Slope fields and solution curves Problem set B One question at a time

For \(\dfrac{dy}{dx}=y^2-x\), complete a table of the gradient at \(x=0, 1, 2\) 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: 1, 0, −1
   • y = 0: 0, −1, −2
   • y = 1: 1, 0, −1
   • y = 2: 4, 3, 2

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

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

Sketch the slope field of \(\dfrac{dy}{dx}=2x+y\) 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: −2, 0, 2, 4, 6
   • y = 1: −3, −1, 1, 3, 5
   • y = 0: −4, −2, 0, 2, 4
   • y = −1: −5, −3, −1, 1, 3
   • y = −2: −6, −4, −2, 0, 2
   • 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 (−1, 2), (0, 0), (1, −2)

Use your calculator to plot the slope field of \(\dfrac{dy}{dx}=8-2y\) for \(0\le x\le 4\) and \(0\le y\le 8\).

1. (a) Sketch the slope field you see. Describe the segments above, on and below the line \(y=4\).

   • From the calculator: dy/dx = 8 − 2y, window 0 to 4 and 0 to 8
   • At y = 1 the gradient is 6 (rising). At y = 4 it is 0. At y = 7 it is −6 (falling)

2. (b) Add the solution curve that starts at \((0, 7)\). Describe what happens to it as \(x\) increases.

   • From the calculator: add the solution curve through (0, 7)
   • The gradient 8 − 2y shrinks in size as y gets near 4, so the curve flattens and approaches y = 4 from above

3. (c) Write down the constant solution.

   • dy/dx = 0 when 8 − 2y = 0
   • The constant solution is y = 4

The slope field of \(\dfrac{dy}{dx}=\dfrac{x+y}{2}\) is shown, with the points P\((−2, 1)\), Q\((0, −2)\) and R\((2, −3)\).

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

   • P: gradient \(\frac{-2+1}{2}\) = −0.5, so the curve starts by falling slightly, then climbs as x increases
   • 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 \(\frac{2+(-3)}{2}\) = −0.5, so the curve starts by falling, but the gradient grows towards 0 and then positive, so it flattens and 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 bacteria \(V\) (in millions) in a culture after \(t\) hours satisfies \(\dfrac{dV}{dt}=0.4V(12-V)\).

1. (a) Find the two constant solutions.

   • Constant solution: dV/dt = 0
   • 0.4V(12 − V) = 0, so V = 0 or V = 12

2. (b) Find the gradient when \(V=3\) and when \(V=15\).

   • V = 3: 0.4 × 3 × (12 − 3) = 10.8
   • V = 15: 0.4 × 15 × (12 − 15) = −18

3. (c) Describe what happens to a culture of 15 million.

   • At V = 15 the gradient is −18, negative
   • So V falls. It cannot cross the constant solution V = 12

A tank holds water with a salt concentration of \(5\text{ g/L}\). Salty water flows in, and the concentration \(c\) after \(t\) minutes satisfies \(\dfrac{dc}{dt}=0.4(25-c)\). The slope field is shown.

1. (a) Sketch the solution curve through \((0, 5)\).

   • Start at (0, 5), where the gradient is 0.4 × 20 = 8
   • Follow the segments: the curve rises and flattens as c gets near 25

2. (b) Estimate \(c\) when \(t=3\).

   • Go along to t = 3 and read up to the curve
   • c ≈ 19 g/L

3. (c) State the concentration the tank approaches.

   • dc/dt = 0 when c = 25
   • The curve flattens towards the constant solution c = 25

The number \(R\) (in hundreds) of rabbits on a farm after \(t\) years satisfies \(\dfrac{dR}{dt}=0.25R\left(1-\dfrac{R}{8}\right)\).

1. (a) Find the two constant solutions.

   • \(\dfrac{dR}{dt}=0\Rightarrow 0.25R\left(1-\dfrac{R}{8}\right)=0\)
   • \(R=0\) or \(R=8\)

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

   • \(R=2\): rate \(=0.375\gt0\) \(R=10\): rate \(=−0.625\lt0\)
   • Above 0 the population rises towards 8. Above 8 it falls towards 8
   • So \(R=8\) is stable and \(R=0\) is unstable

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

   • The rate is a quadratic in \(R\) with zeros at 0 and 8
   • Its greatest value is at \(R=4\)
   • Rate \(=0.25\times4\times\left(1-\dfrac{4}{8}\right)=0.5\)

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

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

   • P: 2 + 0² = 2
   • Q: 2² − 0 = 4
   • R: 0² − 2 = −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 gently
   • Only R has a negative gradient there, so the field belongs to R

The number of users \(V\) (in millions) of a new app after \(t\) months satisfies \(\dfrac{dV}{dt}=0.2V(4-V)\). The slope field is shown.

1. (a) Show that \(V=0\) and \(V=4\) are constant solutions.

   • V = 0: 0.2 × 0 × 4 = 0
   • V = 4: 0.2 × 4 × 0 = 0
   • So dV/dt = 0 on both lines

2. (b) Sketch the solution curve through \((0, 0.5)\) and estimate the time when \(V=2\).

   • Start at (0, 0.5) and follow the segments: the curve rises, steepest near V = 2, and flattens near V = 4
   • Read across at V = 2: t ≈ 2.4 months

3. (c) Sketch the solution curve through \((0, 5)\). Explain why it never goes below \(V=4\).

   • At (0, 5) the gradient is 0.2 × 5 × (−1) = −1, so the curve falls
   • The line V = 4 is itself a solution and curves cannot cross, so the curve stays above it