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?

Answers

Answer 1

Options:

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

Answer:

A. [1,10]

Step-by-step explanation:

Given

[tex]N(t) = 100e^{-0.023t}[/tex]

Required

Interval with the fastest rate of decay

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

[tex]Rate = \frac{N(b) - N(a)}{b - a}[/tex]

So, we have:

A. [1,10]        

[tex]Rate = \frac{N(10)-N(1)}{10 - 1}[/tex]

[tex]Rate = \frac{N(10)-N(1)}{9}[/tex]

Calculate N(10) and N(1)

[tex]N(10) = 100e^{-0.023*10} = 100e^{-0.23} = 79.45[/tex]

[tex]N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73[/tex]

So:

[tex]Rate = \frac{79.45-97.73}{9} = \frac{-18.28}{9} = -2.03[/tex]

B. [10,20]      

[tex]Rate = \frac{N(20)-N(10)}{20 - 10}[/tex]

[tex]Rate = \frac{N(20)-N(10)}{10}[/tex]

Calculate N(20) and N(10)

[tex]N(20) = 100e^{-0.023*20} = 100e^{-0.46} = 63.13[/tex]

[tex]N(10) = 100e^{-0.023*10} = 100e^{-0.23} = 79.45[/tex]

So:

[tex]Rate = \frac{63.13 - 79.45}{10} = \frac{-16.32}{10} = -1.632[/tex]

C. [15,25]    

[tex]Rate = \frac{N(25)-N(15)}{25 - 15}[/tex]

[tex]Rate = \frac{N(25)-N(15)}{20}[/tex]

Calculate N(25) and N(15)

[tex]N(25) = 100e^{-0.023*25} = 100e^{-0.575} = 56.27[/tex]

[tex]N(15) = 100e^{-0.023*15} = 100e^{-0.345} = 70.82[/tex]

So:

[tex]Rate = \frac{56.27 - 70.82}{10} = \frac{-14.55}{10} = -1.455[/tex]

D. [1,30]

[tex]Rate = \frac{N(30)-N(1)}{30 - 1}[/tex]

[tex]Rate = \frac{N(30)-N(1)}{29}[/tex]

Calculate N(30) and N(1)

[tex]N(30) = 100e^{-0.023*30} = 100e^{-0.69} = 50.16[/tex]

[tex]N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73[/tex]

So:

[tex]Rate = \frac{50.16 - 97.73}{29} = \frac{-47.57}{29} = -1.64[/tex]

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

Answer 2

The required average rate of change is the fastest on [tex][1,10][/tex].

The average rate of change :

The formula for the average rate of change on the interval [tex][a,b][/tex]  is,

[tex]\frac{100e^{-0.023b}-100e^{-0.023b}}{b-a}[/tex]

Now, applying the given function into the above function is,

Now, [tex]a=1,b=10[/tex] then,

[tex]\frac{100e^{-0.023b}-100e^{-0.023b}}{10-1} =-2.03[/tex]

Now, [tex]a=10,b=20[/tex] then,

[tex]\frac{100e^{-0.023b}-100e^{-0.023b}}{20-10} =-1.6325[/tex]

Now, [tex]a=15,b=25[/tex] then,

[tex]\frac{100e^{-0.023b}-100e^{-0.023b}}{25-15} =-1.455[/tex]

Now, [tex]a=1,b=30[/tex] then,

[tex]\frac{100e^{-0.023b}-100e^{-0.023b}}{30-1} =-1.64[/tex]

So the average rate of change is fastest on [tex][1,10][/tex]

Learn more about the topic of the average rate of change:

https://brainly.com/question/20884505


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