Further Maths Numerical methods Practice Questions

Free Further Maths Numerical methods practice questions with full step-by-step worked solutions. Covers mid-ordinate-rule, numerical-integration, simpsons-rule, eulers-method. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Use the mid-ordinate rule with 22 strips to estimate ∫0111+x2 dx\int_{0}^{1}\frac{1}{1+x^{2}}\,dx, giving your answer to 44 decimal places.
Show worked solution

Worked solution

  1. Quote the mid-ordinate rule

    ∫abf(x) dx≈h(f(m1)+f(m2)+⋯+f(mn))\int_{a}^{b}f(x)\,dx\approx h\left(f(m_{1})+f(m_{2})+\cdots+f(m_{n})\right)

    Each strip is replaced by a rectangle whose height is the value of ff at the middle of that strip.

  2. Find hh, the mid-ordinates and the function values

    h=0.5,m1=0.25,m2=0.75,f(0.25)=0.94117647,f(0.75)=0.64000000h=0.5,\quad m_{1}=0.25,\quad m_{2}=0.75,\quad f\left(0.25\right)=0.94117647,\quad f\left(0.75\right)=0.64000000

    All the ingredients of the rule in one place.

  3. State the estimate to the required accuracy

    I≈0.5(0.94117647+0.64000000)=0.79058824≈0.7906I\approx0.5\left(0.94117647+0.64000000\right)=0.79058824\approx0.7906

    The question asks for 44 decimal places, so the estimate is rounded at the very end.

Answer
0.79060.7906
Question 2
2 markseasy
The function y(x)y(x) satisfies dydx=y−x\frac{dy}{dx}=y-x with y=2y=2 at x=0x=0. Using Euler's method yr+1=yr+hf(xr,yr)y_{r+1}=y_{r}+hf\left(x_{r},y_{r}\right) with step length h=0.1h=0.1, estimate the value of yy at x=0.1x=0.1, giving your answer to 44 decimal places. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Quote Euler's method

    yr+1=yr+hf(xr,yr)y_{r+1}=y_{r}+hf\left(x_{r},y_{r}\right)

    The gradient at the current point is used to step forward a distance hh in xx.

  2. Write down the step length and the starting values

    h=0.1,x0=0,y0=2h=0.1,\quad x_{0}=0,\quad y_{0}=2

    11 steps of length 0.10.1 take xx from 00 to 0.10.1.

  3. Carry out step 11

    y1=2.00000000+0.1×f(0,2.00000000)=2.00000000+0.1×2.00000000=2.20000000y_{1}=2.00000000+0.1\times f\left(0,2.00000000\right)=2.00000000+0.1\times2.00000000=2.20000000

    The gradient at (0,2.00000000)\left(0,2.00000000\right) is 2.000000002.00000000.

  4. Select the option that matches this estimate

    y(0.1)≈2.2000y\left(0.1\right)\approx2.2000

    The value after the final step, rounded to 44 decimal places.

Answer
2.20002.2000
Question 3
4 marksintermediate
The function y(x)y(x) satisfies dydx=y−x\frac{dy}{dx}=y-x with y=2y=2 at x=0x=0. Using the improved Euler formula yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right) with step length h=0.1h=0.1, estimate the value of yy at x=0.4x=0.4, giving your answer to 44 decimal places. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Quote the improved Euler formula

    yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right)

    The gradients at the start and at the predicted end of the step are averaged, which is far more accurate than plain Euler.

  2. Write down the step length and the starting values

    h=0.1,x0=0,y0=2h=0.1,\quad x_{0}=0,\quad y_{0}=2

    44 steps of length 0.10.1 take xx from 00 to 0.40.4.

  3. Carry out step 11

    k1=hf(0,2.00000000)=0.1×2.00000000=0.20000000,k2=hf(0.1,2.20000000)=0.1×2.10000000=0.21000000,y1=2.00000000+12(0.20000000+0.21000000)=2.20500000k_{1}=hf\left(0,2.00000000\right)=0.1\times2.00000000=0.20000000,\quad k_{2}=hf\left(0.1,2.20000000\right)=0.1\times2.10000000=0.21000000,\quad y_{1}=2.00000000+\frac{1}{2}\left(0.20000000+0.21000000\right)=2.20500000

    One complete improved-Euler step from (0,2.00000000)\left(0,2.00000000\right).

  4. Carry out step 22

    k1=hf(0.1,2.20500000)=0.1×2.10500000=0.21050000,k2=hf(0.2,2.41550000)=0.1×2.21550000=0.22155000,y2=2.20500000+12(0.21050000+0.22155000)=2.42102500k_{1}=hf\left(0.1,2.20500000\right)=0.1\times2.10500000=0.21050000,\quad k_{2}=hf\left(0.2,2.41550000\right)=0.1\times2.21550000=0.22155000,\quad y_{2}=2.20500000+\frac{1}{2}\left(0.21050000+0.22155000\right)=2.42102500

    One complete improved-Euler step from (0.1,2.20500000)\left(0.1,2.20500000\right).

  5. Carry out step 33

    k1=hf(0.2,2.42102500)=0.1×2.22102500=0.22210250,k2=hf(0.3,2.64312750)=0.1×2.34312750=0.23431275,y3=2.42102500+12(0.22210250+0.23431275)=2.64923263k_{1}=hf\left(0.2,2.42102500\right)=0.1\times2.22102500=0.22210250,\quad k_{2}=hf\left(0.3,2.64312750\right)=0.1\times2.34312750=0.23431275,\quad y_{3}=2.42102500+\frac{1}{2}\left(0.22210250+0.23431275\right)=2.64923263

    One complete improved-Euler step from (0.2,2.42102500)\left(0.2,2.42102500\right).

  6. Carry out step 44

    k1=hf(0.3,2.64923263)=0.1×2.34923263=0.23492326,k2=hf(0.4,2.88415589)=0.1×2.48415589=0.24841559,y4=2.64923263+12(0.23492326+0.24841559)=2.89090205k_{1}=hf\left(0.3,2.64923263\right)=0.1\times2.34923263=0.23492326,\quad k_{2}=hf\left(0.4,2.88415589\right)=0.1\times2.48415589=0.24841559,\quad y_{4}=2.64923263+\frac{1}{2}\left(0.23492326+0.24841559\right)=2.89090205

    One complete improved-Euler step from (0.3,2.64923263)\left(0.3,2.64923263\right).

  7. Select the option that matches this estimate

    y(0.4)≈2.8909y\left(0.4\right)\approx2.8909

    The value after the final step, rounded to 44 decimal places.

Answer
2.89092.8909
Question 4
6 markshard
The function y(x)y(x) satisfies dydx=x2+y\frac{dy}{dx}=x^{2}+y with y=1y=1 at x=0x=0. Using Euler's method yr+1=yr+hf(xr,yr)y_{r+1}=y_{r}+hf\left(x_{r},y_{r}\right) with step length h=0.25h=0.25, estimate the value of yy at x=1x=1, giving your answer to 44 decimal places. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Quote Euler's method

    yr+1=yr+hf(xr,yr)y_{r+1}=y_{r}+hf\left(x_{r},y_{r}\right)

    The gradient at the current point is used to step forward a distance hh in xx.

  2. Write down the step length and the starting values

    h=0.25,x0=0,y0=1h=0.25,\quad x_{0}=0,\quad y_{0}=1

    44 steps of length 0.250.25 take xx from 00 to 11.

  3. List the xx-values used

    x0=0,x1=0.25,x2=0.5,x3=0.75,x4=1x_{0}=0,\quad x_{1}=0.25,\quad x_{2}=0.5,\quad x_{3}=0.75,\quad x_{4}=1

    Each step advances xx by h=0.25h=0.25.

  4. Evaluate the gradient at x0=0x_{0}=0

    f(0,1.00000000)=(0)2+(1.00000000)=1.00000000f\left(0,1.00000000\right)=\left(0\right)^{2}+\left(1.00000000\right)=1.00000000

    Substitute the current xx and yy into dydx=x2+y\frac{dy}{dx}=x^{2}+y.

  5. Apply Euler's formula to find y1y_{1}

    y1=y0+hf(x0,y0)=1.00000000+0.25×1.00000000=1.25000000y_{1}=y_{0}+hf\left(x_{0},y_{0}\right)=1.00000000+0.25\times1.00000000=1.25000000

    One step of length hh from (0,1.00000000)\left(0,1.00000000\right).

  6. Evaluate the gradient at x1=0.25x_{1}=0.25

    f(0.25,1.25000000)=(0.25)2+(1.25000000)=1.31250000f\left(0.25,1.25000000\right)=\left(0.25\right)^{2}+\left(1.25000000\right)=1.31250000

    Substitute the current xx and yy into dydx=x2+y\frac{dy}{dx}=x^{2}+y.

  7. Apply Euler's formula to find y2y_{2}

    y2=y1+hf(x1,y1)=1.25000000+0.25×1.31250000=1.57812500y_{2}=y_{1}+hf\left(x_{1},y_{1}\right)=1.25000000+0.25\times1.31250000=1.57812500

    One step of length hh from (0.25,1.25000000)\left(0.25,1.25000000\right).

  8. Evaluate the gradient at x2=0.5x_{2}=0.5

    f(0.5,1.57812500)=(0.5)2+(1.57812500)=1.82812500f\left(0.5,1.57812500\right)=\left(0.5\right)^{2}+\left(1.57812500\right)=1.82812500

    Substitute the current xx and yy into dydx=x2+y\frac{dy}{dx}=x^{2}+y.

  9. Apply Euler's formula to find y3y_{3}

    y3=y2+hf(x2,y2)=1.57812500+0.25×1.82812500=2.03515625y_{3}=y_{2}+hf\left(x_{2},y_{2}\right)=1.57812500+0.25\times1.82812500=2.03515625

    One step of length hh from (0.5,1.57812500)\left(0.5,1.57812500\right).

  10. Evaluate the gradient at x3=0.75x_{3}=0.75

    f(0.75,2.03515625)=(0.75)2+(2.03515625)=2.59765625f\left(0.75,2.03515625\right)=\left(0.75\right)^{2}+\left(2.03515625\right)=2.59765625

    Substitute the current xx and yy into dydx=x2+y\frac{dy}{dx}=x^{2}+y.

  11. Apply Euler's formula to find y4y_{4}

    y4=y3+hf(x3,y3)=2.03515625+0.25×2.59765625=2.68457031y_{4}=y_{3}+hf\left(x_{3},y_{3}\right)=2.03515625+0.25\times2.59765625=2.68457031

    One step of length hh from (0.75,2.03515625)\left(0.75,2.03515625\right).

  12. Select the option that matches this estimate

    y(1)≈2.6846y\left(1\right)\approx2.6846

    The value after the final step, rounded to 44 decimal places.

Answer
2.68462.6846
Question 5
9 markschallenging
The function y(x)y(x) satisfies dydx=x+y\frac{dy}{dx}=\sqrt{x+y} with y=1y=1 at x=0x=0. Using the improved Euler formula yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right) with step length h=0.1h=0.1, estimate the value of yy at x=0.5x=0.5, giving your answer to 44 decimal places. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Quote the improved Euler formula

    yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right)

    The gradients at the start and at the predicted end of the step are averaged, which is far more accurate than plain Euler.

  2. Write down the step length and the starting values

    h=0.1,x0=0,y0=1h=0.1,\quad x_{0}=0,\quad y_{0}=1

    55 steps of length 0.10.1 take xx from 00 to 0.50.5.

  3. List the xx-values used

    x0=0,x1=0.1,x2=0.2,x3=0.3,x4=0.4,x5=0.5x_{0}=0,\quad x_{1}=0.1,\quad x_{2}=0.2,\quad x_{3}=0.3,\quad x_{4}=0.4,\quad x_{5}=0.5

    Each step advances xx by h=0.1h=0.1.

  4. Find k1k_{1} for step 11

    k1=hf(0,1.00000000)=0.1×1.00000000=0.10000000k_{1}=hf\left(0,1.00000000\right)=0.1\times1.00000000=0.10000000

    k1k_{1} uses the gradient at the current point (0,1.00000000)\left(0,1.00000000\right).

  5. Find k2k_{2} for step 11

    k2=hf(0.1,1.10000000)=0.1×1.09544512=0.10954451k_{2}=hf\left(0.1,1.10000000\right)=0.1\times1.09544512=0.10954451

    k2k_{2} uses the gradient at the Euler-predicted point (0.1,1.10000000)\left(0.1,1.10000000\right).

  6. Average the two gradients to find y1y_{1}

    y1=1.00000000+12(0.10000000+0.10954451)=1.10477226y_{1}=1.00000000+\frac{1}{2}\left(0.10000000+0.10954451\right)=1.10477226

    The mean of k1k_{1} and k2k_{2} is added to y0y_{0}.

  7. Find k1k_{1} for step 22

    k1=hf(0.1,1.10477226)=0.1×1.09762118=0.10976212k_{1}=hf\left(0.1,1.10477226\right)=0.1\times1.09762118=0.10976212

    k1k_{1} uses the gradient at the current point (0.1,1.10477226)\left(0.1,1.10477226\right).

  8. Find k2k_{2} for step 22

    k2=hf(0.2,1.21453437)=0.1×1.18934199=0.11893420k_{2}=hf\left(0.2,1.21453437\right)=0.1\times1.18934199=0.11893420

    k2k_{2} uses the gradient at the Euler-predicted point (0.2,1.21453437)\left(0.2,1.21453437\right).

  9. Average the two gradients to find y2y_{2}

    y2=1.10477226+12(0.10976212+0.11893420)=1.21912041y_{2}=1.10477226+\frac{1}{2}\left(0.10976212+0.11893420\right)=1.21912041

    The mean of k1k_{1} and k2k_{2} is added to y1y_{1}.

  10. Find k1k_{1} for step 33

    k1=hf(0.2,1.21912041)=0.1×1.19126841=0.11912684k_{1}=hf\left(0.2,1.21912041\right)=0.1\times1.19126841=0.11912684

    k1k_{1} uses the gradient at the current point (0.2,1.21912041)\left(0.2,1.21912041\right).

  11. Find k2k_{2} for step 33

    k2=hf(0.3,1.33824725)=0.1×1.27994033=0.12799403k_{2}=hf\left(0.3,1.33824725\right)=0.1\times1.27994033=0.12799403

    k2k_{2} uses the gradient at the Euler-predicted point (0.3,1.33824725)\left(0.3,1.33824725\right).

  12. Average the two gradients to find y3y_{3}

    y3=1.21912041+12(0.11912684+0.12799403)=1.34268085y_{3}=1.21912041+\frac{1}{2}\left(0.11912684+0.12799403\right)=1.34268085

    The mean of k1k_{1} and k2k_{2} is added to y2y_{2}.

  13. Find k1k_{1} for step 44

    k1=hf(0.3,1.34268085)=0.1×1.28167112=0.12816711k_{1}=hf\left(0.3,1.34268085\right)=0.1\times1.28167112=0.12816711

    k1k_{1} uses the gradient at the current point (0.3,1.34268085)\left(0.3,1.34268085\right).

  14. Find k2k_{2} for step 44

    k2=hf(0.4,1.47084796)=0.1×1.36778944=0.13677894k_{2}=hf\left(0.4,1.47084796\right)=0.1\times1.36778944=0.13677894

    k2k_{2} uses the gradient at the Euler-predicted point (0.4,1.47084796)\left(0.4,1.47084796\right).

  15. Average the two gradients to find y4y_{4}

    y4=1.34268085+12(0.12816711+0.13677894)=1.47515388y_{4}=1.34268085+\frac{1}{2}\left(0.12816711+0.13677894\right)=1.47515388

    The mean of k1k_{1} and k2k_{2} is added to y3y_{3}.

  16. Find k1k_{1} for step 55

    k1=hf(0.4,1.47515388)=0.1×1.36936258=0.13693626k_{1}=hf\left(0.4,1.47515388\right)=0.1\times1.36936258=0.13693626

    k1k_{1} uses the gradient at the current point (0.4,1.47515388)\left(0.4,1.47515388\right).

  17. Find k2k_{2} for step 55

    k2=hf(0.5,1.61209014)=0.1×1.45330318=0.14533032k_{2}=hf\left(0.5,1.61209014\right)=0.1\times1.45330318=0.14533032

    k2k_{2} uses the gradient at the Euler-predicted point (0.5,1.61209014)\left(0.5,1.61209014\right).

  18. Average the two gradients to find y5y_{5}

    y5=1.47515388+12(0.13693626+0.14533032)=1.61628717y_{5}=1.47515388+\frac{1}{2}\left(0.13693626+0.14533032\right)=1.61628717

    The mean of k1k_{1} and k2k_{2} is added to y4y_{4}.

  19. Select the option that matches this estimate

    y(0.5)≈1.6163y\left(0.5\right)\approx1.6163

    The value after the final step, rounded to 44 decimal places.

Answer
1.61631.6163

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