AI HL · 12.7
0 of 36 skills done
Sinusoidal and logistic models
12.7.4
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
Change L, C and k. The dashed line is the carrying capacity. The orange point is where the population grows fastest. Say which of the three values it depends on before you test it.
Move a slider.
These are depth readings from a harbour. Before you move anything, work out the mean d and the amplitude a from the highest and lowest readings. Then drag the orange points on the curve, or use the sliders, until the curve passes close to the readings. The teal dots are the readings that are already close.
A rider goes round a Ferris wheel. Choose where the rider gets on and write down the model first. Then drag the orange point along the curve: the rider on the wheel is at the same height. Drag the centre or the top of the wheel to change d and a.
Move a slider.
The depth of water in a harbour is modelled by \(D=a\sin(b(t-c))+d\), where t is in hours. The mean depth is 4.5 m, the amplitude is 1.8 m and the period is 12 hours. At t = 2 the depth is the mean depth and rising. (a) Find a, b, c and d. (b) Find the depth at t = 4.
Have a go first. Then open the working one line at a time.
a = 1.8, d = 4.5
The amplitude is a. The mean is d.
\(b=2\pi\div12=0.524\)
Radians. The period is 12 hours.
c = 2
The curve is on the mean and rising at t = 2. That is the phase shift.
\(D=1.8\sin(0.524(t-2))+4.5\)
Check by graphing: the maximum is 6.3 and the minimum is 2.7.
\(D=1.8\sin(0.524\times2)+4.5=6.06\)
About 6.06 m.
A water wave has amplitude 0.6 m. The still-water level is 2.5 m above the sea bed. The wavelength is 8 m and the phase shift is 1 m. The height of the surface is \(h=a\sin(b(x-c))+d\) where x is the horizontal distance in metres. (a) Write the model. (b) Find h when x = 3.
Have a go first. Then open the working one line at a time.
a = 0.6, d = 2.5
The central line is the still-water level.
\(b=2\pi\div8=0.785\)
For a wave, the wavelength is the length of one period.
\(h=0.6\sin(0.785(x-1))+2.5\)
The shift is 1, so (x − 1).
\(h=0.6\sin(0.785\times2)+2.5=3.1\)
The surface is at its crest: 2.5 + 0.6 = 3.1 m.
The temperature in a greenhouse varies between a maximum of 31°C and a minimum of 13°C. The period is 24 hours and the phase shift is 6 hours. Model it as \(T=a\sin(b(t-c))+d\). (a) Find the central line d. (b) Write the model. (c) Find T when t = 10.
Have a go first. Then open the working one line at a time.
\(d=\dfrac{31+13}{2}=22\)
The central line is halfway between the max and the min.
\(a=\dfrac{31-13}{2}=9\)
Half the range.
\(b=2\pi\div24=0.262\)
The period is 24 hours.
\(T=9\sin(0.262(t-6))+22\)
At t = 10: 29.8°C.
A population of deer on a reserve is modelled by \(P=\dfrac{L}{1+Ce^{-kt}}\), t in years. The reserve can support 800 deer. There are 20 deer at t = 0 and 65 deer at t = 3. Find C and k.
Have a go first. Then open the working one line at a time.
L = 800. At t = 0: \(20=\dfrac{800}{1+C}\), so C = 800 ÷ 20 − 1 = 39
At t = 0, \(e^{0}=1\).
\(65=\dfrac{800}{1+39e^{-3k}}\)
Put in the second population.
Solve for k: k = 0.413
Use the calculator's solver.
\(P=\dfrac{800}{1+39e^{-0.413t}}\)
Check: at t = 3 it gives 65.
A school's flu cases are modelled by \(P=\dfrac{1200}{1+15e^{-0.6t}}\), where t is in days. (a) Find P when t = 5. (b) Find when P reaches 1000.
Have a go first. Then open the working one line at a time.
\(P=\dfrac{1200}{1+15e^{-0.6\times5}}=687\)
About 687 cases.
\(\dfrac{1200}{1+15e^{-0.6t}}=1000\)
Set P = 1000.
Solve for t: t = 7.20
About 7.20 days. The model never passes 1200, so 1000 is reachable.
The graph shows a logistic model \(P=\dfrac{300}{1+14e^{-0.4t}}\) (solid) and an exponential model with the same start (dashed). (a) Write down the carrying capacity. (b) Find the initial population. (c) Find when the logistic population is growing fastest. (d) Compare the two models for large t.
Have a go first. Then open the working one line at a time.
L = 300
The dashed line is where the logistic curve levels off.
\(P(0)=\dfrac{300}{1+14}=20\)
L ÷ (1 + C).
\(t=\dfrac{\ln14}{0.4}=6.60\), when P = 150
The steepest point is at half the carrying capacity.
Logistic levels off at 300. Exponential keeps growing.
Both start at 20 and look alike at first. The exponential has no limit, so it is only a good model while resources are plentiful.