AI HL · AHL 9
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Euler's method: accuracy and contexts
AHL 9.7
Screenshot and video to come
Screenshot and video to come
Choose an equation and a step length h. Drag the orange start point if you like. Press Next: the gradient at the point you are leaving gives one straight piece. Tick the box to compare with the true curve.
| n | x | y | dy/dx | new y |
|---|
The table shows 4 decimal places; the calculation keeps more. Dots and ink line: Euler's method. Orange: the start point and the latest piece. Teal: the true curve, with its error at the last point.
Press Next to take the first step.
\(\dfrac{dy}{dx}=4x\) and \(y=1\) when \(x=0\). Use Euler's method with step length \(h=0.5\) and then \(h=0.25\) to estimate \(y(1)\) and \(y(2)\).
Have a go first. Then open the working one line at a time.
\(x_{n+1}=x_n+h,\quad y_{n+1}=y_n+h\,f(x_n)\)
f(x) = 4x. Each step uses the gradient at the start of the step.
\(y_1=1+0.5(0)=1,\ y_2=1+0.5(2)=2,\ y_3=2+0.5(4)=4,\ y_4=4+0.5(6)=7\)
x goes 0, 0.5, 1, 1.5, 2: four steps.
\(y(1)\approx2,\quad y(2)\approx7\)
y(1) is the value of y at x = 1, after two steps.
\(y(1)\approx2.5,\quad y(2)\approx8\)
Eight steps now. Both estimates are larger than before.
\(\dfrac{dy}{dx}=0.5y\) and \(y=4\) when \(x=0\). (a) Use Euler's method with \(h=1\) and then \(h=0.5\) to estimate \(y(2)\). (b) Solve the equation exactly. (c) Find the percentage error of each estimate, and say which is further from the true value.
Have a go first. Then open the working one line at a time.
\(y_1=4+1(2)=6,\ y_2=6+1(3)=9\)
The gradient is 0.5y, so use the y-value at the start of each step.
\(y: 4, 5, 6.25, 7.8125, 9.7656\quad y(2)\approx9.766\)
Four steps of 0.5.
\(\int\frac1y\,dy=\int0.5\,dx\Rightarrow\ln y=0.5x+c\Rightarrow y=4e^{0.5x}\)
y = 4 when x = 0 gives the constant.
\(y(2)=4e^{1}=10.87\)
This is the value to compare with.
\(h=1:\ \dfrac{10.87-9}{10.87}\times100=17.2\%,\quad h=0.5:\ 10.2\%\)
The h = 1 estimate is further from the true value.
A smaller step length gives a smaller error.
Halving h roughly halved the error here.
\(\dfrac{dy}{dx}=xe^{-y}\) and \(y=0.5\) when \(x=1\). Use Euler's method with \(h=0.25\) to estimate \(y(2)\).
Have a go first. Then open the working one line at a time.
\(x_0=1,\ y_0=0.5,\ h=0.25;\quad y_{n+1}=y_n+0.25\,x_ne^{-y_n}\)
The start is x = 1, not 0. Four steps reach x = 2.
\(y_1=0.5+0.25(1)e^{-0.5}=0.6516\)
x_1 = 1.25.
\(y_2=0.6516+0.25(1.25)e^{-0.6516}=0.8145\)
Use the unrounded y in the calculator.
\(y_3=0.9806,\quad y_4=1.145\)
x_4 = 2.
\(y(2)\approx1.14\)
Give 3 significant figures.
\(\dfrac{dy}{dx}=xy\) and \(y=2\) when \(x=0\). (a) Use Euler's method with \(h=0.5\) to estimate \(y(2)\). (b) The exact solution is \(y=2e^{x^2/2}\). Explain, using the graph, why the Euler estimate is smaller than the true value.
Have a go first. Then open the working one line at a time.
\(y_1=2,\ y_2=2.5,\ y_3=3.75,\ y_4=6.5625\)
y_{n+1} = y_n + 0.5 x_n y_n.
\(y(2)=2e^{2}=14.78\)
The estimate is far below.
\(\dfrac{d^2y}{dx^2}=y+x\dfrac{dy}{dx}=y+x^2y>0\)
The solution curve is concave up for y > 0.
A tangent to a concave up curve lies below it.
Each Euler step follows a tangent, so each new point is below the curve, and the error builds up.
\(\dfrac{dy}{dx}=3\cos x-y\) with \(y=0\) when \(x=0\) (\(x\) in radians). Use Euler's method with \(h=0.1\) for \(0\le x\le6\) to sketch the solution, and estimate the largest value of \(y\) in this interval to 1 decimal place.
Have a go first. Then open the working one line at a time.
x in column A, y in column B; start row: 0 and 0
Sixty steps of 0.1 reach x = 6, so use technology.
Next y = y + 0.1 × (3cos(x) − y), using the cells of the row above
Fill down for 60 rows. Check the first rows by hand: y_1 = 0.3, y_2 = 0.5685.
Plot column B against column A, points joined
The curve rises, turns, falls below the axis and then oscillates.
\(y_{\max}\approx1.615\ \text{at}\ x=1.1\)
Read the largest value in column B.
\(y_{\max}\approx1.6\)
1 decimal place, as asked.
The volume \(V\) m³ of water in a tank, \(t\) minutes after a valve is opened, is modelled by \(\dfrac{dV}{dt}=-0.25\sqrt{V}-0.1t\). At the start \(V=12\). Use Euler's method with a step length you choose to estimate the time, to the nearest minute, when the tank is empty.
Have a go first. Then open the working one line at a time.
\(h=1:\ V_{n+1}=V_n+1(-0.25\sqrt{V_n}-0.1t_n)\)
The question leaves the step to you. A step of 1 minute is sensible.
t: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11; V: 12, 11.1, 10.2, 9.2, 8.14, 7.03, 5.87, 4.66, 3.42, 2.16, 0.892, −0.344
Keep going until V changes sign.
\(V(10)=0.892>0,\quad V(11)=-0.344<0\)
The tank empties between 10 and 11 minutes, very close to 11.
The estimate is still 11 minutes (V first negative at t = 11)
A smaller step gives the same answer to the nearest minute, so the estimate is reliable.
The numbers of perch \(x\) and pike \(y\) in a lake, \(t\) years after monitoring starts, are modelled by \(\dfrac{dx}{dt}=0.6x-0.01xy\) and \(\dfrac{dy}{dt}=0.004xy-0.3y\). At the start there are 120 perch and 30 pike. Use Euler's method with \(h=0.1\) to estimate the numbers of perch and pike after 0.4 years.
Have a go first. Then open the working one line at a time.
\(x_{n+1}=x_n+h\,f_1,\quad y_{n+1}=y_n+h\,f_2,\quad t_{n+1}=t_n+h\)
Both new values use the old x and the old y.
\(x_1=120+0.1(0.6(120)-0.01(120)(30))=123.6,\ y_1=30.54\)
Work out both before moving on.
\(x_2=127.24,\ y_2=31.134\)
Use the unrounded values.
\(x_3=130.91,\ y_3=31.784;\ x_4=134.61,\ y_4=32.495\)
t = 0.4 after four steps.
About 135 perch and 32.5 pike
Perch to 3 significant figures.
The table shows the speed of a cyclist as she accelerates from rest, measured every 2 seconds. Use Euler's method to estimate the distance travelled in the first 10 seconds.
Have a go first. Then open the working one line at a time.
\(\dfrac{ds}{dt}=v\ \Rightarrow\ s_{n+1}=s_n+h\,v_n\)
The distance gained in each interval is about h times the speed at its start.
\(s=2(0+3.1+5.4+7.2+8.3)\)
Five intervals of 2 s. Leave out the last speed (8.9): it is the start of no interval.
\(s\approx48\text{ m}\)
The cyclist is speeding up, so these start-of-interval speeds are too low: this is an underestimate.