Mathematical models: exam review Exam review: second paper Worked solutions

The mass M grams of a radioactive isotope in a sample, t days after the sample is prepared, is modelled by \(M=M_0e^{-kt}\). [7]

1. (a) The half-life of the isotope is 8 days. Find the value of k. Give your answer to 3 significant figures. [2]

   • M1 for e^(−8k) = 1/2, so k = ln 2 ÷ 8
   • A1 0.0866

2. (b) After 12 days the mass is 35 g. Find \(M_0\). Use the unrounded value of k. [2]

   • M1 for 35 = M₀e^(−k×12), so M₀ = 35e^(12k)
   • A1 99.0

3. (c) Find the time for the mass to fall to 5 g. Give your answer to 3 significant figures. [2]

   • M1 for solving 99.0e^(−kt) = 5, using the unrounded values
   • A1 34.5

4. (d) State one assumption made by this model. [1]

   • R1 The isotope decays at a rate proportional to the mass present, and none is added.

The height H cm of a seedling t days after planting is modelled by \(H=a+b\ln t\). After 3 days the height is 8. After 12 days the height is 15. [6]

1. (a) Use the information to find a and b. Give each value to 3 significant figures. [3]

   • M1 for two equations: a + b ln 3 = 8 and a + b ln 12 = 15
   • M1 for solving the pair, using the calculator
   • A1 a = 2.45, b = 5.05

2. (b) Use the model to predict H when t = 30. [1]

   • A1 19.6

3. (c) Find the value of t when H = 20. [2]

   • M1 for solving 2.45 + 5.05 ln t = 20, or t = e^((20 − a)/b)
   • A1 32.3

A metal pan is taken out of an oven at 200°C into a kitchen at 24°C. Its temperature T°C, t minutes later, is modelled by \(T=B+Ae^{-kt}\). [7]

1. (a) Write down the value of B and the value of A. [2]

   • A1 B = 24 (the temperature of the surroundings)
   • A1 A = 176 (200 − 24)

2. (b) After 10 minutes the temperature is 120°C. Find the value of k. Give your answer to 3 significant figures. [2]

   • M1 for 120 = 24 + 176e^(−10k), so e^(−10k) = 96/176
   • A1 0.0606

3. (c) Find the time for the temperature to reach 50°C. Give your answer to 3 significant figures. [2]

   • M1 for solving 24 + 176e^(−kt) = 50
   • A1 31.6

4. (d) Find the half-life of the difference between the temperature and the surroundings. Give your answer to 3 significant figures. [1]

   • A1 11.4 (ln 2 ÷ k)

A pond is filled with water. The volume V litres after t minutes is modelled by \(V=pt\) for \(0\le t\le 5\), and \(V=qt^2\) for \(t>5\). After 4 minutes the volume is 180 litres. [5]

1. (a) Find the value of p. [1]

   • A1 p = 180 ÷ 4 = 45.0

2. (b) The model is continuous at t = 5. Find the value of q. [2]

   • M1 for pt = qt² at t = 5: 225 = q×5²
   • A1 q = 9.00

3. (c) Find the time at which the volume reaches 1600 litres. Give your answer to 3 significant figures. [2]

   • M1 for qt² = 1600, using t > 5
   • A1 13.3

In each part, the function is continuous at the point where the rule changes. [8]

1. (a) \(f(x)=\begin{cases}2x^2+a & x\le 3\\ 4x+1 & x>3\end{cases}\). Find the value of a. [2]

   • M1 for 2×3² + a = 4×3 + 1
   • A1 a = −5

2. (b) \(g(x)=\begin{cases}x^a & x\le 8\\ 3x-8 & x>8\end{cases}\). Find the value of a. Give your answer to 3 significant figures. [3]

   • M1 for 8^a = 3×8 − 8 = 16
   • M1 for a ln 8 = ln 16
   • A1 a = 1.33

3. (c) \(h(x)=\begin{cases}e^{ax} & x\le 3\\ 2x+6 & x>3\end{cases}\). Find the value of a. Give your answer to 3 significant figures. [3]

   • M1 for e^(3a) = 2×3 + 6 = 12
   • M1 for 3a = ln 12
   • A1 a = 0.828

In each part, give answers that are not exact to 3 significant figures. [9]

1. (a) The time t days taken to build a wall is inversely proportional to the number n of workers, so \(t=\dfrac{a}{n}\). With 5 workers the wall is built in 12 days. Find the value of a, and the time when n = 8. [3]

   • M1 for substituting: 12 = a/5
   • A1 a = 60
   • A1 time = 60 ÷ 8 = 7.5

2. (b) Describe what the model predicts for the time as n gets close to 0, and say why this is not reasonable. [1]

   • A1 t increases without limit as n approaches 0; there cannot be 0 workers, or part of a worker

3. (c) A metal rod is left to cool in a room that is kept at 20 °C. Its temperature T °C after t minutes is modelled by \(T=c+k(0.9)^t\), where c and k are constants. At t = 0 the temperature is 180 °C. Write down the equation of the horizontal asymptote of the graph of T, and find the value of k. [3]

   • A1 T = 20 (the curve levels off at c)
   • M1 for substituting t = 0: 180 = 20 + k
   • A1 k = 160

4. (d) Find the time for the temperature to reach 50 °C. Give your answer to 3 significant figures. [2]

   • M1 for solving 20 + 160(0.9)^t = 50
   • A1 15.9

In each part, the amount is modelled by \(y=y_0e^{kt}\), where t is in years. [5]

1. (a) The value of a car falls by 9% each year. Find the value of k. Give your answer to 3 significant figures. [2]

   • M1 for \(e^k=1-0.09\), or k = ln 0.91
   • A1 k = -0.0943

2. (b) A bone has 62% of the carbon-14 that a living animal has. The half-life of carbon-14 is 5730 years. Find the age of the bone. Give your answer to 3 significant figures. [3]

   • M1 for k = −ln 2 ÷ 5730 = -0.000121 (half-life)
   • M1 for solving \(e^{kt}=0.62\)
   • A1 3950 years

A planter is made from a rectangular sheet of plastic 40 cm by 24 cm. A square of side x cm is cut from each corner and the sides are folded up. The volume of the planter is \(V=x(40-2x)(24-2x)\) cm³. [3]

1. (a) Write down the domain of x for which V makes sense. [1]

   • A1 the shorter side gives 24 − 2x > 0, so 0 < x < 12

2. (b) Use your calculator to find the greatest volume, and the value of x that gives it. Give your answers to 3 significant figures. [2]

   • M1 for graphing V and finding the maximum with the calculator
   • A1 2100 cm³ at x = 4.85

The temperature C°C at a weather station, t hours after midnight, can be modelled by a sinusoidal function. The greatest temperature is 17°C and the least temperature is 3°C. The temperature repeats every 24 hours. [12]

1. (a) Find the amplitude and the mean value of C. [2]

   • A1 amplitude = (17 − 3) ÷ 2 = 7.00
   • A1 mean = (17 + 3) ÷ 2 = 10.0

2. (b) The period is 24 hours. Find the value of b. Give your answer to 3 significant figures. [2]

   • M1 for b = 2π ÷ 24
   • A1 0.262

3. (c) The graph of C first crosses its mean value, going upwards, at t = 8. Write down a model of the form \(C=a\sin(b(t-c))+d\). Use b to 3 significant figures. [1]

   • A1 C = 7.00 sin(0.262(t − 8)) + 10.0

4. (d) Use your model to find C when t = 15. [2]

   • M1 for substituting t = 15, with the calculator in radians
   • A1 16.8

5. (e) Find the total time, in the first 24 hours, for which C is greater than 15. Give your answer to 3 significant figures. [3]

   • M1 for solving C = 15
   • A1 t = 11.0 and t = 17.0
   • A1 5.92 (17.0 − 11.0)

6. (f) Find the first time at which C is a maximum. [2]

   • M1 for a quarter of a period after t = 8: 8 + 24/4
   • A1 14.0

The number P of members of a new online club, t months after it is launched, is modelled by \(P=\dfrac{L}{1+Ce^{-kt}}\). The club can have at most 5000 members, so L = 5000. At launch there are 400 members. [12]

1. (a) Find the value of C. [2]

   • M1 for L ÷ (1 + C) = 400 at t = 0
   • A1 C = 11.5

2. (b) After 2 months the number of members is 1100. Find the value of k. Give your answer to 3 significant figures. [3]

   • M1 for 5000 ÷ (1 + 11.5e^(−2k)) = 1100
   • M1 for solving for k using the calculator, or by rearranging
   • A1 k = 0.588

3. (c) Find the number of members after 5 months. [2]

   • M1 for substituting t = 5 with the unrounded k and C
   • A1 3110

4. (d) Find the time at which the number of members is half of 5000. Give your answer to 3 significant figures. [2]

   • M1 for 5000 ÷ (1 + Ce^(−kt)) = 2500, or Ce^(−kt) = 1
   • A1 t = ln C ÷ k = 4.15

5. (e) Write down the limiting value of the number of members as t increases. [1]

   • A1 5000 (L, the carrying capacity)

6. (f) Give one reason why a logistic model is more realistic than an exponential model for this situation. [2]

   • R1 the logistic model levels off at the carrying capacity
   • R1 an exponential model grows without limit, which cannot continue (limited space, food or users)

A scientist studies two cultures. Culture X starts with 650 units and loses 5% of its amount each day. Culture Y starts with 25 units. [8]

1. (a) Write down an expression for the amount of culture X after n days. [1]

   • A1 650 × 0.95^n

2. (b) Find the least whole number of days after which there is less than 80 units of culture X. [3]

   • M1 for 650×0.95^n < 80
   • M1 for n > ln(80/650) ÷ ln 0.95; the inequality turns round because ln 0.95 is negative
   • A1 n = 41 (n > 40.8)

3. (c) The amount of culture Y is modelled by \(Y=B_0e^{kt}\) with \(B_0=25\). It doubles every 9 days. Find the value of k. Give your answer to 3 significant figures. [2]

   • M1 for e^(9k) = 2
   • A1 k = ln 2 ÷ 9 = 0.0770

4. (d) Find the time for the amount of culture Y to reach 700 units. Give your answer to 3 significant figures. [2]

   • M1 for solving 25e^(kt) = 700
   • A1 43.3

A cinema sells n tickets for a film when the ticket price is £p. When the price is £9, 270 tickets are sold. When the price is £15, 180 tickets are sold. The cinema uses a linear model for n in terms of p. [9]

1. (a) Find a model of the form \(n=mp+c\). [3]

   • M1 for the gradient: (180 − 270) ÷ (15 − 9) = -15
   • M1 for substituting a point: 270 = -15(9) + c
   • A1 n = -15p + 405

2. (b) State what the value of m means in this context. [1]

   • A1 Each £1 increase in the ticket price reduces the number of tickets sold by 15

3. (c) The income from ticket sales is \(R=p\,n\) pounds. Find the ticket price that gives the greatest income, and that income. Give your answers to 3 significant figures. [2]

   • M1 for R = p(-15p + 405) and finding the maximum with the calculator
   • A1 p = 13.5, R = 2730

4. (d) Find the range of ticket prices for which the income is more than £2500. Give your answers to 3 significant figures. [2]

   • M1 for solving -15p² + 405p = 2500 (graph R and the line R = 2500)
   • A1 9.55 < p < 17.4

5. (e) State a reasonable domain for p, and say why the model should not be used for a price of £40. [1]

   • A1 n ≥ 0 gives p ≤ 405/15 = 27.0, so 0 ≤ p ≤ 27.0; at £40 n would be negative

A water wheel has its lowest point 0.5 m above the stream and its highest point 6.5 m above the stream. It makes one complete turn every 10 seconds. A bucket is at the lowest point at t = 0, where t is in seconds. The height of the bucket above the stream is modelled by \(h=a\cos(bt)+d\) metres. [9]

1. (a) Find the values of a, b and d in the model \(h=a\cos(bt)+d\). Give b to 3 significant figures. [3]

   • M1 for a = −(greatest − least) ÷ 2 = -3, negative because it starts at the lowest point, and d = (6.5 + 0.5) ÷ 2 = 3.5
   • M1 for b = 2π ÷ 10
   • A1 a = -3, b = 0.628, d = 3.5

2. (b) Find the height of the bucket at t = 4. Give your answer to 3 significant figures. [1]

   • A1 h = 3.5 − 3cos(0.628 × 4) = 5.93

3. (c) Find the total time in one turn for which the height is more than 5 m. Give your answer to 3 significant figures. [3]

   • M1 for solving -3cos(0.628t) + 3.5 = 5 (graph the model and the line h = 5)
   • A1 t = 3.33 and t = 6.67
   • A1 3.33

4. (d) State what the value of |a| represents in this context. [1]

   • A1 |a| = 3 is the radius of the wheel (half the difference between the greatest and least heights)

5. (e) Give one reason why the model may not be reasonable. [1]

   • A1 The wheel is assumed to turn at a constant speed; in reality the speed of the stream changes, so the wheel does not turn at a constant speed.