Question

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

Answer 1

Exponential model

`y=Ae^(rt)`

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:

`\begin{aligned}y & =Ae^(rt)\\\implies 100 & = Ae^(r * 0)\\100 & = Ae^0\\100 & = A(1)\\A & = 100\end{aligned}`

Part (b)

Given:

• A = 100 g
• y = 50 g when t = 30.17

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

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

Therefore, the final equation is:

`y=100e^{\left(-(\ln 2)/(30.17)\right)t}`

Question 1

Part (a)

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

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

Part (b)

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

10% of 100 g = 10 g

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

Question 2

Part (a)

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

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

Part (b)

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

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