N2023 P1 Q10

N2023 P1 Q10

13 marks
Free

The mass of a person depends both on daily rate of energy intake and on a daily rate of energy expenditure.

In this question, mass is in kg, time is in days, and energy intake and energy expenditure are measured in Calories per day.

The rate of change of a person's mass with respect to time is proportional to the difference between energy intake and energy expenditure.

Andrew has a mass of \(M\) kg and his energy intake is fixed at \(C\) Calories per day. For every kg of his mass, he expends \(30\) Calories per day.

  1. Show that \(\frac{\mathrm{d}M}{\mathrm{d}t} = k(C - 30M)\), where \(t\) is time and \(k\) is a constant.[1]

Andrew's initial mass is \(110\) kg.

  1. Find the energy intake such that he maintains his mass at \(110\) kg.[1]

As part of a new health plan, Andrew fixes his energy intake at \(80\)% of the value found in part (b).

  1. By solving the differential equation in part (a), show that Andrew's mass while he is on the plan satisfies the equation \(M = 88 + 22\mathrm{e}^{-30kt}\).[4]

Andrew's mass after \(75\) days on the plan is \(100\) kg.

  1. Find the number of additional days required on the plan for Andrew's mass to fall below \(96\) kg.[4]
    1. Sketch a graph of Andrew's mass while on this plan. Explain why Andrew cannot achieve a mass of \(80\) kg using this plan.[2]
    2. State the range of possible values of energy intake for which Andrew could achieve a mass of \(80\) kg.[1]

Video Solution:

Default solution

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Answer:(a) \(\dfrac{\mathrm dM}{\mathrm dt}=k(C-30M)\) (b) \(3300\text{ Calories per day}\) (c) \(M=88+22e^{-30kt}\) (d) \(51\text{ additional days}\) (e)(i) Sketch shown, and \(M>88\) so \(80\text{ kg}\) is impossible (ii) \(0<C<2400\text{ Calories per day}\)

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