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Precalculus. Please look at the picture

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Radioactive Cesium 137 was released into the waters off the coast of Japan and continues to leak into the Pacific Ocean. Considering that the half-life of Cesium is λ =30.17 years, create an exponential model, y=aert use it to make some predictions

y= Aert

Precalculus Please look at the picture Show steps Radioactive Cesium 137 was released into the waters off the coast of Japan and continues to leak into the Paci class=

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Answer:

Exponential model

[tex]y=Ae^{rt}[/tex]

where:

  • y = value at "t" time
  • A = initial value
  • r = rate of growth/decay
  • t = time (in years)

Part (a)

Given:

  • y = 100 g
  • t = 0 years

Substituting given values into the formula and solving for A:

[tex]\begin{aligned}y & =Ae^{rt}\\\implies 100 & = Ae^{r \times 0}\\100 & = Ae^0\\100 & = A(1)\\A & = 100\end{aligned}[/tex]

Part (b)

Given:

  • A = 100 g
  • y = 50 g when t = 30.17

Substituting the given values into the equation and solving for r:

[tex]\begin{aligned}y& =Ae^{rt}\\\\\implies 50 & =100e^{30.17r}\\\\\dfrac{1}{2} & = e^{30.17r}\\\\ln \dfrac{1}{2} & = \ln e^{30.17r}\\\\\ln 1-\ln2 & =30.17r \ln e\\\\0-\ln 2 & =30.17r(1)\\\\-\ln 2 & =30.17r\\\\r & = \dfrac{-\ln 2}{30.17}\end{aligned}[/tex]

Therefore, the final equation is:

[tex]y=100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}[/tex]

Question 1

Part (a)

Q:   From 100g how much remains in 80 years?

[tex]\begin{aligned}t=80 \implies y & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)80}\\& = 15.91389949 \: \sf g\end{aligned}[/tex]

Part (b)

Q:  How long will it take to have 10% remaining?

10% of 100 g = 10 g

[tex]\begin{aligned}y=10 \implies 10 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{10} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{10} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 10}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 100.2225706\: \sf years\end{aligned}[/tex]

Question 2

Part (a)

Q:   How much remains after 50 years (time)?

[tex]\begin{aligned}t=50 \implies y & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)50}\\& = 31.70373153 \: \sf g\end{aligned}[/tex]

Part (b)

Q:   How long to reach 20 g (amount remaining)?

[tex]\begin{aligned}y=20 \implies 20 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{5} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{5} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 5}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 70.05257062\: \sf years\end{aligned}[/tex]

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