Exponential Models

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