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. a. the radioactive substance decays exponentially with a half - life of approximately 1.22 billion years. use the fact that after 1.22 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 = a_0e^{-0.56815t} ), where t is in billions of years. to show that the decay model for the radioactive substance is ( a = a_0e^{-0.56815t} ), find the decay rate k for a substance. substitute the values of a and t in the exponential decay model, ( a = a_0e^{kt} ). ( a = a_0e^{kt} ) ( \frac{a_0}{2}=a_0e^{1.22k} ) substitute. ( \frac{1}{2}=e^{1.22k} ) divide both sides of the equation by ( a_0 ). ( lnleft(\frac{1}{2}
ight)=1.22k ) take the natural logarithm on both sides and simplify. ( - 0.56815=k ) solve for k. (round to five decimal places as needed.) b. analysis of the rocks surrounding the dinosaur bones indicated that 94.3% of the original amount of radioactive substance was still present. let ( a = 0.943a_0 ) in the model in part (a) and estimate the age of the bones of the dinosaur. the estimated age of these dinosaur bones is about ( square ) billion years. (round to four decimal places as needed.)

Explanation:

Step1: Substitute \(A = 0.943A_0\) into the decay model

Given the decay model \(A=A_0e^{- 0.56815t}\), substitute \(A = 0.943A_0\) into it. We get \(0.943A_0=A_0e^{-0.56815t}\).

Step2: Simplify the equation

Divide both sides of the equation \(0.943A_0 = A_0e^{-0.56815t}\) by \(A_0\) (since \(A_0
eq0\)). The equation simplifies to \(0.943=e^{-0.56815t}\).

Step3: Take the natural logarithm of both sides

Take the natural - logarithm of both sides: \(\ln(0.943)=\ln(e^{-0.56815t})\). Using the property \(\ln(e^{x})=x\), the right - hand side simplifies to \(-0.56815t\). So, \(\ln(0.943)=- 0.56815t\).

Step4: Solve for \(t\)

We know that \(\ln(0.943)\approx - 0.0589\). Then, \(t=\frac{\ln(0.943)}{-0.56815}\). Substitute \(\ln(0.943)\approx - 0.0589\) into the formula: \(t=\frac{-0.0589}{-0.56815}\approx0.1037\).

Answer:

\(0.1037\)