exp growth_decay

Y t 100001 1 t 2 20000 100001 1 t 2 2 1 1 t 2 ln 2 ln

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
y ( t ) = 10000(1 . 1) t/ 2 20000 = 10000(1 . 1) t/ 2 2 = 1 . 1 t/ 2 ln 2 = ln 1 . 1 t/ 2 = t 2 (ln 1 . 1) t = 2 ln 2 ln 1 . 1 14 . 55 yrs
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Example 2a Radioactive Decay. Find rate constant k given a fraction (or percentage) of y 0 at time t . The rate at which a radioactive element decays is approximately proportional to the number of radioactive nuclei present. A radioactive element’s half-life is the time required for half of the nuclei to decay. Radioactive carbon-14 has a half-life of 5700 years. Find the relative growth rate k . Solution: We wish to find the carbon-14 rate constant given y (5700) = y 0 / 2 . y ( t ) = y 0 e kt y (5700) = y 0 e k (5700) y 0 2 = y 0 e 5700 k 1 2 = e 5700 k ln 1 2 = ln e 5700 k = 5700 k k = ln 1 2 5700 = - ln 2 5700 (Note that ln 1 2 = ln 2 ( - 1) = - ln 2 .) Example 2b Find time t given k and a fraction (or percentage) of y 0 . A fossilized animal bone contains 30% of its original carbon-14 content. Approximately how many years ago did the animal die? Solution: Use the carbon-14 rate constant k = - ln 2 5700 to find t when y ( t ) = 0 . 3 y 0 . y ( t ) = y 0 e kt 0 . 3 y 0 = y 0 e kt 0 . 3 = e kt ln 0 . 3 = ln e kt = kt t = ln 0 . 3 k = - 5700 ln 0 . 3 ln 2 9901 years Example 3 Newton’s Law of Cooling. Find time t given T 0 , T S and temperature T at time t . A hot bowl of soup cools from 70 C to 60 C after 10 minutes inside a re- frigerator that measures - 5 C . How many more minutes will it take for the soup to reach 40 C ?
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern