Sinusoidal and logistic models Problem set B Worked solutions

A quantity P is modelled by \(P=a\sin(b(t-c))+d\). The amplitude is 7, the central line is P = 25, the period is 20 and the phase shift is 3.

1. (a) Write down the model.

   • a = 7, d = 25, c = 3, \(b=2\pi\div20=0.314\)
   • \(P=7\sin(0.314(t-3))+25\)

2. (b) Find P when t = 8.

   • \(P=7\sin(0.314\times5)+25=32\)

A quantity P is modelled by \(P=a\sin(b(t-c))+d\). Its maximum is 48 and its minimum is 12. The period is 8 and the phase shift is 1.

1. (a) Write down the model.

   • \(a=(48-12)\div2=18\), \(d=(48+12)\div2=30\), \(b=2\pi\div8=0.785\)

2. (b) Find P when t = 2.5.

   • \(P=18\sin(0.785(2.5-1))+30=46.6\)

A quantity P is modelled by \(P=a\sin(b(t-c))+d\). The mean is 55, the amplitude is 15 and the period is 10. At t = 2 the quantity is at its mean and rising.

1. (a) Write down the model.

   • d = 55, a = 15, \(b=2\pi\div10=0.628\)
   • The curve is on the mean and rising at t = 2, so c = 2
   • \(P=15\sin(0.628(t-2))+55\)

2. (b) Find P when t = 3.5.

   • \(P=15\sin(0.628\times1.5)+55=67.1\)

A population is modelled by \(P=\dfrac{L}{1+Ce^{-kt}}\), with t in years. The carrying capacity is 350. The population is 14 at t = 0 and 60 at t = 3.

1. (a) Find C.

   • At t = 0: \(14=\dfrac{350}{1+C}\), so C = 350 ÷ 14 − 1 = 24

2. (b) Find k.

   • \(60=\dfrac{350}{1+24e^{-3k}}\)
   • Solve for k: k = 0.534

A population is modelled by \(P=\dfrac{640}{1+9e^{-0.8t}}\), where t is in hours.

1. (a) Find P when t = 3.

   • \(P=\dfrac{640}{1+9e^{-0.8\times3}}=352\)

2. (b) Find t when P = 400.

   • \(\dfrac{640}{1+9e^{-0.8t}}=400\)
   • Solve for t: t = 3.39

The pedal of a bicycle has a mean height of 0.45 m above the ground and an amplitude of 0.17 m. One turn takes 0.8 seconds. At t = 0.1 seconds the pedal is at its mean height and rising. Model the height as \(h=a\sin(b(t-c))+d\).

1. (a) Write a model for the height h m of the pedal, t seconds after the start.

   • d = 0.45, a = 0.17, \(b=2\pi\div0.8=7.85\), c = 0.1
   • \(h=0.17\sin(7.85(t-0.1))+0.45\)

2. (b) Find the height at t = 0.35.

   • \(h=0.17\sin(7.85\times0.25)+0.45=0.607\)

The electricity demand in a city has a maximum of 52 GW and a minimum of 28 GW. The period is 24 hours and the phase shift is 6 hours. Model it as \(E=a\sin(b(t-c))+d\).

1. (a) Write a model for the demand \(E\) GW, t hours after midnight.

   • \(a=(52-28)\div2=12\), \(d=(52+28)\div2=40\), \(b=2\pi\div24=0.262\)
   • \(E=12\sin(0.262(t-6))+40\)

2. (b) Find the demand at t = 10.

   • \(E=12\sin(0.262\times4)+40=50.4\)

The number of fish in a lake is modelled by \(P=\dfrac{L}{1+Ce^{-kt}}\), with t in months. The lake can support 1200 fish. There are 30 at t = 0 and 100 at t = 3.

1. (a) Find C and k.

   • \(C=1200\div30-1=39\)
   • \(100=\dfrac{1200}{1+39e^{-3k}}\): k = 0.422

2. (b) Predict the population at t = 6.

   • \(P=\dfrac{1200}{1+39e^{-0.422\times6}}=293\)

A mass hangs on a spring and bobs up and down. Its amplitude is 0.12 m, its mean height is 1.5 m, the period is 0.9 s and the phase shift is 0.1 s. Model the height as \(h=a\sin(b(x-c))+d\).

1. (a) Write a model for the height h m of the mass at time x s.

   • a = 0.12, d = 1.5, \(b=2\pi\div0.9=6.98\), c = 0.1
   • \(h=0.12\sin(6.98(x-0.1))+1.5\)

2. (b) Find h when x = 0.4.

   • \(h=0.12\sin(6.98\times0.3)+1.5=1.60\)

The graph shows a logistic model \(P=\dfrac{400}{1+7e^{-0.8t}}\) (solid) and an exponential model with the same start (dashed), for a population of frogs.

1. (a) Write down the carrying capacity.

   • The logistic curve levels off at the dashed line: L = 400

2. (b) Find the initial population.

   • \(P(0)=\dfrac{400}{1+7}=50\)

3. (c) Find when the logistic population grows fastest, and the population then.

   • Fastest growth at \(P=L\div2=200\)
   • \(t=\ln7\div0.8=2.43\)

4. (d) Say how the two models differ for large t.

   • The exponential model has no limit. The logistic model levels off at 400.

A Ferris wheel carriage has height \(h=9\sin(b(t-2))+10\) metres, with period 20 seconds, where t is in seconds.

1. (a) Find the first time after t = 0 when the carriage is 14 m above the ground.

   • \(9\sin(0.314(t-2))+10=14\)
   • Solve on the calculator, or graph both sides: first crossing t = 3.47

2. (b) Find the second time in the same turn when the carriage is 14 m above the ground.

   • The curve falls back to the same value at the second crossing: t = 10.5