1 / 19

Exponential Models

Exponential Models. Thursday, April 24 th. Derivative Rules Warm Up. To make sure that no one is wobbly on derivative rules, we have a page of practice. Half-life. 6mg of a radioactive isotope of gold, Au-198, decays into 4.6mg after 1day. What is the decay constant for Au-198?

marly
Télécharger la présentation

Exponential Models

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Exponential Models Thursday, April 24th

  2. Derivative Rules Warm Up • To make sure that no one is wobbly on derivative rules, we have a page of practice

  3. Half-life 6mg of a radioactive isotope of gold, Au-198, decays into 4.6mg after 1day. • What is the decay constant for Au-198? • What is the half-life of Au-198? • What is the decay rate after 3 days?

  4. Half-life 6mg of a radioactive isotope of gold, Au-198, decays into 4.6mg after 1day. • What is the decay constant for Au-198? N = N0e–λt

  5. Half-life 6.0mg of a radioactive isotope of gold, Au-198, decays into 4.6mg after 1day. • What is the decay constant for Au-198? 4.6 = 6.0e–λ(1) 4.6/6 = e–λ(1) ln(4.6/6) = –λ 0.266 = λ

  6. Half-life 6mg of a radioactive isotope of gold, Au-198, decays into 4.6mg after 1day. • What is the decay constant for Au-198? • What is the half-life of Au-198? N = N0e–0.266t N/N0 = e–0.266t ½ = e–0.266t ln(½) = –0.266t t½ = 2.6 days

  7. Half-life 6mg of a radioactive isotope of gold, Au-198, decays into 4.6mg after 1day. • What is the decay constant for Au-198? • What is the half-life of Au-198? • What is the decay rate after 3 days? N = N0e–0.266t N’(t) = –0.266N0 e–0.266t N’(3) = –0.266N0 e–0.266(3) N’(3) = –0.72 mg/day

  8. Damped Oscillators On a whiteboard, make a guess for what this function might look like: f(t) = e-0.5t sin(10t)

  9. Damped Oscillators On a whiteboard, make a guess for what this function might look like: f(t) = e-0.5t sin(10t)

  10. Damped Oscillators On a whiteboard, make a guess for what this function might look like: f(t) = e-0.5t sin(5t)

  11. Damped Oscillators On a whiteboard, make a guess for what this function might look like: f(t) = e-0.5t sin(5t)

  12. Damped Oscillators How are these two functions the same? Different? f(t) = e-0.5t sin(10t) f(t) = e-0.5t sin(5t)

  13. Damped Oscillators • A damped oscillator is a sinusoidal function trapped inside an exponential • The exponential coefficient is controlling the amplitude of the sinusoidal function inside

  14. Shock Absorbers The vertical displacement of a car driving over potholes in Chatham can be modelled by the function (in cm for h and seconds for t): h(t) = e-0.5t sin(t) • Graph this function. • Determine the maximum displacement of the car from it’s equilibrium position.

  15. Shock Absorbers The vertical displacement of a car driving over potholes in Chatham can be modelled by the function (in cm for h and seconds for t): h(t) = e-0.5t sin(t) • Graph this function. On a whiteboard, make a guess for the shape of this function

  16. Shock Absorbers The vertical displacement of a car driving over potholes in Chatham can be modelled by the function (in cm for h and seconds for t): h(t) = e-0.5t sin(t) • Graph this function.

  17. Shock Absorbers h(t) = e-0.5t sin(t) • Graph this function. • Determine the maximum displacement of the car from it’s equilibrium position. h(t) = e-0.5t sin(t) h’(t) = -0.5e-0.5t sin(t) + e-0.5tcos(t) 0= -0.5e-0.5t sin(t) + e-0.5tcos(t) 0.5e-0.5t sin(t) = e-0.5tcos(t) 0.5sin(t) = cos(t) 0.5tan(t) = 1 tan(t) = 2

  18. Shock Absorbers h(t) = e-0.5t sin(t) • Graph this function. • Determine the maximum displacement of the car from it’s equilibrium position. tan(t) = 2 t = 1.1 seconds h(t) = e-0.5t sin(t) h(1.1) = e-0.5(1.1) sin(1.1) h(1.1) = 0.51 cm

  19. Assignment Check! • Homework: Page 289 #1 – 3, 7 – 10, 18

More Related