AI HL · AHL 9
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Slope fields and solution curves
AHL 9.4
Screenshot and video to come
Choose an equation dy/dx = f(x, y). Each segment shows the gradient at its point. Drag the orange point to choose a starting condition, predict where the solution curve goes, then press "Show the solution curve".
The short teal line is the gradient at your point. For y cos x, x is in radians.
Drag the orange point to choose a starting condition.
A solution curve of \(\dfrac{dy}{dx}=y-x\) passes through some points. Find the gradient at the point \((−1, 2)\). Then make a table of the gradients at the integer points with \(-2\le x\le 2\) and \(-2\le y\le 2\).
Have a go first. Then open the working one line at a time.
\(\dfrac{dy}{dx}=2-(-1)=3\)
Put x = −1 and y = 2 into the equation. Keep the bracket round the negative number.
Row by row: put each x and y into \(y-x\)
Along a row y is fixed, so the gradient falls by 1 each time x goes up by 1. Use that to check your row.
Gradient 0 at \((−2,−2),(−1,−1),(0,0),(1,1),(2,2)\)
The gradient is 0 wherever y = x. Those points are on one diagonal of the table.
Use the table of gradients for \(\dfrac{dy}{dx}=y-x\) to sketch the slope field on the integer points. What is the gradient at the point \((2, 2)\)?
Have a go first. Then open the working one line at a time.
gradient 0: flat. Gradient 1: up at 45°. Gradient −1: down at 45°. Gradient 3 or −3: steep
At each point draw a short line segment with the gradient from the table. The segment is a tiny piece of tangent.
25 short segments, one at each integer point
Draw them all the same length and centre each one on its point.
On \(y=x\) the gradient is 0. On \(y=x+1\) the gradient is 1
Points with the same gradient lie along lines. Spotting them saves drawing time.
The slope field of \(\dfrac{dy}{dx}=y-x\) is drawn on the integer points. Sketch the solution curves through \((−1, 2)\), \((0, 1)\) and \((1, −1)\). Estimate \(y\) when \(x=2\) on the curve through \((0, 1)\).
Have a go first. Then open the working one line at a time.
Mark \((−1,2)\), \((0,1)\), \((1,−1)\)
A solution curve has to pass through its starting point.
Stay tangent to the lines. Never cross another curve
At each point the curve points the way the segment points. Look ahead to the next segment and bend smoothly.
\(y=x+1\) gives \(y-x=1\), the gradient of that line
Every segment on this line has gradient 1, so the curve is the straight line itself.
\(y(2)\approx 3\)
Go along to x = 2 and read up to the curve.
Use your calculator to plot the slope field of \(\dfrac{dy}{dx}=y\cos x\) for \(0\le x\le 7\) and \(0\le y\le 4\). Add the solution curve through \((0, 1)\). Then check one segment by hand: find the gradient at \((\pi, 2)\).
Have a go first. Then open the working one line at a time.
From the calculator: dy/dx = y cos x, angles in radians
Set radians first. In degrees the segments come out wrong.
x from 0 to 7, y from 0 to 4
The window decides which part of the field you see.
From the calculator: start point (0, 1)
The curve rises while cos x is positive, levels off where cos x = 0, then falls.
\(y\cos x=2\cos \pi=2(-1)=-2\)
The picture should show a segment falling at that point. Always check one.
For \(\dfrac{dy}{dx}=y^2-4\), find the constant solutions. State the one with \(y>0\).
Have a go first. Then open the working one line at a time.
\(\dfrac{dy}{dx}=0\) for every x
The curve is a horizontal line, so its gradient is 0 everywhere on it.
\(y^2-4=0\), so \(y=2\) or \(y=-2\)
There is no x in the equation, so each answer is a plain number. The lines y = 2 and y = −2 are solutions.
Flat segments along \(y=2\) and \(y=-2\)
Other curves bend towards these lines or away from them, but never cross them.
The number of deer \(P\) (in hundreds) on an island after \(t\) years satisfies \(\dfrac{dP}{dt}=0.6P\left(1-\dfrac{P}{12}\right)\). Find the constant solutions and say which is stable. Find the value of \(P\) at which the population grows fastest, and the greatest rate of growth.
Have a go first. Then open the working one line at a time.
\(0.6P\left(1-\dfrac{P}{12}\right)=0\ \Rightarrow\ P=0\) or \(P=12\)
The population does not change when dP/dt = 0.
\(P=3:\ \dfrac{dP}{dt}=1.35\gt0\) \(P=15:\ \dfrac{dP}{dt}=−2.25\lt0\)
Between 0 and 12 the population rises. Above 12 it falls.
P = 12 is stable. P = 0 is unstable.
Near 12 the curves move towards it. Near 0 they move away.
\(\dfrac{dP}{dt}=0.6P-0.05P^2\), greatest at \(P=\dfrac{0.6}{2\times0.05}=6\)
A quadratic in P: its greatest value is halfway between the two constant solutions. The calculator's maximum of the graph gives the same.
\(0.6\times6\times\left(1-\dfrac{6}{12}\right)=1.8\)
1.8 hundred per year, at 600 deer.