) = 100e^{-0.023t}$

Required

Interval with the fastest rate of decay

The rate of change over [a,b] is calculated as:

$Rate = \frac{N(b) - N(a)}{b - a}$

So, we have:

A. [1,10]        

$Rate = \frac{N(10)-N(1)}{10 - 1}$

$Rate = \frac{N(10)-N(1)}{9}$

Calculate N(10) and N(1)

$N(10) = 100e^{-0.023*10} = 100e^{-0.23} = 79.45$

$N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73$

So:

$Rate = \frac{79.45-97.73}{9} = \frac{-18.28}{9} = -2.03$

B. [10,20]      

$Rate = \frac{N(20)-N(10)}{20 - 10}$

$Rate = \frac{N(20)-N(10)}{10}$

Calculate N(20) and N(10)

$N(20) = 100e^{-0.023*20} = 100e^{-0.46} = 63.13$

$N(10) = 100e^{-0.023*10} = 100e^{-0.23} = 79.45$

So:

$Rate = \frac{63.13 - 79.45}{10} = \frac{-16.32}{10} = -1.632$

C. [15,25]    

$Rate = \frac{N(25)-N(15)}{25 - 15}$

$Rate = \frac{N(25)-N(15)}{20}$

Calculate N(25) and N(15)

$N(25) = 100e^{-0.023*25} = 100e^{-0.575} = 56.27$

$N(15) = 100e^{-0.023*15} = 100e^{-0.345} = 70.82$

So:

$Rate = \frac{56.27 - 70.82}{10} = \frac{-14.55}{10} = -1.455$

D. [1,30]

$Rate = \frac{N(30)-N(1)}{30 - 1}$

$Rate = \frac{N(30)-N(1)}{29}$

Calculate N(30) and N(1)

$N(30) = 100e^{-0.023*30} = 100e^{-0.69} = 50.16$

$N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73$

So:

$Rate = \frac{50.16 - 97.73}{29} = \frac{-47.57}{29} = -1.64$

By comparing the values of the calculated rates,

-2.03 is the smallest, in other words; the fastest rate

Hence, the interval [1,10] has the fastest rate of decay