Further Numerical Methods - IB Mathematics Applications and Interpretation

1. Using a starting value of $x_0 = 1.5$, find the solution to the equation correct to four significant figures.
   Answer: 1.649 The iterative process is performed as follows: $x_0 = 1.5$ $x_1 = \sqrt[3]{5 - \ln(1.5)} \approx 1.6626$ $x_2 = \sqrt[3]{5 - \ln(1.6626)} \approx 1.6481$ $x_3 = \sqrt[3]{5 - \ln(1.6481)} \approx 1.6493$ $x_4 = \sqrt[3]{5 - \ln(1.6493)} \approx 1.6492$ $x_5 = \sqrt[3]{5 - \ln(1.6492)} \approx 1.6492$ The sequence converges to 1.6492. To four significant figures, the solution is 1.649.
2. Using the formula for arc length, $L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx$, calculate the total length of this section of the trail.
   Answer: 1.32 First, we need to find the derivative of $y = \ln(\sec x)$. Using the chain rule, $\frac{dy}{dx} = \frac{1}{\sec x} \cdot (\sec x \tan x) = \tan x$. Now, substitute this into the arc length formula: $L = \int_0^{\pi/3} \sqrt{1 + (\tan x)^2} dx$. Using the identity $1 + \tan^2 x = \sec^2 x$, this simplifies to $L = \int_0^{\pi/3} \sqrt{\sec^2 x} dx = \int_0^{\pi/3} \sec x dx$, since $\sec x$ is positive on the interval. Using a calculator to evaluate this definite integral gives $L \approx 1.317$ km. Rounded to two decimal places, the length is 1.32 km.
3. Using a starting value of $x_0=1$, find the solution correct to five significant figures.
   Answer: 0.85193 The iterative process is performed in radians mode: $x_0 = 1$ $x_1 = \arctan(2-1) = \arctan(1) \approx 0.78540$ $x_2 = \arctan(2-0.78540) \approx 0.88143$ $x_3 = \arctan(2-0.88143) \approx 0.83925$ $x_4 = \arctan(2-0.83925) \approx 0.85869$ Continuing the process, the sequence converges. $x_8 \approx 0.85185$, $x_9 \approx 0.85198$, $x_{10} \approx 0.85193$, $x_{11} \approx 0.85195$. The value stabilizes at approximately 0.85193.
4. Find the smallest integer $n \ge 1$ such that for all integers $k \ge n$, the value of $x_k$ rounded to three decimal places is constant.
   Answer: 4 We compute the sequence of iterates, rounded to 4 decimal places for intermediate steps: $x_0 = 0.8$ $x_1 = \cos(0.8^2) \approx 0.8021 \implies 0.802$ $x_2 = \cos(0.8021^2) \approx 0.8010 \implies 0.801$ $x_3 = \cos(0.8010^2) \approx 0.8016 \implies 0.802$ $x_4 = \cos(0.8016^2) \approx 0.8012 \implies 0.801$ $x_5 = \cos(0.8012^2) \approx 0.8014 \implies 0.801$ $x_6 = \cos(0.8014^2) \approx 0.8013 \implies 0.801$ The value rounded to three decimal places is 0.802, then 0.801, then 0.802. From $n=4$ onwards, the value remains 0.801. Therefore, the smallest integer $n$ that satisfies the condition is 4.
5. The two fixed points are $x=1$ and $x=3$. For which of the following starting values $x_0$ does the iterative sequence converge to the smaller fixed point, $x=1$?
   Answer: 2.5 Let $g(x) = \frac{1}{4}(x^2+3)$. The derivative is $g'(x) = \frac{x}{2}$. For convergence to a fixed point $\alpha$, we generally need $|g'(\alpha)| < 1$. At $\alpha=1$, $g'(1) = 0.5$, which satisfies $|0.5| < 1$, so it is a stable fixed point. At $\alpha=3$, $g'(3) = 1.5$, which satisfies $|1.5| > 1$, so it is an unstable fixed point. The sequence will only converge to 1. Testing the starting values: $x_0=2.5$ gives $x_1 = 2.3125$, $x_2 = 2.087...,$ which approaches 1. $x_0=-3.5$ and $x_0=3.5$ diverge. $x_0=3.0$ is the unstable fixed point itself; any small perturbation would cause it to move away.
6. Using the starting value $x_0=1$, find the value of the root correct to three significant figures.
   Answer: 0.745 We perform the iteration: $x_0 = 1$ $x_1 = \frac{1^3 + 5}{7} = \frac{6}{7} \approx 0.85714$ $x_2 = \frac{(0.85714)^3 + 5}{7} \approx 0.75623$ $x_3 = \frac{(0.75623)^3 + 5}{7} \approx 0.74711$ $x_4 = \frac{(0.74711)^3 + 5}{7} \approx 0.74567$ $x_5 = \frac{(0.74567)^3 + 5}{7} \approx 0.74545$ The sequence converges to 0.745. The other two roots of the equation are approximately -2.93 and 2.18, which are provided as distractors.
7. What is the total amount of water, in litres, that leaks from the tank during the first 3 hours? Give your answer correct to two decimal places.
   Answer: 34.66 The total amount of water leaked is the definite integral of the rate function over the time interval. We need to calculate $\int_0^3 R(t) dt = \int_0^3 15e^{-0.1t^2} dt$. Using a calculator's numerical integration function, the value is approximately 34.66 litres. Distractor A is the rate at t=3, $R(3)$. Distractor B is $3 \times R(3)$. Distractor D is the initial rate multiplied by time, $R(0) \times 3$.
8. Assuming the population starts at a reasonable non-zero level, it will stabilize at a long-term carrying capacity. What is the value of this non-zero stable population?
   Answer: 2000 The long-term stable population is a stable fixed point of the iterative formula. A fixed point $P$ satisfies the equation $P = f(P)$, where $f(P) = \frac{5 P}{1 + (P/1000)^2}$. For a non-zero solution, we can divide by $P$: $1 = \frac{5}{1 + (P/1000)^2}$. Rearranging gives $1 + (P/1000)^2 = 5$, so $(P/1000)^2 = 4$. Since population must be positive, $P/1000 = 2$, which means $P = 2000$. To check for stability, we evaluate the derivative of $f(P)$ at $P=2000$. A calculator shows $f'(2000) = -0.6$. Since $|-0.6| < 1$, this fixed point is stable.
9. Calculate the work done, in Joules, in stretching this spring from $x = 0.1$ m to $x = 0.5$ m.
   Answer: 7.87 The work done $W$ is the definite integral of the force function from the initial position to the final position. $W = \int_{0.1}^{0.5} (50x + 120x^3) dx$. This integral can be solved analytically or numerically. Analytically: $W = [25x^2 + 30x^4]_{0.1}^{0.5} = (25(0.5)^2 + 30(0.5)^4) - (25(0.1)^2 + 30(0.1)^4) = (6.25 + 1.875) - (0.25 + 0.003) = 8.125 - 0.253 = 7.872$. Using a calculator's numerical integration function will yield the same result, which to two decimal places is 7.87 J.
10. What is the mean temperature over this 24-hour period? Give your answer correct to one decimal place.
    Answer: 20.6 The mean value of a function $f(x)$ on an interval $[a, b]$ is given by $\frac{1}{b-a}\int_a^b f(x) dx$. Here, we need to calculate the average temperature $\bar{T} = \frac{1}{24-0} \int_0^{24} \left(20 - 8\cos(\frac{\pi t}{12}) + 3\sin(\frac{\pi t}{8})\right) dt$. Using a calculator's numerical integration function, the integral $\int_0^{24} T(t) dt \approx 495.278$. Dividing by 24 gives the average temperature: $\bar{T} \approx \frac{495.278}{24} \approx 20.6366$. To one decimal place, the mean temperature is 20.6°C.
11. The architect needs to calculate the area of the garden bed. The area is the region enclosed between the two intersection points of the curves for $x > 0$. To three significant figures, what is the area of the garden bed in square metres?
    Answer: 5.99 First, find the intersection points of $f(x)$ and $g(x)$ by solving $2\sin(\frac{\pi x}{5}) = 0.5x - 1$ using a graphing calculator. The positive intersection points are approximately at $x_1 = 0.748$ and $x_2 = 4.545$. The area $A$ is the definite integral of the upper function minus the lower function between these points: $A = \int_{0.748}^{4.545} \left( 2\sin(\frac{\pi x}{5}) - (0.5x - 1) \right) dx$. Using a calculator's numerical integration function, this integral evaluates to approximately 5.99 m².
12. Calculate the volume of the material needed to create the vase, in cm³, correct to one decimal place.
    Answer: 290.6 The volume $V$ of a solid of revolution is given by the formula $V = \pi \int_a^b [f(x)]^2 dx$. Here, $f(x) = 0.1e^{0.5x} + 2$, $a = 0$, and $b = 5$. So we must calculate $V = \pi \int_0^5 (0.1e^{0.5x} + 2)^2 dx$. Using a calculator's numerical integration feature, $\int_0^5 (0.1e^{0.5x} + 2)^2 dx \approx 92.485$. Therefore, the volume is $V \approx \pi \times 92.485 \approx 290.55$ cm³. Rounded to one decimal place, the volume is 290.6 cm³.
13. Calculate the volume of this component.
    Answer: 7.39 The base of the solid extends from $x = -2$ to $x = 2$. At any point $x$, the side length $s$ of the equilateral triangle is given by the height of the parabola, $s = y = 4 - x^2$. The area of an equilateral triangle with side $s$ is $A = \frac{\sqrt{3}}{4}s^2$. So, the cross-sectional area at $x$ is $A(x) = \frac{\sqrt{3}}{4}(4 - x^2)^2$. The volume is the integral of this area function: $V = \int_{-2}^2 \frac{\sqrt{3}}{4}(4 - x^2)^2 dx$. Using a calculator, $V = \frac{\sqrt{3}}{4} \int_{-2}^2 (16 - 8x^2 + x^4) dx = \frac{\sqrt{3}}{4} \left[ \frac{256}{15} \right] = \frac{64\sqrt{3}}{15} \approx 7.39$.
14. Find the total distance travelled by the particle. Give your answer to three significant figures.
    Answer: 3.85 This is a two-step problem. First, find $\alpha$ by solving $\ln(x) = \cos(x)$. Using a GDC's equation solver or graphical intersection feature, the smallest positive solution is $\alpha \approx 1.30296$. Second, calculate the distance travelled, which is the definite integral of the velocity function: $D = \int_0^2 v(t) dt = \int_0^2 \alpha \sqrt{t^2+1} dt = \alpha \int_0^2 \sqrt{t^2+1} dt$. Using the GDC, $\int_0^2 \sqrt{t^2+1} dt \approx 2.9579$. Therefore, the total distance is $D \approx 1.30296 \times 2.9579 \approx 3.854$. To three significant figures, the distance is 3.85.
15. Let $x_0 = 0.4$ and $y_0 = 0.4001$. Calculate the values of $x_{10}$ and $y_{10}$ and find the absolute difference $|x_{10} - y_{10}|$. Give your answer correct to two decimal places.
    Answer: 0.68 We need to iterate the formula 10 times for each starting value. For $x_0 = 0.4$, the sequence is: 0.4, 0.936, 0.2396, 0.710..., eventually yielding $x_{10} \approx 0.8805$. For $y_0 = 0.4001$, the sequence is: 0.4001, 0.9361, 0.2392, 0.709..., eventually yielding $y_{10} \approx 0.1983$. The absolute difference is $|x_{10} - y_{10}| = |0.8805 - 0.1983| = 0.6822$. Rounded to two decimal places, this is 0.68. This demonstrates the sensitive dependence on initial conditions characteristic of chaos.
16. Engineers approximate the MTTF by evaluating the integral over a finite interval where the reliability becomes negligible. Estimate the MTTF by calculating $\int_0^5 e^{-0.1t^3} dt$. Give your answer to three significant figures.
    Answer: 1.08 The question requires the evaluation of a definite integral that does not have a simple analytical solution. We must use a numerical integration tool. Using a graphing calculator, we evaluate $\int_0^5 e^{-0.1t^3} dt$. The result is approximately 1.077 years. Rounded to three significant figures, the MTTF is 1.08 years. Distractor A corresponds to $R(5)$. The other distractors come from miscalculation or integrating a different function.
17. Calculate the producer surplus at the market equilibrium point. Give your answer to the nearest whole number.
    Answer: 813 First, find the market equilibrium by setting demand equal to supply: $100e^{-0.05q} = 10 + 2q$. Using a GDC solver, we find the equilibrium quantity $q_e \approx 28.52$. The equilibrium price is $p_e = 10 + 2(28.52) \approx 67.04$. The producer surplus (PS) is given by the integral $PS = \int_0^{q_e} (p_e - S(q)) dq$. So, $PS = \int_0^{28.52} (67.04 - (10 + 2q)) dq = \int_0^{28.52} (57.04 - 2q) dq$. Evaluating this integral with a calculator gives $PS \approx 813.4$. To the nearest whole number, this is 813.

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