Unit 4 · Topic 3.3 · Rates of change and differential equations

Slope fields

A slope field turns a differential equation into a picture. You don't need to solve anything to read one, which is why slope field questions are usually quick marks. This guide is mostly figures: read each one, then try the explorer until you can predict the solution curve before it appears. Tick off each skill once you can do it without help.

Skills in this topic

Only tick a skill when you could do it in an exam.

Find the gradient at a point by substituting both coordinates into the differential equation. Part 1

Sketch the solution curve through a given point, following the segments in both directions. Part 1

Sketch the segments of a slope field from its differential equation. Part 2

Match a slope field to its differential equation using the axes, the quadrants, and the rows and columns. Part 2

Solve the differential equation and check your solution against the slope field. Part 2

Part 1

Reading a slope field

A slope field, also called a direction field, shows a first-order differential equation as a grid of short segments. Each segment has the gradient that the differential equation gives at that point.

To find the gradient at a point, substitute both coordinates into the differential equation. For $\frac{dy}{dx}=\frac{4-y}{2}$ at $(-4,-2)$:

$$\frac{dy}{dx}=\frac{4-(-2)}{2}=3$$

This equation has no $x$ in it, so every point at height $y=-2$ has a gradient of $3$.

Solution curves

A solution curve follows the segments. Wherever it goes, its tangent lines up with the segments nearby. Through a given point there is normally only one solution curve, so a starting point picks out one curve, as an initial condition does in Topic 3.2.

Sketch the curve in both directions from the starting point. If a whole row of segments is flat, that line is a solution too, and nearby curves get closer and closer to it without crossing it.

Try it

Choose a differential equation, then tap the field or use the sliders to move the starting point. Before you look at the orange curve, picture where it should go.

Start Start

The row of flat segments at $y=4$ is a solution on its own. Every other curve gets closer to it.

Every curve is the same parabola, moved up or down.

The curve stops at the $y$-axis, because the gradient is undefined when $x=0$.

Each curve is half of a circle. It stops at the $x$-axis, where the segments are vertical.

This is the field from 2021 Paper 1, Question 9.

The gradient is undefined here, so there is no solution curve through this point. Move the starting point off the axis.

Practise

Exam questions: reading slope fields

2021 Paper 1, Question 9

Multiple choice · Technology-free · [1 mark]

The slope field for the differential equation $\dfrac{dy}{dx}=y-x^2$ is shown.

The solution curve to the differential equation that passes through the point $(3,-5)$ also passes through point

Start at $(3,-5)$ and follow the segments to the left. The gradient there is $-5-9=-14$, so going left the curve climbs steeply. It passes well above $S$, dips slightly just left of the $y$-axis, then rises again through $P$.

To the right of $(3,-5)$, the curve keeps falling steeply, so it can't reach $Q$. $R$ sits on a different curve, below this one. You can check this in the explorer above with $\frac{dy}{dx}=y-x^2$.

Answer

(A)

2022 Paper 1, Question 14

Short response · Technology-free · [4 marks]

The slope field for the differential equation $\dfrac{dy}{dx}=\dfrac{-0.5(y-4)}{x}$, $x\ne0$ using $-6\le x\le6$ and $-6\le y\le6$ is shown.

a)Determine the value of the slope at point $A$.[2 marks]

b)Use the slope field to sketch the solution curve for $\dfrac{dy}{dx}=\dfrac{-0.5(y-4)}{x}$ given that when $x=-6$, $y=3.5$.[2 marks]

Part a)

$$A(-4,-2):\quad\frac{dy}{dx}=\frac{-0.5(-2-4)}{-4}$$

1 markCorrectly substitutes the coordinates of $A$ into the differential equation.

$$\frac{dy}{dx}=\frac{3}{-4}=-0.75$$

1 markDetermines a value of the slope at $A$.

Part b)

The curve starts at $(-6,3.5)$ with a small negative gradient, and falls slowly across the second quadrant.

1 markCorrectly shows a solution curve that starts at $(-6,3.5)$ and has a negative slope in the second quadrant.

As $x\to0^-$ the segments get steeper, so the curve plunges down into the third quadrant. The differential equation is undefined at $x=0$, so the curve never reaches the $y$-axis, and none of it is in the first or fourth quadrant.

1 markCorrectly shows a solution curve that is asymptotic in the third quadrant as $x\to0^-$, with no solution curve in the first or fourth quadrant.

Point $A$ is the only labelled point, so read its coordinates off the grid carefully. A common slip is $(-2,-4)$ in place of $(-4,-2)$. Solving the equation gives $y=4-\frac{\sqrt6}{2\sqrt{-x}}$ for $x\lt0$. You don't need it here, but it confirms that $y\to-\infty$ as $x\to0^-$.

Answer

a) $-0.75$

Part 2

Sketching and matching

Sketching the segments

Work out the gradient at each point and draw a short segment through the point with that gradient. When the differential equation only contains $y$, every point in a row has the same gradient, so work row by row. When it only contains $x$, work column by column.

For $\frac{dy}{dx}=\frac{4-y}{2}$:

$y$	$6$	$5$	$4$	$3$	$2$	$1$	$0$
$\frac{dy}{dx}$	$-1$	$-0.5$	$0$	$0.5$	$1$	$1.5$	$2$

Draw every segment about the same length, centred on its point. Its steepness comes from the gradient: one step across, then $m$ steps up, or down if $m$ is negative.

$m=0$: flat

$m=1$: up one for every one across

$m=-2$: down two for every one across

Patterns to look for

The variables in the differential equation leave a pattern in the field. The blue curves are solutions.

$\frac{dy}{dx}=\frac{x}{2}$ only contains $x$. Every column is the same, and the solutions are vertical shifts of each other.

$\frac{dy}{dx}=\frac{y}{2}$ only contains $y$. Every row is the same, and the solutions are horizontal shifts of each other. The flat row, $y=0$, is a solution on its own.

$\frac{dy}{dx}=-\frac{x}{y}$ contains both. The segments are flat on the $y$-axis and vertical on the $x$-axis, and the solutions are circles.

Matching a field to its equation

When you are given a slope field and asked which equation it shows, test each option against the picture.

1The $x$-axis: substitute $y=0$. Flat segments mean the gradient is $0$ there.
2The $y$-axis: substitute $x=0$. Vertical segments mean the gradient is undefined, so $x$ is in a denominator.
3The quadrants: decide whether the gradient is positive or negative in each one, and rule out options with the wrong signs.
4Rows and columns: if every row matches, the equation only contains $y$. If every column matches, it only contains $x$.

The idea

Four checks on one field

Practise

Exam questions: sketching and matching

2020 Paper 1, Question 7

Multiple choice · Technology-free · [1 mark]

The diagram shows a slope field.

The differential equation represented by the slope field is

On the $x$-axis the segments are flat, so the gradient is $0$ when $y=0$. All four options pass this check.

Above the $x$-axis the gradient is positive on both sides of the $y$-axis, and below it the gradient is negative on both sides. So the sign depends on $y$ only.

$$\frac{5y}{x^2}:\quad x^2\gt0,\ \text{so the sign of }\frac{dy}{dx}\text{ is the sign of }y$$

(A) and (B) change sign across the $y$-axis. (D) is never negative.

Answer

(C)

2024 Paper 1, Question 15

Short response · Technology-free · [6 marks]

A sketch of a partially completed slope field for the differential equation $\dfrac{dy}{dx}=1-y$ is shown.

a)Complete the slope field by sketching the slopes at the 10 points indicated.[2 marks]

b)Use the completed slope field to sketch the solution curve for $\dfrac{dy}{dx}=1-y$, given the point $(-1,0)$ lies on this curve.[1 mark]

The differential equation can be solved by rearranging it into the form $\dfrac{1}{1-y}\,\dfrac{dy}{dx}=1$.

c)Determine the equation of the solution curve sketched in Question 15b) by solving the differential equation, given the point $(-1,0)$ lies on this curve. Leave your answer in the form $y=f(x)$.[3 marks]

Part a)

At $y=1$, $\frac{dy}{dx}=1-1=0$, so the five segments are flat.

1 markCorrectly sketches the slope as $0$ for all five points where $y=1$.

At $y=2$, $\frac{dy}{dx}=-1$, so the segments go down one for every one across.

1 markCorrectly sketches the slope as $-1$ for all five points where $y=2$.

Part b)

Start at $(-1,0)$ and follow the segments both ways. To the right, the curve rises and flattens out towards $y=1$. To the left, it falls more and more steeply.

1 markSketches a solution curve that passes through $(-1,0)$ and is asymptotic to the line $y=1$.

Part c)

$$\int\frac{1}{1-y}\,dy=\int1\,dx\ \Rightarrow\ -\ln|1-y|=x+c$$

1 markCorrectly determines a general solution to the equation.

$$(-1,0):\quad-\ln(1)=-1+c\ \Rightarrow\ c=1$$

1 markDetermines an appropriate constant of integration.

$$\ln|1-y|=-x-1\ \Rightarrow\ 1-y=e^{-x-1}\ \Rightarrow\ y=1-e^{-x-1}$$

1 markExpresses the solution in the form $y=f(x)$.

The curve stays below $y=1$, so $1-y\gt0$ and the absolute value signs can go. Check the answer against your sketch: as $x\to\infty$, $e^{-x-1}\to0$ and $y\to1$. The guide also accepts $y=1-\frac{e^{-x}}{e}$.

Answer

c) $y=1-e^{-x-1}$

Game plan

Working with slope fields

For the whole topic on one page, see Game plan 7 · Rates of change and differential equations in the External Game plan.

Sketching a solution curve

1Mark the starting point clearly.
2Follow the segments to the right, then go back and follow them to the left.
3Bend the curve gradually between rows and columns of segments.
4Near a row of flat segments, approach it without crossing. Stop at any line where the equation is undefined.

Matching a field to an equation

1Check the $x$-axis by substituting $y=0$.
2Check the $y$-axis by substituting $x=0$.
3Check the sign of the gradient in each quadrant.
4Substitute one point to confirm your choice.

Mistakes I see a lot

sketching the solution curve in one direction only
letting the curve cross a row of flat segments
drawing the curve across the segments instead of along them
swapping $x$ and $y$ when substituting a point
drawing every segment with the same steepness
continuing the curve across a line where the equation is undefined

Practice

Practice questions

Ten questions of my own, worth 51 marks in total. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Check whether each one is Technology-free or Technology-active before you start.

Question 1 of 10

Completing a field; sketching and solving · Technology-free · Simple familiar · [6 marks]

A sketch of a partially completed slope field for the differential equation $\dfrac{dy}{dx}=2-y$ is shown.

a)Complete the slope field by sketching the slopes at the 10 points indicated.[2 marks]

b)Use the completed slope field to sketch the solution curve for $\dfrac{dy}{dx}=2-y$, given the point $(1,1)$ lies on this curve.[1 mark]

The differential equation can be solved by rearranging it into the form $\dfrac{1}{2-y}\,\dfrac{dy}{dx}=1$.

c)Determine the equation of the solution curve sketched in Question 1b) by solving the differential equation, given the point $(1,1)$ lies on this curve. Leave your answer in the form $y=f(x)$.[3 marks]

Part a)

At every point where $y=2$: $\frac{dy}{dx}=2-2=0$ (horizontal).

At every point where $y=3$: $\frac{dy}{dx}=2-3=-1$.

1 markCorrectly sketches the slope as 0 for all five points where $y=2$.

1 markCorrectly sketches the slope as $-1$ for all five points where $y=3$.

Part b)

The solution curve is the blue curve in the figure above.

1 markSketches a solution curve that passes through $(1,1)$ and is asymptotic to the line $y=2$.

Part c)

$\displaystyle \int\frac{1}{2-y}\,dy=\int1\,dx\Rightarrow-\ln|2-y|=x+c$

1 markCorrectly determines a general solution to the equation.

At $(1,1)$: $-\ln(1)=1+c\Rightarrow c=-1$

1 markDetermines an appropriate constant of integration.

$\ln(2-y)=1-x\Rightarrow y=2-e^{1-x}$ ($y<2$ on this curve, so $|2-y|=2-y$)

1 markExpresses solution in the form $y=f(x)$.

Dropping the negative sign when you integrate $\frac{1}{2-y}$ gives $y=2-e^{x-1}$. It still passes through $(1,1)$, but it falls as $x$ increases, which doesn't match your sketch in part b). That loses the first mark of part c), but the other marks can still follow through.

Answer

a) $0$ on $y=2$, $-1$ on $y=3$c) $y=2-e^{1-x}$

Question 2 of 10

Reading a slope; a curve with an asymptote · Technology-free · Simple familiar · [4 marks]

The slope field for the differential equation $\dfrac{dy}{dx}=\dfrac{-(y-2)}{2x}$, $x\neq0$, using $-6\le x\le6$ and $-6\le y\le6$ is shown.

a)Determine the value of the slope at point $A$.[2 marks]

b)Use the slope field to sketch the solution curve for $\dfrac{dy}{dx}=\dfrac{-(y-2)}{2x}$ given that when $x=-6$, $y=1$.[2 marks]

Part a)

$\displaystyle \frac{dy}{dx}=\frac{-(-1-2)}{2(4)}$

1 markCorrectly substitutes the coordinates of $A$ into the differential equation.

$\displaystyle =\frac38$

1 markDetermines a value of the slope at $A$.

Part b)

(Exact curve: $y=2-\sqrt{\tfrac{6}{-x}}$, crossing the $x$-axis at $x=-1.5$.)

1 markCorrectly shows a solution curve that starts at $(-6,1)$ and has a negative slope in the 2nd quadrant.

1 markCorrectly shows a solution curve with an asymptotic nature in the 3rd quadrant as $x\to0^-$ and there is no solution curve in the 1st or 4th quadrant.

Leaving out the negative sign in part a) gives $-\frac38$ and only the first mark. In part b), stop the curve at the $y$-axis. Carrying it on into $x\gt0$ loses the second mark.

Answer

a) $\dfrac38$

Question 3 of 10

Matching a field to its equation · Technology-free · Simple familiar · [4 marks]

The diagram shows a slope field.

The slope field corresponds to one of the following differential equations.

(i)$\dfrac{dy}{dx}=\dfrac{x}{2y}$

(ii)$\dfrac{dy}{dx}=\dfrac{x^2}{2y}$

(iii)$\dfrac{dy}{dx}=\dfrac{x}{2y^2}$

(iv)$\dfrac{dy}{dx}=\dfrac{x^2}{2y^2}$

a)State which of the differential equations (i), (ii), (iii) or (iv) corresponds to the slope field.[1 mark]

b)Justify your choice in Question 3a) by referring to features of the slope field.[2 marks]

c)Use your result from Question 3a) to determine the slope of the field at the point $(-2,-1)$.[1 mark]

Part a)

(ii) $\displaystyle \frac{dy}{dx}=\frac{x^2}{2y}$

1 markStates the correct differential equation.

Part b)

In the 1st and 2nd quadrants all slopes are positive, so the sign of the slope does not change when $x$ changes sign. This rules out (i) and (iii), which give negative slopes in the 2nd quadrant.

1 markUses the sign of the slopes for $x<0$ and $x>0$ to eliminate (i) and (iii).

Below the $x$-axis all slopes are negative, so the sign of the slope changes when $y$ changes sign. This rules out (iv), which gives positive slopes everywhere.

1 markUses the sign of the slopes for $y<0$ to eliminate (iv).

Part c)

$\displaystyle \frac{dy}{dx}=\frac{(-2)^2}{2(-1)}=-2$

1 markDetermines the slope at $(-2,-1)$.

All four options have flat segments on the $y$-axis, so that feature can't justify your choice. Use the sign of the slopes in each quadrant. If you chose the wrong equation in part a), part c) can still get the mark, e.g. (i) gives $1$.

Answer

a) (ii)c) $-2$

Question 4 of 10

Choosing a field; equilibrium solutions · Technology-free · Simple familiar · [5 marks]

Three slope fields, A, B and C, are shown.

Slope field A

Slope field B

Slope field C

a)State which of the slope fields A, B or C corresponds to the differential equation $\dfrac{dy}{dx}=y(2-y)$. Justify your choice.[2 marks]

b)State the equations of the two equilibrium solutions of $\dfrac{dy}{dx}=y(2-y)$.[1 mark]

c)On the slope field you chose in Question 4a), sketch the solution curve that passes through the point $(0,1)$.[1 mark]

d)Describe the behaviour of the solution curve that passes through the point $(0,-0.5)$ as $x$ increases.[1 mark]

Part a)

$\frac{dy}{dx}=y(2-y)=0$ when $y=0$ or $y=2$ for every $x$, so the field has whole rows of horizontal segments along $y=0$ and $y=2$. Field A has columns of horizontal segments (along $x=0$ and $x=2$), so it is not A.

1 markCorrectly uses the location of the horizontal segments to eliminate field A.

At $(0,1)$: $\frac{dy}{dx}=1(2-1)=1>0$. Field C has positive slopes between $y=0$ and $y=2$; field B has negative slopes there. The field is C.

1 markCorrectly uses the sign of the slope between $y=0$ and $y=2$ to select field C.

Part b)

$y=0$ and $y=2$

1 markStates both equilibrium solutions.

Part c)

1 markSketches an increasing curve through $(0,1)$ that is asymptotic to $y=2$ as $x$ increases and to $y=0$ as $x$ decreases.

Part d)

Below $y=0$ the slopes are negative and become steeper as $y$ decreases, so the curve moves away from $y=0$ and $y$ decreases without bound.

1 markDescribes $y$ decreasing without bound (moving away from the equilibrium $y=0$).

In part d), saying the curve approaches $y=0$ or $y=2$ doesn't get the mark. Below $y=0$ the slopes get steeper as you go down, so the curve moves away from $y=0$. Part c) is marked on whichever field you chose in part a).

Answer

a) Cb) $y=0$ and $y=2$d) $y$ decreases without bound

Question 5 of 10

A field with vertical segments · Technology-free · Simple familiar · [5 marks]

A sketch of a partially completed slope field for the differential equation $\dfrac{dy}{dx}=-\dfrac{x}{y}$ is shown. Where $y=0$ and $x\neq0$, the slope is undefined and the segments are vertical.

a)Complete the slope field by sketching the slopes at the 8 points indicated.[2 marks]

b)Use the completed slope field to sketch the solution curve that passes through the point $(0,2)$.[1 mark]

c)Solve the differential equation, using separation of variables, to determine the equation of the solution curve sketched in Question 5b). Express your answer in the form $y=f(x)$.[2 marks]

Part a)

$x>0$: slope $-1$ at $(1,1)$ and $(2,2)$; $-2$ at $(2,1)$; $-\tfrac12$ at $(1,2)$.

$x<0$: slope $1$ at $(-1,1)$ and $(-2,2)$; $2$ at $(-2,1)$; $\tfrac12$ at $(-1,2)$.

1 markCorrectly sketches the slopes at the four points where $x>0$.

1 markCorrectly sketches the slopes at the four points where $x<0$.

Part b)

The blue curve in the figure above is an arc through $(0,2)$, horizontal at $(0,2)$, becoming vertical as it approaches $(\pm2,0)$, with $y>0$.

1 markSketches a solution curve through $(0,2)$ that follows the field and approaches $(-2,0)$ and $(2,0)$ with $y>0$.

Part c)

$\displaystyle \int y\,dy=-\int x\,dx\Rightarrow\frac{y^2}{2}=-\frac{x^2}{2}+c$

1 markCorrectly determines a general solution to the equation.

At $(0,2)$: $c=2\Rightarrow x^2+y^2=4$

The curve passes through $(0,2)$, so $y>0$: $y=\sqrt{4-x^2}$

1 markDetermines the particular solution in the form $y=f(x)$, choosing the positive root.

Writing $y=\pm\sqrt{4-x^2}$ loses the second mark of part c). The curve passes through $(0,2)$, so take the positive root. In part b), draw only the top half of the circle, because the field is undefined on $y=0$. A full circle loses that mark.

Answer

c) $y=\sqrt{4-x^2}$

Question 6 of 10

A logistic model from its field · Technology-active · Simple familiar · [5 marks]

The number of fish, $P$, in a lake $t$ years after the lake is stocked is modelled by the differential equation

$\frac{dP}{dt}=0.2P\left(1-\frac{P}{400}\right)$

for $t\ge0.$.

The slope field for this differential equation is shown.

a)State the equations of the equilibrium solutions of the differential equation.[1 mark]

b)On the slope field, sketch the solution curves for $P(0)=100$ and $P(0)=550$.[2 marks]

The solution curve for $P(0)=100$ is $P(t)=\dfrac{400}{1+3e^{-0.2t}}$.

c)Determine the time at which the number of fish is increasing most rapidly. Give your answer rounded to two decimal places.[2 marks]

Part a)

$P=0$ and $P=400$

1 markStates both equilibrium solutions.

Part b)

1 markSketches the curve for $P(0)=100$ increasing, steepest near $P=200$, and asymptotic to $P=400$.

1 markSketches the curve for $P(0)=550$ decreasing and asymptotic to $P=400$.

Part c)

$\frac{dP}{dt}=0.2P-0.0005P^2$ is a quadratic in $P$ with maximum at $P=\dfrac{0.2}{2(0.0005)}=200$ (the steepest segments lie along $P=200$).

1 markDetermines the value of $P$ at which $\frac{dP}{dt}$ is greatest.

$\displaystyle \frac{400}{1+3e^{-0.2t}}=200\Rightarrow e^{-0.2t}=\frac13\Rightarrow t=5\ln(3)\approx5.49$ years

1 markDetermines the time, rounded to two decimal places.

Setting $P''(t)=0$ also works in part c). The fastest growth is at half of 400, which is $P=200$. Solving $P(t)=300$ instead gets neither mark.

Answer

a) $P=0$ and $P=400$c) $t=5\ln(3)\approx5.49$ years

Question 7 of 10

A field with matching columns; a local maximum · Technology-free · Simple familiar · [5 marks]

The slope field for a differential equation of the form $\dfrac{dy}{dx}=x^2-k$, where $k$ is a constant, is shown.

a)Explain how the slope field shows that $\dfrac{dy}{dx}$ does not depend on $y$.[1 mark]

b)Use the slope field to determine the value of $k$.[1 mark]

c)Determine the equation of the solution curve that passes through the point $(0,1)$.[2 marks]

d)Use your result from Question 7c) to determine the coordinates of the local maximum of this solution curve.[1 mark]

Part a)

In each vertical column (fixed $x$) every segment is parallel, so the slope is the same for every value of $y$.

1 markExplains using the parallel segments in each column.

Part b)

The segments are horizontal along $x=-2$ and $x=2$, so $2^2-k=0\Rightarrow k=4$.

1 markDetermines the value of $k$ from the field.

Part c)

$\displaystyle y=\int\left(x^2-4\right)dx=\frac{x^3}{3}-4x+c$

1 markCorrectly determines a general solution.

At $(0,1)$: $c=1$, so $y=\dfrac{x^3}{3}-4x+1$

1 markDetermines the particular solution.

Part d)

$\frac{dy}{dx}=0$ at $x=\pm2$; the slopes change from positive to negative at $x=-2$.

Local maximum $\left(-2,\tfrac{19}{3}\right)$

1 markDetermines the coordinates of the local maximum.

$\left(2,-\frac{13}{3}\right)$ is the local minimum, so it doesn't get the mark in part d). If your value of $k$ was wrong, parts c) and d) can still get follow-through marks.

Answer

b) $k=4$c) $y=\dfrac{x^3}{3}-4x+1$d) $\left(-2,\dfrac{19}{3}\right)$

Question 8 of 10

Finding the equation from the field · Technology-free · Simple familiar · [5 marks]

The slope field for the differential equation $\dfrac{dy}{dx}=ax+by$, where $a$ and $b$ are constants, is shown.

a)Use the slope field to state the equation of the line through the origin on which every segment is horizontal.[1 mark]

b)The slope of the field at the point $(0,1)$ is $-1$. Use this information and your result from Question 8a) to determine the values of $a$ and $b$.[2 marks]

c)Show that $y=2x-2$ is a solution of the differential equation.[1 mark]

d)Use the slope field and your result from Question 8c) to describe the behaviour of the solution curve that passes through the point $(0,2)$ as $x$ increases.[1 mark]

Part a)

The segments at $(0,0)$, $(1,2)$ and $(-1,-2)$ are horizontal: $y=2x$

1 markStates the equation of the line.

Part b)

On $y=2x$: $ax+b(2x)=0$ for all $x\Rightarrow a=-2b$

At $(0,1)$: $a(0)+b(1)=-1\Rightarrow b=-1$

1 markForms two equations in $a$ and $b$.

$a=2$, $b=-1$, so $\frac{dy}{dx}=2x-y$

1 markDetermines the values of $a$ and $b$.

Part c)

$y=2x-2\Rightarrow\frac{dy}{dx}=2$

$2x-y=2x-(2x-2)=2=\frac{dy}{dx}$, so $y=2x-2$ is a solution.

1 markCorrectly shows the mathematical reasoning needed to complete the proof.

Part d)

The segments on both sides point towards the line $y=2x-2$, so the curve through $(0,2)$ approaches the line $y=2x-2$ (from above) as $x$ increases.

1 markDescribes the curve approaching the line $y=2x-2$.

You don't need the exact solution in part d), although it is $y=2x-2+4e^{-x}$. Saying the curve approaches $y=2x$ doesn't get the mark. That is the line of flat segments, and it isn't a solution.

Answer

a) $y=2x$b) $a=2$, $b=-1$d) it approaches the line $y=2x-2$

Question 9 of 10

From the field to a particular solution · Technology-free · Complex familiar · [6 marks]

The slope field for a differential equation of the form $\dfrac{dy}{dx}=\dfrac{k(x-a)}{y}$, $y\neq0$, where $k$ and $a$ are constants, is shown. Where $y=0$, the segments are vertical.

The slope of the field at the point $(0,2)$ is $2$.

Determine the equation of the solution curve that passes through the point $(0,2)$. Express your answer in the form $y=f(x)$.

The segments are horizontal along $x=2$, so $a=2$.

1 markDetermines the value of $a$ from the field.

At $(0,2)$: $\displaystyle \frac{k(0-2)}{2}=2\Rightarrow k=-2$, so $\frac{dy}{dx}=\dfrac{-2(x-2)}{y}$

1 markDetermines the value of $k$.

$\displaystyle \int y\,dy=\int(4-2x)\,dx\Rightarrow\frac{y^2}{2}=4x-x^2+c$

1 markCorrectly determines a general solution to the equation.

At $(0,2)$: $2=c\Rightarrow y^2=4+8x-2x^2$

1 markDetermines an appropriate constant of integration.

The curve passes through $(0,2)$ and cannot cross $y=0$, so $y=\sqrt{4+8x-2x^2}$

1 markExpresses the solution in the form $y=f(x)$, choosing the positive root.

$y=\sqrt{4+8x-2x^2}$

1 markShows logical organisation, having fully attempted the question.

A sign slip that gives $k=2$ leads to $y^2=2x^2-8x+4$. That loses the second mark, but the rest can still follow through. You don't need to state the domain, $2-\sqrt6\le x\le2+\sqrt6$.

Answer

$y=\sqrt{4+8x-2x^2}$

Question 10 of 10

Newton's law of cooling; evaluating a claim · Technology-active · Complex familiar · [6 marks]

On a hot day, a drink is left outdoors to cool. The temperature of the drink, $T^{\circ}$C, $t$ minutes after it is poured is modelled by a differential equation of the form

$\frac{dT}{dt}=-k\left(T-T_s\right)$

where $k>0,$, and $T_s^{\circ}$C is the constant temperature of the surroundings. The slope field for this differential equation is shown. The drink is poured at $90^{\circ}$C, and the slope of the field at the point $(0,90)$ is $-6$.

A student notes that the drink is initially cooling at $6^{\circ}$C per minute and claims that the drink will be below $50^{\circ}$C within 10 minutes of being poured.

Evaluate the reasonableness of the student's claim. Justify your decision using mathematical reasoning.

The segments are horizontal along $T=30$, so $T_s=30$.

At $(0,90)$: $-k(90-30)=-6\Rightarrow k=0.1$

1 markDetermines $T_s$ and $k$ from the field and the given slope.

$\displaystyle \int\frac{1}{T-30}\,dT=-\int0.1\,dt\Rightarrow\ln|T-30|=-0.1t+c$

1 markCorrectly determines a general solution to the equation.

At $(0,90)$: $c=\ln(60)$, so $T=30+60e^{-0.1t}$

1 markDetermines the particular solution.

$T(10)=30+60e^{-1}\approx52.07^{\circ}$C

(equivalently, $T=50$ when $t=10\ln(3)\approx10.99$ minutes)

1 markDetermines the temperature after 10 minutes, or the time to reach $50^{\circ}$C.

After 10 minutes the drink is still about $52.07^{\circ}$C, above $50^{\circ}$C. The slopes flatten as $T$ approaches 30, so the drink does not keep cooling at $6^{\circ}$C per minute.

1 markUses mathematical reasoning to justify the decision.

The student's claim is not reasonable.

1 markStates an appropriate evaluation of the reasonableness of the claim.

If you assume the drink keeps cooling at $6^{\circ}$C per minute, you get $90-6(10)=30^{\circ}$C after 10 minutes. That linear reasoning only gets the first mark. The slopes flatten as $T$ gets close to 30, so the rate drops. If your $k$ or $T_s$ was wrong, the last three marks can still follow through.

Answer

$T(10)\approx52.07^{\circ}$C, which is above $50^{\circ}$Cthe claim is not reasonable

Keep practising

Ready for something harder?

Every past exam question on slope fields has been simple familiar. The harder questions in this part of the course solve the differential equation, as in 2024 Paper 1, Question 15c). Those are in Topic 3.2, Differential equations, and in the complex unfamiliar collection.

All complex unfamiliar questions

Past questions and marking-guide steps are from the 2020–2024 QCAA Specialist Mathematics external assessments, © State of Queensland (QCAA) 2020–2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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