Revise Modelling with logistic, piecewise and other functions for IB Mathematics: Applications and Interpretation (2.6)

(a) Here $w=1.5$. Since $0<1.5\le 2$, the first rule applies. [M1: correct interval]\n$C(1.5)=10$, so the cost is $10. [A1]\n\n**(b)** Here $w=5$. Since $5>2$, the second rule applies. [M1: correct interval]\n$C(5)=10+4(5-2)=10+12=22$, so the cost is $22. [A1]\n\n**(c)** A charge of $24 exceeds $10, so the package is heavier than 2 kg and the second rule applies. Solve\n$24=10+4(w-2)\;\Rightarrow\;14=4(w-2)\;\Rightarrow\;w-2=3.5\;\Rightarrow\;w=5.5.$\nThe package weighed 5.5 kg. [A1]\n\nThe whole skill is choosing the interval first; once the right rule is selected the arithmetic is routine.

(a) At $t=0$: $N(0)=\frac{800}{1+79e^{0}}=\frac{800}{1+79}=\frac{800}{80}=10$. [M1: substitute $t=0$] [A1]\nSo 10 students started the rumour.\n\n**(b)** The maximum is the carrying capacity, the number on top of the fraction: $L=800$ students. [A1]\n\n**(c)** At $t=2$: $N(2)=\frac{800}{1+79e^{-0.9(2)}}=\frac{800}{1+79e^{-1.8}}$. [M1: substitute $t=2$]\nBy GDC, $N(2)=61.1\ldots$, so about 61 students. [A1]\n\n**(d)** Solve $400=\frac{800}{1+79e^{-0.9t}}$. Graph $y_1=\frac{800}{1+79e^{-0.9t}}$ and $y_2=400$ and use the intersect tool. [M1: correct equation set up]\nThe intersection is at $t=4.85$ days (3 s.f.). This is when the rumour reaches half its carrying capacity — the moment of fastest spread. [A1]

(a) The counts rise quickly at first but the growth slows as $P$ nears about 1000 — the lake's ceiling. An exponential model would keep growing without limit and would soon exceed what the lake can support, so it is unsuitable. A logistic model levels off at a carrying capacity, matching the biology. [A1: growth slows toward a ceiling] [A1: exponential has no limit / would overshoot]\n\n**(b)** At $t=4$: $P(4)=\frac{1000}{1+24e^{-0.6(4)}}=\frac{1000}{1+24e^{-2.4}}$. [M1: substitute $t=4$]\nBy GDC, $P(4)=316$ (3 s.f.), close to the recorded 340, so the model fits well here. [A1]\n\n**(c)** At $t=10$: $P(10)=\frac{1000}{1+24e^{-6}}=985$ (3 s.f.), against a recorded 960. The model sits just above the data and is approaching its carrying capacity $L=1000$, so the fit remains good and behaves sensibly at the ceiling. [A1]\n\nThe model is judged by comparing it with the data across the range and confirming it respects the known limit of 1000.

Model answer — full working.\n\nCarrying capacity. The limiting value is the number on top of the fraction:\n$L=5000.$ The population approaches 5000 but never exceeds it.\n\nPopulation at $t=0$. Substitute $t=0$, using $e^{0}=1$:\n$P(0)=\frac{5000}{1+9e^{0}}=\frac{5000}{1+9}=\frac{5000}{10}=500.$\n\nTime to reach 4000. Set $P(t)=4000$:\n$4000=\frac{5000}{1+9e^{-0.4t}}.$\nGraph $y_1=\frac{5000}{1+9e^{-0.4t}}$ and $y_2=4000$ on the GDC and use the intersect tool (or solve the equation numerically):\n$t=8.96$ years (3 s.f.).\n\nInterpretation. The population starts at 500, climbs along an S-curve, and passes 4000 after about 8.96 years, on its way to the ceiling of 5000.\n\n---\nHow our marking engine awards the 5 marks:\n\n- A1 — carrying capacity. Awarded for stating $L=5000$, read correctly from the numerator. This is a stand-alone accuracy mark.\n- M1 — substitute $t=0$. A method mark for putting $t=0$ into the model, i.e. writing $P(0)=\frac{5000}{1+9e^{0}}=\frac{5000}{1+9}$. It is the substitution that is rewarded, so it survives an arithmetic slip in the division.\n- A1 — value 500. Awarded for $P(0)=500$. This accuracy mark depends on the M1 above; FT applies, so a candidate who substitutes correctly but slips arithmetically still earns credit on their own figure if consistent.\n- M1 — set $P=4000$ and solve. A method mark for forming the equation $4000=\frac{5000}{1+9e^{-0.4t}}$ and solving it (GDC intersect is expected on Paper 2). The engine checks that the target 4000 was equated to the model, not confused with the carrying capacity.\n- A1 — value 8.96. Awarded for $t=8.96$ years (3 s.f.). This is FT on the candidate's own equation: a student whose earlier value differed but who set up and solved their equation correctly keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts the carrying capacity written as $5000$ or 'the model levels off at 5000', accepts $P(0)$ as $\frac{5000}{10}$ or $500$, and accepts the time as $8.96$, $8.959$ or a correctly-rounded equivalent. Once a correct final value appears, ISW means later restatements do not lose marks.\n\nBottom line: of the 5 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through — but only if the substitution and the equation are written on the page. A student who writes just '8.96' with no equation risks the final method mark; a student who shows $L$, the $t=0$ substitution, and the equation $4000=\frac{5000}{1+9e^{-0.4t}}$ keeps the method regardless of a slip in the final number.