IB Mathematics: Applications and Interpretation Quiz: Further Numerical Methods

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Further Numerical MethodsQuestion 1 of 20

The equation $x = \frac{1}{4}(x^2 + 3)$ has two solutions, which are the fixed points of the iterative formula $x_{n+1} = \frac{1}{4}(x_n^2 + 3)$ .

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, $x=1$ ?

-3.5

2.5

3.0

3.5

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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Further Numerical Methods

Practice Further Numerical Methods in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Further Numerical Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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Question 1

The equation $x = \frac{1}{4}(x^2 + 3)$ has two solutions, which are the fixed points of the iterative formula $x_{n+1} = \frac{1}{4}(x_n^2 + 3)$ .

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, $x=1$ ?

-3.5
2.5 (correct answer)
3.0
3.5

Explanation: 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.

Question 2

The population $P_n$ (in thousands) of a species of fish in a lake in year $n$ is modelled by the discrete logistic equation $P_{n+1} = \frac{5 P_n}{1 + (P_n/1000)^2}$ .

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?

1000
2000 (correct answer)
4000
5000

Explanation: 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)

P=2000

. A calculator shows

f'(2000) = -0.6

. Since

|-0.6| < 1

, this fixed point is stable.

Question 3

A non-linear spring requires a force $F(x) = 50x + 120x^3$ Newtons to stretch it by a distance of $x$ meters from its natural length. The work done in stretching the spring from $x=a$ to $x=b$ is given by the integral $W = \int_a^b F(x) dx$ .

Calculate the work done, in Joules, in stretching this spring from $x = 0.1$ m to $x = 0.5$ m.

7.87 (correct answer)
10.0
15.0
15.6

Explanation: 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.

Question 4

A physical system is described by two key parameters. The first parameter, $\alpha$ , is the smallest positive solution to the equation $\ln(x) = \cos(x)$ . The second parameter is the total distance travelled by a particle from $t=0$ to $t=2$ , whose velocity is given by $v(t) = \alpha \sqrt{t^2+1}$ .

Find the total distance travelled by the particle. Give your answer to three significant figures.

1.30
2.96
3.85 (correct answer)
5.15

Explanation: 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.

Question 5

To find the solution to the equation $\tan(x) = 2-x$ in the interval $0 < x < \frac{\pi}{2}$ , the iterative formula $x_{n+1} = \arctan(2-x_n)$ is used.

Using a starting value of $x_0=1$ , find the solution correct to five significant figures.

0.78540
0.83925
0.85193 (correct answer)
0.88143

Explanation: 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.

Question 6

An algorithm uses the iterative formula $x_{n+1} = \cos(x_n^2)$ with a starting value $x_0 = 0.8$ to approximate a solution to $x = \cos(x^2)$ . The process is stopped when the result is considered stable.

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.

2
3
4 (correct answer)
6

Explanation: 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.

Question 7

The rate at which water leaks from a storage tank is modelled by the function $R(t) = 15e^{-0.1t^2}$ , where $R(t)$ is in litres per hour and $t$ is the time in hours from the moment the leak began.

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.

6.69
20.08
34.66 (correct answer)
45.00

Explanation: 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

Question 8

In a competitive market, the price $p$ of a product is related to the quantity demanded $q$ by the demand function $p = 100e^{-0.05q}$ . The supply function is given by $p = 10 + 2q$ . The producer surplus is the area of the region bounded by the horizontal line of the equilibrium price, the supply curve, and the p-axis.

Calculate the producer surplus at the market equilibrium point. Give your answer to the nearest whole number.

813 (correct answer)
1098
1911
2724

Explanation: 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.

Question 9

The equation $x = \frac{1}{4}(x^2 + 3)$ has two solutions, which are the fixed points of the iterative formula $x_{n+1} = \frac{1}{4}(x_n^2 + 3)$ .

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, $x=1$ ?

-3.5
2.5 (correct answer)
3.0
3.5

Explanation: 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.

Question 10

The rate at which water leaks from a storage tank is modelled by the function $R(t) = 15e^{-0.1t^2}$ , where $R(t)$ is in litres per hour and $t$ is the time in hours from the moment the leak began.

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.

6.69
20.08
34.66 (correct answer)
45.00

Explanation: 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

Question 11

To find the solution to the equation $\tan(x) = 2-x$ in the interval $0 < x < \frac{\pi}{2}$ , the iterative formula $x_{n+1} = \arctan(2-x_n)$ is used.

Using a starting value of $x_0=1$ , find the solution correct to five significant figures.

0.78540
0.83925
0.85193 (correct answer)
0.88143

Explanation: 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.

Question 12

A hiking trail follows a path on a map that can be modelled by the function $y = \ln(\sec x)$ for $0 \le x \le \frac{\pi}{3}$ . The units are in kilometres.

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.

0.44
1.05
1.32 (correct answer)
1.44

Explanation: 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.

Question 13

Calculate the producer surplus at the market equilibrium point. Give your answer to the nearest whole number.

813 (correct answer)
1098
1911
2724

Explanation: 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.

Question 14

The population $P_n$ (in thousands) of a species of fish in a lake in year $n$ is modelled by the discrete logistic equation $P_{n+1} = \frac{5 P_n}{1 + (P_n/1000)^2}$ .

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?

1000
2000 (correct answer)
4000
5000

Explanation: 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)

P=2000

. A calculator shows

f'(2000) = -0.6

. Since

|-0.6| < 1

, this fixed point is stable.

Question 15

The reliability of an electronic component, which is the probability that it is still functioning at time $t$ (in years), is modelled by the function $R(t) = e^{-0.1t^3}$ . The mean time to failure (MTTF) is given by the improper integral $\int_0^\infty R(t) dt$ .

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.

0.00
1.08 (correct answer)
1.97
3.55

Explanation: 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.

Question 16

Find the total distance travelled by the particle. Give your answer to three significant figures.

1.30
2.96
3.85 (correct answer)
5.15

Explanation: 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.

Question 17

A landscape architect is designing a garden bed whose shape is defined by the region enclosed between the curves of two functions: an upper boundary given by $f(x) = 2\sin(\frac{\pi x}{5})$ and a lower boundary by $g(x) = 0.5x - 1$ . The variable $x$ represents the horizontal distance in metres.

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?

3.80
5.99 (correct answer)
7.34
9.12

Explanation: 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.

Question 18

An engineering model is described by the equation $x^3 + \ln(x) = 5$ . To solve this equation for $x > 0$ , an iterative formula is proposed: $x_{n+1} = \sqrt[3]{5 - \ln(x_n)}$ .

Using a starting value of $x_0 = 1.5$ , find the solution to the equation correct to four significant figures.

1.649 (correct answer)
1.663
1.671
1.705

Explanation: 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.

Question 19

A decorative vase is designed by rotating the curve with equation $y = 0.1e^{0.5x} + 2$ about the x-axis for $0 \le x \le 5$ . The units are in centimetres.

Calculate the volume of the material needed to create the vase, in cm³, correct to one decimal place.

92.5
188.7
290.6 (correct answer)
354.1

Explanation: 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.

Question 20

A hiking trail follows a path on a map that can be modelled by the function $y = \ln(\sec x)$ for $0 \le x \le \frac{\pi}{3}$ . The units are in kilometres.

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.

0.44
1.05
1.32 (correct answer)
1.44

Explanation: 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.