Slope fields and solution curves Problem set A Worked solutions

For \(\dfrac{dy}{dx}=x^2-y\), complete a table of the gradient at \(x=−1, 0, 1\) and \(y=−1, 0, 1, 2\).

1. (a) Complete the table of gradients.

   • Put each x and y into the equation, keeping brackets round negatives
   • y = −1: 2, 1, 2
   • y = 0: 1, 0, 1
   • y = 1: 0, −1, 0
   • y = 2: −1, −2, −1

2. (b) Which points in your table have a flat segment?

   • A flat segment has gradient 0
   • The gradient is 0 at (−1, 1), (0, 0), (1, 1)

Sketch the slope field of \(\dfrac{dy}{dx}=x-2y\) at the integer points with \(-2\le x\le 2\) and \(-2\le y\le 2\), on the grid shown. Make a table of gradients first.

1. (a) Draw the slope field.

   • Substitute each integer point into the equation
   • y = 2: −6, −5, −4, −3, −2
   • y = 1: −4, −3, −2, −1, 0
   • y = 0: −2, −1, 0, 1, 2
   • y = −1: 0, 1, 2, 3, 4
   • y = −2: 2, 3, 4, 5, 6
   • Draw a short segment with each gradient at its point

2. (b) Which points on the grid have a flat segment?

   • A flat segment has gradient 0
   • The gradient is 0 at (−2, −1), (0, 0), (2, 1)

Use your calculator to plot the slope field of \(\dfrac{dy}{dx}=3-y\) for \(0\le x\le 4\) and \(0\le y\le 6\).

1. (a) Sketch the slope field you see. Describe the segments above, on and below the line \(y=3\).

   • From the calculator: dy/dx = 3 − y, window 0 to 4 and 0 to 6
   • At y = 1 the gradient is 3 − 1 = 2 (rising). At y = 3 it is 0. At y = 5 it is −2 (falling)

2. (b) Add the solution curve that starts at \((0, 1)\). Describe what happens to it as \(x\) increases.

   • From the calculator: add the solution curve through (0, 1)
   • The gradient 3 − y shrinks as y gets near 3, so the curve flattens and approaches y = 3 from below

3. (c) Write down the constant solution.

   • dy/dx = 0 when 3 − y = 0
   • The constant solution is y = 3

The slope field of \(\dfrac{dy}{dx}=\dfrac{x-y}{2}\) is shown, with the points P\((−2, 1)\), Q\((0, −2)\) and R\((2, 2)\).

1. (a) Sketch the solution curve through each of P, Q and R.

   • P: gradient \(\frac{-2-1}{2}\) = −1.5, so the curve starts by falling
   • Q: gradient \(\frac{0-(-2)}{2}\) = 1, and the segments on y = x − 2 all have gradient 1, so the curve is that line
   • R: gradient \(\frac{2-2}{2}\) = 0, so the curve starts flat, then rises

2. (b) Explain why your curves do not cross.

   • A point has one gradient, so a curve through it has one direction
   • Two curves meeting would have to be the same curve

The number of fish \(P\) (in thousands) in a farm pond after \(t\) years satisfies \(\dfrac{dP}{dt}=0.5P(8-P)\).

1. (a) Find the two constant solutions.

   • Constant solution: dP/dt = 0
   • 0.5P(8 − P) = 0, so P = 0 or P = 8

2. (b) Find the gradient when \(P=2\) and when \(P=10\).

   • P = 2: 0.5 × 2 × (8 − 2) = 6
   • P = 10: 0.5 × 10 × (8 − 10) = −10

3. (c) Describe what happens to a population of 10 thousand.

   • At P = 10 the gradient is −10, negative
   • So P falls. It cannot cross the constant solution P = 8

A mug of tea at \(50^\circ\text{C}\) cools in a room. Its temperature \(T\) after \(t\) minutes satisfies \(\dfrac{dT}{dt}=-0.5(T-20)\). The slope field is shown.

1. (a) Sketch the solution curve through \((0, 50)\).

   • Start at (0, 50), where the gradient is −0.5 × 30 = −15
   • Follow the segments: the curve falls and flattens as T gets near 20

2. (b) Estimate \(T\) when \(t=2\).

   • Go along to t = 2 and read up to the curve
   • T ≈ 31 °C

3. (c) State the temperature the tea approaches.

   • dT/dt = 0 when T = 20
   • The curve flattens towards the constant solution T = 20

The number \(N\) (in thousands) of bacteria in a dish after \(t\) hours satisfies \(\dfrac{dN}{dt}=0.4N\left(1-\dfrac{N}{15}\right)\).

1. (a) Find the two constant solutions.

   • \(\dfrac{dN}{dt}=0\Rightarrow 0.4N\left(1-\dfrac{N}{15}\right)=0\)
   • \(N=0\) or \(N=15\)

2. (b) Say which constant solution is stable. Give a reason using the sign of the rate of change on each side.

   • \(N=3.75\): rate \(=1.125\gt0\) \(N=18.75\): rate \(=−1.875\lt0\)
   • Above 0 the population rises towards 15. Above 15 it falls towards 15
   • So \(N=15\) is stable and \(N=0\) is unstable

3. (c) Find the value of \(N\) at which the population grows fastest, and the greatest rate of growth.

   • The rate is a quadratic in \(N\) with zeros at 0 and 15
   • Its greatest value is at \(N=7.5\)
   • Rate \(=0.4\times7.5\times\left(1-\dfrac{7.5}{15}\right)=1.5\)

A slope field is shown. It belongs to one of these equations: \(\text{P: }\dfrac{dy}{dx}=x^2+y\), \(\text{Q: }\dfrac{dy}{dx}=y-x^2\), \(\text{R: }\dfrac{dy}{dx}=x-y\).

1. (a) Work out the gradient of each equation at \((2, 0)\).

   • P: 2² + 0 = 4
   • Q: 0 − 2² = −4
   • R: 2 − 0 = 2

2. (b) Which equation gives the field shown? Use the segment at \((2, 0)\) to decide.

   • At (2, 0) the segment in the picture falls steeply
   • Only Q has a large negative gradient there, so the field belongs to Q

The number of rabbits \(P\) (in hundreds) on an island after \(t\) years satisfies \(\dfrac{dP}{dt}=0.1P(6-P)\). The slope field is shown.

1. (a) Show that \(P=0\) and \(P=6\) are constant solutions.

   • P = 0: 0.1 × 0 × 6 = 0
   • P = 6: 0.1 × 6 × 0 = 0
   • So dP/dt = 0 on both lines

2. (b) Sketch the solution curve through \((0, 1)\) and estimate the time when \(P=3\).

   • Start at (0, 1) and follow the segments: the curve rises, steepest near P = 3, and flattens near P = 6
   • Read across at P = 3: t ≈ 2.7 years

3. (c) Sketch the solution curve through \((0, 7)\). Explain why it never goes below \(P=6\).

   • At (0, 7) the gradient is 0.1 × 7 × (−1) = −0.7, so the curve falls
   • The line P = 6 is itself a solution and curves cannot cross, so the curve stays above it