Question

the bones of a newly discovered dinosaur weigh 170 pounds and measure 9 feet, with a 6 - inch claw on one toe of each hind foot. the age of the dinosaur was estimated using a radioactive substance dating of rocks surrounding the bones. complete parts a and b. the radioactive substance decays exponentially with a half - life of approximately 1.34 billion years. use the fact that after 1.34 billion years a given amount of the radioactive substance will have decayed to half the original amount to show that the decay model for the radioactive substance is given by a = a₀e⁻⁰.⁵¹⁷²⁷ᵗ, where t is in billions of years. to show the decay model for the radioactive substance, find the decay rate k for a substance. substitute the values of a and t in the exponential decay model a = a₀eᵏᵗ. substitute: a = a₀eᵏᵗ, (\frac{a₀}{2}=a₀e^{1.34k}). divide both sides of the equation by a₀: (\frac{1}{2}=e^{1.34k})

Explanation:

Step1: Recall Decay Model

The exponential decay model is \( A = A_0e^{kt} \), and for half - life, when \( t = 1.34 \) (billion years), \( A=\frac{A_0}{2} \).

Step2: Substitute Values

Substitute \( A=\frac{A_0}{2} \), \( t = 1.34 \) into \( A = A_0e^{kt} \), we get \( \frac{A_0}{2}=A_0e^{k\times1.34} \).

Step3: Divide by \( A_0 \)

Divide both sides of the equation \( \frac{A_0}{2}=A_0e^{1.34k} \) by \( A_0 \) (assuming \( A_0
eq0 \)), we have \( \frac{1}{2}=e^{1.34k} \).

Step4: Take Natural Logarithm

Take the natural logarithm of both sides: \( \ln(\frac{1}{2})=\ln(e^{1.34k}) \). Since \( \ln(e^{x}) = x \), this simplifies to \( \ln(\frac{1}{2})=1.34k \).

Step5: Solve for \( k \)

We know that \( \ln(\frac{1}{2})=-\ln(2)\approx - 0.6931 \). So, \( k=\frac{\ln(\frac{1}{2})}{1.34}=\frac{- 0.6931}{1.34}\approx - 0.51727 \), which matches the given \( k=-0.51727 \).

Answer:

The decay rate \( k\approx - 0.51727 \) is derived by substituting \( A = \frac{A_0}{2} \) and \( t = 1.34 \) into the exponential decay model \( A = A_0e^{kt} \), dividing by \( A_0 \), taking the natural logarithm, and solving for \( k \), which confirms the given decay model.