Numerical methods – Year 13 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(f(1)=1+2-7=-4<0\) and \(f(2)=8+4-7=5>0\). The function is continuous and there is a change of sign, so there is a root in \([1,2]\).
   2. \(f(1.5)=3.375+3-7=-0.625<0\) and \(f(1.6)=4.096+3.2-7=0.296>0\). Change of sign and continuity give a root in \([1.5,1.6]\).
   3. \(g(0)=-1\) and \(g(2)=1\), but \(g\) is not continuous on \([0,2]\) because it is undefined at \(x=1\). The graph jumps from \(-\infty\) to \(+\infty\) and never crosses the axis, so there is no root.

2. Answer to Question 2

   Method (fixed-point iteration): rearrange \(f(x)=0\) as \(x=g(x)\), then use \(x_{n+1}=g(x_n)\), keeping full calculator accuracy and rounding only when writing the values. The iteration converges near the root \(\alpha\) when \(|g'(\alpha)|<1\).

   1. \(x_1=1.5874\), \(x_2=1.5639\), \(x_3=1.5703\), \(x_4=1.5686\), \(x_5=1.5690\).
   2. The values alternate above and below the root and get closer together (a cobweb pattern, the gradient of \(g\) is negative).
   3. The root is \(1.57\) to 2 decimal places.

   Answer: the iterates oscillate around and close in on the root, which is 1.57 to 2 decimal places.

3. Answer to Question 3

   Method (Newton-Raphson): \(x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\). Each step follows the tangent at \(x_n\) down to the \(x\)-axis. Use the ANS key on the calculator and stop when two values agree to the required accuracy.

   1. Differentiate term by term. \(f\prime(x)=3x^2+2\).
   2. First \(f(1.5)=3.375+3-7=-0.625\) and \(f'(1.5)=8.75\). \(x_1=1.5-\dfrac{-0.625}{3(2.25)+2}=1.5+\dfrac{0.625}{8.75}=1.57143\). \(x_2=1.57143-\dfrac{f(1.57143)}{f\prime(1.57143)}=1.56895\). The two values already agree to 2 decimal places, which shows how fast the method converges.
   3. A further step gives \(x_3=1.568946\), which agrees with \(x_2\) to 4 decimal places. The root is \(1.5689\) (4 d.p.). Check: \(f(1.56885)<0\) and \(f(1.56895)>0\), so the root is 1.5689 to 4 decimal places.

4. Answer to Question 4

   Method (trapezium rule): \(\displaystyle\int_a^by\,dx\approx\dfrac h2\left[y_0+y_n+2(y_1+\dots+y_{n-1})\right]\) with strip width \(h=\dfrac{b-a}n\). The estimate is too large when the curve is convex and too small when it is concave.

   1. \(h=0.25\): \(x=1,1.25,1.5,1.75,2\) gives \(y=1,0.8,0.6667,0.5714,0.5\).
   2. \(\dfrac{0.25}{2}\left[1+0.5+2(0.8+0.6667+0.5714)\right]=0.125\times5.5762=0.6970\).
   3. The estimate \(0.6970\) is too large. The curve \(y=\dfrac1x\) is convex (bends upwards), so the straight chord tops lie above the curve.

5. Answer to Question 5

   Method (fixed-point iteration): rearrange \(f(x)=0\) as \(x=g(x)\), then use \(x_{n+1}=g(x_n)\), keeping full calculator accuracy and rounding only when writing the values. The iteration converges near the root \(\alpha\) when \(|g'(\alpha)|<1\).

   1. Use radian mode and press \(\cos\) repeatedly on the previous answer. \(x_1=0.5403\), \(x_2=0.8576\), \(x_3=0.6543\), \(x_4=0.7935\), \(x_5=0.7014\). The values jump above and below 0.7391.
   2. Differentiate \(\cos x\). \(g\prime(x)=-\sin x\), so \(g\prime(0.7391)=-0.6736\).
   3. Compare the size of the gradient at the root with 1. \(|g\prime(\alpha)|=0.6736<1\), so the iteration converges, and since \(g\prime<0\) it alternates (cobweb). So each step multiplies the error by about \(-0.67\), and the values converge slowly while alternating.

6. Answer to Question 6

   1. \(f(0.6)=1.8221-1.8=0.0221>0\) and \(f(0.7)=2.0138-2.1=-0.0862<0\); continuous, so a root lies in \([0.6,0.7]\).
   2. \(f(1.5)=4.4817-4.5=-0.0183<0\) and \(f(1.6)=4.9530-4.8=0.1530>0\); continuous, so a root lies in \([1.5,1.6]\).
   3. The line \(y=3x\) starts below the curve at the origin (\(e^0=1>0\)), crosses it, falls below it again and then the exponential curve overtakes the line permanently. That gives two intersections.

7. Answer to Question 7

   1. The base is a square of side \(30-2x\) and the height is \(x\), so \(V=x(30-2x)^2\).
   2. \(f(1)=784-1000=-216<0\) and \(f(1.5)=1.5\times729-1000=93.5>0\). Change of sign, continuous, so the root lies in \([1,1.5]\).
   3. \(f\prime(x)=(30-2x)^2-4x(30-2x)\). \(x_1=1.2-\dfrac{f(1.2)}{f\prime(1.2)}=1.3365\), \(x_2=1.3397\), \(x_3=1.33975\). So \(x=1.340\) cm to 3 d.p.
   4. \(x=10\) gives a base of side 10 cm and height 10 cm, a deep narrow box, whereas a tray needs low sides: it is a valid solution of the equation but gives a different shape.

8. Answer to Question 8

   Method (change of sign): if \(f\) is continuous on \([a,b]\) and \(f(a)\), \(f(b)\) have opposite signs, then \(f(x)=0\) has a root between \(a\) and \(b\). Always state both the change of sign and the continuity.

   1. Evaluate \(f\) at both ends of the interval. \(f(0.07)=250\times4.1002-1000=25.05>0\) and \(f(0.08)=250\times3.9927-1000=-1.82<0\). Change of sign and \(f\) continuous for \(r>0\). So the rate lies between 7% and 8%.
   2. Evaluate \(f\) at both values; the sign changes, so the root lies between them. \(f(0.079)=0.81>0\) and \(f(0.0795)=-0.51<0\) so \(r\) lies in \([0.079,0.0795]\), giving \(r=0.079\) to 3 d.p. (more precisely 0.0793). Every value in this interval rounds to 0.079.
   3. Multiply the decimal by 100. About 7.9% per year, which is typical for a personal loan.

9. Answer to Question 9

   1. \(h=2\). Area \(\approx\dfrac22\left[0+0+2(1.2+2.5+3.1+2.8+1.6)\right]=2\times11.2=22.4\) m\(^2\).
   2. Volume per second \(=22.4\times0.8=17.9\) m\(^3\)/s.
   3. For a bed with the shape of a bowl, the cross-section is concave (bends downwards when depth is plotted upward), so the chords lie below the curve and the rule underestimates the area.

10. Answer to Question 10

    Method (trapezium rule): \(\displaystyle\int_a^by\,dx\approx\dfrac h2\left[y_0+y_n+2(y_1+\dots+y_{n-1})\right]\) with strip width \(h=\dfrac{b-a}n\). The estimate is too large when the curve is convex and too small when it is concave.

    1. The strip width is \(h=10\) minutes; the first and last ordinates are counted once and the others twice. \(\dfrac{10}2\left[0.05+0.04+2(0.08+0.11+0.12+0.10+0.07)\right]=5\times(0.09+0.96)=5.25\) litres. So the van uses about 5.25 litres of fuel.
    2. Divide the fuel used by the capacity of the tank. \(\dfrac{5.25}{20}=26.25\%\approx26\%\). So the van uses about a quarter of its tank.
    3. Think about the units: litres per minute times minutes. The area is the total volume of fuel used, because rate \(\times\) time gives volume.

11. Answer to Question 11

    Method (Newton-Raphson): \(x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\). Each step follows the tangent at \(x_n\) down to the \(x\)-axis. Use the ANS key on the calculator and stop when two values agree to the required accuracy.

    1. Set the volume equal to 3 and multiply both sides by \(\dfrac3\pi\). \(\dfrac{\pi h^2(4.5-h)}3=3\) gives \(h^2(4.5-h)=\dfrac9\pi=2.8648\). This is the required equation.
    2. Use the change of sign method: if \(f\) is continuous on \([a,b]\) and \(f(a)\), \(f(b)\) have opposite signs, then \(f(x)=0\) has a root between \(a\) and \(b\). Always state both the change of sign and the continuity. \(f(0.89)=0.7921\times3.61-2.8648=-0.0053<0\) and \(f(0.9)=0.81\times3.6-2.8648=0.0512>0\). Change of sign, continuous: root in \([0.89,0.9]\). So the depth is between 0.89 m and 0.9 m.
    3. Expand \(f(h)=4.5h^2-h^3-\tfrac9\pi\) to differentiate it. \(f\prime(h)=9h-3h^2\). \(h_1=1-\dfrac{0.6352}{6}=0.8941\), \(h_2=0.8909\), \(h_3=0.8909\). So \(h=0.891\) m to 3 d.p. (89.1 cm). The last two values agree, so the depth is about 0.891 m.

12. Answer to Question 12

    Method (fixed-point iteration): rearrange \(f(x)=0\) as \(x=g(x)\), then use \(x_{n+1}=g(x_n)\), keeping full calculator accuracy and rounding only when writing the values. The iteration converges near the root \(\alpha\) when \(|g'(\alpha)|<1\).

    1. \(f(x)=x+\ln x-3\): \(f(2)=2+0.6931-3=-0.3069<0\) and \(f(3)=3+1.0986-3=1.0986>0\). Continuous, so there is a root in \([2,3]\).
    2. \(x+\ln x=3\Rightarrow x=3-\ln x\).
    3. \(x_1=2.3069\), \(x_2=2.1641\), \(x_3=2.2280\), \(x_4=2.1989\), \(x_5=2.2120\), \(x_6=2.2061\).
    4. The root is \(2.21\) to 2 d.p. (2.2079). \(g\prime(x)=-\dfrac1x\) and \(|g\prime(2.21)|=0.45<1\), so the iteration converges; the negative sign gives the alternating pattern.

13. Answer to Question 13

    1. \(f(-2)=-8+4+2=-2<0\) and \(f(-1.5)=-3.375+3+2=1.625>0\), continuous: a root in \([-2,-1.5]\).
    2. \(f\prime(x)=3x^2-2\). \(x_1=0-\dfrac{2}{-2}=1\); \(x_2=1-\dfrac{1}{1}=0\); \(x_3=1\); \(x_4=0\).
    3. The iteration cycles between 0 and 1 and never approaches the root, because the tangent at \(x=1\) sends it back to 0. Choose a start closer to the root, such as \(x_0=-1.8\), which converges to \(-1.7693\).

14. Answer to Question 14

    1. \(f(2.3)=12.167-9.2-3=-0.033<0\) and \(f(2.31)=12.326-9.24-3=0.086>0\). Continuous, so there is a root in \([2.3,2.31]\).
    2. \(x_1=\sqrt[3]{11}=2.2240\), \(x_2=\sqrt[3]{11.896}=2.2828\), \(x_3=2.2977\), \(x_4=2.3015\) (approaching 2.3028 from below).
    3. \(f\prime(x)=3x^2-4\). \(x_1=2.3-\dfrac{-0.033}{11.87}=2.3027801\), \(x_2=2.3027756\). So \(\alpha=2.3028\) (4 d.p.).
    4. \(f(2.3025)=-0.0033<0\) and \(f(2.3035)=0.0086>0\). Continuous with a change of sign, so \(\alpha\) lies in \([2.3025,2.3035]\), and every number in this interval rounds to 2.303. So \(\alpha=2.303\) to 3 d.p.

15. Answer to Question 15

    Method (fixed-point iteration): rearrange \(f(x)=0\) as \(x=g(x)\), then use \(x_{n+1}=g(x_n)\), keeping full calculator accuracy and rounding only when writing the values. The iteration converges near the root \(\alpha\) when \(|g'(\alpha)|<1\).

    1. (A): \(x_1=2.375\), \(x_2=12.396\), \(x_3=1904.0\). (B): \(x_1=1.3572\), \(x_2=1.3309\), \(x_3=1.3259\), \(x_4=1.3249\), \(x_5=1.3248\).
    2. (A): \(g\prime(x)=3x^2\), \(g\prime(\alpha)=3(1.3247)^2=5.26\). (B): \(g\prime(x)=\dfrac1{3(x+1)^{2/3}}\), \(g\prime(\alpha)=0.19\).
    3. (B) converges because \(|g\prime(\alpha)|=0.19<1\). (A) diverges because \(|g\prime(\alpha)|=5.26>1\) so the errors grow each step.
    4. In (B) the values decrease towards \(\alpha\) from above: \(0\lt g\prime<1\) gives a staircase. Each step reduces the error by about 81%.

16. Answer to Question 16

    1. \(h=0.5\): \(\dfrac{0.5}2\left[1+0.3679+2(0.7788)\right]=0.25\times2.9255=0.7314\).
    2. \(h=0.25\): \(\dfrac{0.25}2\left[1+0.3679+2(0.9394+0.7788+0.5698)\right]=0.125\times5.9439=0.7430\).
    3. Errors: \(0.7468-0.7314=0.0154\) and \(0.7468-0.7430=0.0038\). Doubling the number of strips divides the error by about 4, which is typical of the trapezium rule.
    4. Both estimates are too small: on most of \([0,1]\) the curve is concave (bends downwards), so the chords lie below the curve.

17. Answer to Question 17

    1. \(x_{n+1}=x_n-\dfrac{e^{x_n}-3x_n}{e^{x_n}-3}\).
    2. \(x_1=0.61006\), \(x_2=0.61900\), \(x_3=0.61906\) (root \(0.619\)).
    3. \(x_1=1.68352\), \(x_2=1.54348\), \(x_3=1.51349\), \(x_4=1.51214\) (root \(1.512\)).
    4. \(f\prime(1)=e-3=-0.2817\) and \(f(1)=e-3=-0.2817\), so \(x_1=1-\dfrac{-0.2817}{-0.2817}=0\). Then \(f\prime(0)=-2\), \(f(0)=1\), so \(x_2=0+\dfrac12=0.5\). The start \(x_0=1\) is close to the minimum of \(f\) (where \(f\prime=0\) at \(x=\ln3=1.0986\)), so the tangent is almost horizontal and the first step is poor: it throws the iteration to the other root region.

18. Answer to Question 18

    Method (Newton-Raphson): \(x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\). Each step follows the tangent at \(x_n\) down to the \(x\)-axis. Use the ANS key on the calculator and stop when two values agree to the required accuracy.

    1. Area of segment \(=\tfrac12r^2(\theta-\sin\theta)=12.5(\theta-\sin\theta)=12\), so \(\theta-\sin\theta=\dfrac{12}{12.5}=0.96\).
    2. \(g(\theta)=\theta-\sin\theta-0.96\): \(g(1.9)=1.9-0.9463-0.96=-0.0063<0\) and \(g(2.0)=2-0.9093-0.96=0.1307>0\). Continuous with a change of sign, so \(\theta\in[1.9,2.0]\).
    3. \(g\prime=1-\cos\theta\). \(\theta_1=1.8-\dfrac{-0.13385}{1.22720}=1.9091\), \(\theta_2=1.90476\), \(\theta_3=1.90475\). So \(\theta=1.905\) (3 d.p.).
    4. Chord \(=2r\sin\dfrac\theta2=10\sin(0.95238)=8.15\) dm.

    Answer: \(\theta\approx1.905\) radians and the chord is about 8.15 dm long.

19. Answer to Question 19

    1. \(x_{n+1}=x_n-\dfrac{x_n^2-a}{2x_n}=\dfrac{2x_n^2-x_n^2+a}{2x_n}=\dfrac12\left(x_n+\dfrac a{x_n}\right)\).
    2. \(x_{n+1}-\sqrt a=\dfrac{x_n^2+a-2x_n\sqrt a}{2x_n}=\dfrac{(x_n-\sqrt a)^2}{2x_n}\).
    3. \(x_1=\tfrac12(2.5+2.8)=2.65\), \(x_2=\tfrac12(2.65+2.641509)=2.645755\), \(x_3=2.645751\) (6 d.p.), and \(\sqrt7=2.645751\). The error is squared (and divided by about \(2x_n\)) at each step, so if the error is \(10^{-k}\) the next error is about \(10^{-2k}\): the number of correct digits doubles.

20. Answer to Question 20

    1. \(h=\dfrac1n\) and \(T_n=\dfrac h2\left[0+1+2\sum_{k=1}^{n-1}\dfrac{k^2}{n^2}\right]\). Using \(\sum_{k=1}^{n-1}k^2=\dfrac{(n-1)n(2n-1)}6\): \(T_n=\dfrac1{2n}\left[1+\dfrac{(n-1)(2n-1)}{3n}\right]=\dfrac1{2n}\cdot\dfrac{3n+2n^2-3n+1}{3n}=\dfrac{2n^2+1}{6n^2}=\dfrac13+\dfrac1{6n^2}\).
    2. \(T_4=\dfrac13+\dfrac1{96}=0.34375\) and \(T_8=\dfrac13+\dfrac1{384}=0.33594\). The errors are \(\dfrac1{96}\) and \(\dfrac1{384}\): the ratio is 4.
    3. \(y=x^2\) is convex, so each chord lies above the curve and every trapezium has more area than the strip under the curve.