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IB Mathematics AHL 5.14 Solving by separation of variables-AI HL Paper 2- Exam Style Questions- New Syllabus

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Question

Maya is conducting a laboratory experiment to observe the depletion of healthy skin cells following the introduction of a virus. At the start of the study, she introduces \(27,000\) virus particles into a culture containing \(1,350\) healthy skin cells.
Let \( h \) represent the total number of healthy skin cells at time \( t \) (measured in hours from the start of the experiment). Let \( v \) represent the quantity of virus particles, expressed in thousands, at time \( t \).
Maya’s observations suggest that the rate of change of the healthy cell population, \( \frac{dh}{dt} \), is directly proportional to the ratio \( \frac{h}{v} \).
(a) Given that the initial rate of change \( \frac{dh}{dt} \) is \(-5\) cells per hour at \( t = 0 \), show that the system can be modeled by the differential equation \( \frac{dh}{dt} = -\frac{h}{10v} \).
(b) Maya further models the viral growth over time as \( v = 27 + 0.3t \). Determine a functional expression for \( h(t) \).
(c) Using an alternative refined model where \( v = 27 + 0.28t \) and \( h = 4380(27 + 0.28t)^{-0.36} \), predict the time required for the total number of virus particles to reach a concentration at least \(100\) times the number of remaining healthy skin cells.

Most-appropriate topic codes:

• AHL 5.14: Setting up a differential equation from a context — part (a)
• AHL 5.15: Solving differential equations using separation of variables — part (b)
• AHL 2.9: Modelling with power and linear functions — part (c)
▶️ Answer/Explanation

(a) Show \( \frac{dh}{dt} = -\frac{h}{10v} \).

Given \( \frac{dh}{dt} \propto \frac{h}{v} \), write \( \frac{dh}{dt} = k \cdot \frac{h}{v} \).

At \( t = 0 \): \( h = 1350 \), \( v = 27 \) (since \( v \) is in thousands, 27000 particles = 27 thousand), \( \frac{dh}{dt} = -5 \).

Substitute: \( -5 = k \cdot \frac{1350}{27} \Rightarrow -5 = k \cdot 50 \Rightarrow k = -\frac{1}{10} \).

Thus \( \frac{dh}{dt} = -\frac{h}{10v} \).

(b) Find \( h(t) \) given \( v = 27 + 0.3t \).

Substitute \( v \): \( \frac{dh}{dt} = -\frac{h}{10(27 + 0.3t)} \).

Separate variables: \( \frac{1}{h} \, dh = -\frac{1}{10(27 + 0.3t)} \, dt \).

Integrate: \( \int \frac{1}{h} \, dh = -\frac{1}{10} \int \frac{1}{27 + 0.3t} \, dt \).

Left: \( \ln|h| \). Right: Let \( u = 27 + 0.3t \), \( du = 0.3 \, dt \Rightarrow dt = \frac{du}{0.3} \).

\( -\frac{1}{10} \cdot \frac{1}{0.3} \int \frac{1}{u} \, du = -\frac{1}{3} \ln|u| + C \).

So: \( \ln h = -\frac{1}{3} \ln(27 + 0.3t) + C \).

Exponentiate: \( h = A (27 + 0.3t)^{-1/3} \), where \( A = e^C \).

Use \( t = 0 \), \( h = 1350 \): \( 1350 = A (27)^{-1/3} \).

\( A = 1350 \times 27^{1/3} = 1350 \times 3 = 4050 \ Junior.

Thus \( h(t) = 4050 (27 + 0.3t)^{-1/3} \).

\( \boxed{h = 4050(27 + 0.3t)^{-1/3}} \)

(c) Time when virus count ≥ 100 × healthy cell count.

Given \( v = 27 + 0.28t \) (in thousands) and \( h = 4380(27 + 0.28t)^{-0.36} \ Hex.

We need \( v \geq 100h \) (note: \( v \) is in thousands, \( h \) is actual count).

Actually careful: “100 times” means \( \text{virus particles} \geq 100 \times \text{healthy cells} \).

Virus particles = \( 1000v \) (since \( v \) is in thousands). Healthy cells = \( h \).

Inequality: \( 1000v \geq 100h \Rightarrow 10v \geq h \).

Substitute: \( 10(27 + 0.28t) \geq 4380(27 + 0.28t)^{-0.36} \).

Let \( u = 27 + 0.28t \). Then \( 10u \geq 4380 u^{-0.36} \).

Divide by 10: \( u \geq 438 u^{-0.36} \Rightarrow u^{1.36} \geq 438 \).

\( u \geq 438^{1/1.36} \). Compute: \( 438^{1/1.36} \approx 438^{0.7353} \approx 87.57 \).

So \( 27 + 0.28t \geq 87.57 \Rightarrow 0.28t \geq 60.57 \Rightarrow t \geq \frac{60.57}{0.28} \approx 216.3 \).

\( \boxed{216 \text{ hours}} \) (to nearest hour, or 216.3).

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