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14 Calculus

Before you start

The gradient of a straight line, and its equation in the form $y - y_{1} = m (x - x_{1})$ .
Perpendicular lines: their gradients multiply to $- 1$ , so a line perpendicular to gradient $m$ has gradient $- \frac{1}{m}$ .
The laws of indices, and writing roots and fractions as powers: $\sqrt{x} = x^{\frac{1}{2}}$ , $\frac{1}{x^{3}} = x^{- 3}$ .
Radians, and the exact values of $\sin$ , $\cos$ and $\tan$ at $\frac{π}{6}$ , $\frac{π}{4}$ and $\frac{π}{3}$ .
$e^{x}$ and $\ln x$ as inverse functions, and the laws of logarithms (Topic 6).
Composite functions such as $f (g (x))$ , solving quadratics by factorising, and the area and circumference of a circle and the volume of a cylinder.

Syllabus objectives covered (15)

Understand the idea of a derived function.
Use the notations f′(x), f″(x), dy/dx, d²y/dx² [= d/dx(dy/dx)] δx, δx → 0, δy/δx.
Know and use the derivatives of the standard functions xⁿ (for any rational n), sin x, cos x, tan x, eˣ, ln x.
Differentiate products and quotients of functions.
Use differentiation to find gradients, tangents and normals.
Use differentiation to find stationary points.
Apply differentiation to connected rates of change, small increments and approximations.
Apply differentiation to practical problems involving maxima and minima.
Use the first and second derivative tests to discriminate between maxima and minima.
Understand integration as the reverse process of differentiation.
Integrate sums of terms in powers of x, including 1/x and 1/(ax + b).
Integrate functions of the form: • (ax + b)ⁿ for any rational n • sin (ax + b) • cos (ax + b) • sec² (ax + b) • eᵃˣ⁺ᵇ
Evaluate definite integrals and apply integration to the evaluation of plane areas.
Apply differentiation and integration to kinematics problems that involve displacement, velocity and acceleration of a particle moving in a straight line with variable or constant acceleration.
Make use of the relationships in 14.14 to draw and use the following graphs: • displacement–time • distance–time • velocity–time • speed–time • acceleration–time.

Part 1 of 12

The derived function: a gradient at every point

A straight line has one gradient everywhere. A curve does not: $y = x^{2}$ is nearly flat near the origin and steep further out. The gradient of a curve at a point is the gradient of the tangent there, the straight line that just touches the curve at that point.
To measure it, start with a chord. Take a point $P$ on the curve and a second point $Q$ a little further along. Call the small step in $x$ from $P$ to $Q$ $δ x$ , and the matching step in $y$ $δ y$ . The chord $P Q$ has gradient $\frac{δ y}{δ x}$ . Read $δ x$ as 'delta $x$ ': it is one symbol, not $δ$ multiplied by $x$ .
Now slide $Q$ along the curve towards $P$ , so that $δ x \to 0$ (' $δ x$ tends to zero'). The chord turns until it lies along the tangent at $P$ . The value that $\frac{δ y}{δ x}$ gets closer and closer to is the gradient of the curve at $P$ , and it is written $\frac{d y}{d x}$ .
$\frac{d y}{d x} = \lim_{δ x \to 0} \frac{δ y}{δ x}$
graph diagram — not drawn yet
Axes for 0 ≤ x ≤ 4.5 and 0 ≤ y ≤ 20 with the curve y = x² drawn through (0, 0), (1, 1), (2, 4), (3, 9), (4, 16) and (4.4, 19.36). Mark P(3, 9) and three points Q at (4, 16), (3.5, 12.25) and (3.2, 10.24), each joined to P by a thin straight chord. Draw the tangent at P, y = 6x − 9, as a thicker line from (1.5, 0) to (4.5, 18), labelled 'tangent, gradient 6'. On the chord to (4, 16), mark δx = 1 as a horizontal step from P and δy = 7 as a vertical step up to Q.
As $Q$ moves towards $P$ the chord gradients $7$ , $6.5$ and $6.2$ fall towards the tangent gradient $6$ .
See it with numbers on $y = x^{2}$ at $P (3, 9)$ . With $δ x = 0.1$ , $Q$ is $(3.1, 9.61)$ and the chord gradient is $\frac{0.61}{0.1} = 6.1$ . With $δ x = 0.01$ it is $\frac{0.0601}{0.01} = 6.01$ . The chord gradients close in on $6$ , so the gradient of the curve at $(3, 9)$ is $6$ .
Do this at every $x$ and you get a new function that gives the gradient anywhere on the curve. This is the derived function, or derivative. For $y = x^{2}$ it is $2 x$ : at $x = 3$ that gives $6$ , which matches the chords. You will not be asked to work a derivative out from chords (that is 'first principles', and it is not required). You need the idea, and then the rules in the next sections.
If the curve is $y = f (x)$ , the derived function has two names, and they mean the same thing.
$\frac{d y}{d x} = f^{'} (x)$
Differentiate again and you get the second derivative. It tells you how fast the gradient itself is changing. Read $\frac{d^{2} y}{d x^{2}}$ as 'd two $y$ by d $x$ squared'.
$f^{''} (x) = \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x})$
For $y = x^{3}$ : $\frac{d y}{d x} = 3 x^{2}$ and $\frac{d^{2} y}{d x^{2}} = 6 x$ . Notice that $\frac{d^{2} y}{d x^{2}}$ is not ${(\frac{d y}{d x})}^{2}$ , which here would be $9 x^{4}$ .

Part 2 of 12

The standard derivatives

No calculus formulas are printed on the List of formulas, so every result in this section must be recalled. There are only six, and each one is used everywhere in this topic.
For any rational power $n$ (positive, negative or a fraction), multiply by the power, then reduce the power by one.
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$
The rule only works on a power of $x$ , so rewrite roots and fractions first: $\sqrt{x} = x^{\frac{1}{2}}$ , $\frac{1}{x^{2}} = x^{- 2}$ , $\frac{1}{\sqrt{x}} = x^{- \frac{1}{2}}$ . A constant on its own differentiates to $0$ , because its graph is a horizontal line.
Constant multiples stay where they are, and a sum is differentiated term by term. For $y = 4 \sqrt{x} - \frac{6}{x^{2}}$ , first write $y = 4 x^{\frac{1}{2}} - 6 x^{- 2}$ .
$\frac{d y}{d x} = 2 x^{- \frac{1}{2}} + 12 x^{- 3} = \frac{2}{\sqrt{x}} + \frac{12}{x^{3}}$
Watch the second term: $- 6 \times (- 2) = + 12$ , and the power goes from $- 2$ down to $- 3$ , not up to $- 1$ .
The other five standard derivatives. The three trigonometric results are true only when $x$ is in radians, which is why calculus questions always use radians.
$\frac{d}{d x} (\sin x) = \cos x \frac{d}{d x} (\cos x) = - \sin x \frac{d}{d x} (\tan x) = \sec^{2} x \frac{d}{d x} (e^{x}) = e^{x} \frac{d}{d x} (\ln x) = \frac{1}{x}$
The minus sign belongs to cosine only: sine differentiates to plus cosine, and cosine to minus sine. And $e^{x}$ is its own derivative, so at every point on $y = e^{x}$ the gradient equals the $y$ -value.

Part 3 of 12

The chain rule for a function of a function

$y = {(3 x^{2} + 4)}^{\frac{1}{3}}$ is a composite function: an inner function $u = 3 x^{2} + 4$ inside an outer function $y = u^{\frac{1}{3}}$ . You cannot expand a power of $\frac{1}{3}$ , so you need the chain rule.
$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
Why it works: if $u$ changes $6$ times as fast as $x$ , and $y$ changes $2$ times as fast as $u$ , then $y$ changes $2 \times 6 = 12$ times as fast as $x$ . Rates along a chain multiply.
Apply it. $\frac{d y}{d u} = \frac{1}{3} u^{- \frac{2}{3}}$ and $\frac{d u}{d x} = 6 x$ . Multiply them and put $u$ back in terms of $x$ .
$\frac{d y}{d x} = \frac{1}{3} {(3 x^{2} + 4)}^{- \frac{2}{3}} \times 6 x = 2 x {(3 x^{2} + 4)}^{- \frac{2}{3}}$
With practice it is one line: differentiate the outside, leaving the inside untouched, then multiply by the derivative of the inside.
When the inside is linear, $a x + b$ , the derivative of the inside is just $a$ . So every standard result simply gains a factor $a$ .
$\frac{d}{d x} {(a x + b)}^{n} = a n {(a x + b)}^{n - 1} \frac{d}{d x} e^{a x + b} = a e^{a x + b} \frac{d}{d x} \ln (a x + b) = \frac{a}{a x + b}$
The same for the trigonometric functions.
$\frac{d}{d x} \sin (a x + b) = a \cos (a x + b) \frac{d}{d x} \cos (a x + b) = - a \sin (a x + b) \frac{d}{d x} \tan (a x + b) = a \sec^{2} (a x + b)$
For $\ln$ of any function, the chain rule gives 'the derivative of the inside over the inside': $\frac{d}{d x} \ln (x^{2} + 1) = \frac{2 x}{x^{2} + 1}$ . A useful check: $\frac{d}{d x} \ln 5 x = \frac{5}{5 x} = \frac{1}{x}$ . That agrees with the log law $\ln 5 x = \ln 5 + \ln x$ , because $\ln 5$ is a constant.
Powers of trigonometric functions are composites too. $\sin^{2} x$ means ${(\sin x)}^{2}$ , so the inside is $\sin x$ , and $\frac{d}{d x} \sin^{2} x = 2 \sin x \cos x$ .

Part 4 of 12

Products and quotients

The derivative of a product is not the product of the derivatives. For $y = u v$ , where $u$ and $v$ are both functions of $x$ :
$\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$
Write down $u$ , $v$ , $\frac{d u}{d x}$ and $\frac{d v}{d x}$ separately before you combine them. Four parts held in your head are easy to mix up, and written parts show your method.
For $y = x^{2} e^{3 x}$ : $u = x^{2}$ , $v = e^{3 x}$ , $\frac{d u}{d x} = 2 x$ , and $\frac{d v}{d x} = 3 e^{3 x}$ by the chain rule.
$\frac{d y}{d x} = 3 x^{2} e^{3 x} + 2 x e^{3 x} = x e^{3 x} (3 x + 2)$
Factorise the answer by taking out every common factor. A factorised derivative is easy to set equal to zero later, and in a Show that question you must reach the printed form exactly.
For $y = \frac{u}{v}$ , use the quotient rule.
$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$
The order in the numerator matters because of the minus sign: start with $v$ , the bottom. A numerator written the other way round gives the right size with the wrong sign.
For $y = \frac{\ln x}{x}$ : $u = \ln x$ , $v = x$ , $\frac{d u}{d x} = \frac{1}{x}$ and $\frac{d v}{d x} = 1$ .
$\frac{d y}{d x} = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^{2}} = \frac{1 - \ln x}{x^{2}}$
Any valid method is allowed. $\frac{\ln x}{x}$ is also the product $x^{- 1} \ln x$ , and the product rule gives $- x^{- 2} \ln x + x^{- 1} \cdot \frac{1}{x}$ , which is the same answer. Use whichever you make fewer slips with.

Part 5 of 12

Gradients, tangents and normals

The gradient of the curve $y = f (x)$ at $x = a$ is $f^{'} (a)$ . Differentiate first, then substitute. If you substitute first, the expression becomes a number, and a number differentiates to $0$ .
The tangent at the point $(a, f (a))$ is the straight line through that point with gradient $m = f^{'} (a)$ .
$y - f (a) = m (x - a)$
The normal is the line through the same point at right angles to the tangent. Perpendicular gradients multiply to $- 1$ , so the normal has gradient $- \frac{1}{m}$ .
Example: $y = x^{3} - 4 x + 1$ at $x = 2$ . The point is $y = 8 - 8 + 1 = 1$ , so $(2, 1)$ . Then $\frac{d y}{d x} = 3 x^{2} - 4$ , which is $8$ at $x = 2$ .
Tangent: $y - 1 = 8 (x - 2)$ , so $y = 8 x - 15$ . Normal: gradient $- \frac{1}{8}$ , so $y - 1 = - \frac{1}{8} (x - 2)$ . Multiply by $8$ to clear the fraction: $8 y - 8 = - x + 2$ , which is $x + 8 y = 10$ .
If the gradient is $0$ , the tangent is horizontal, $y = f (a)$ , and the normal is vertical, $x = a$ . Do not try to work out $- \frac{1}{0}$ .
Questions also run backwards: find the points where the gradient is $k$ . Solve $f^{'} (x) = k$ , then get each $y$ from the curve's equation, not from the derivative. On $y = x^{3} - 4 x + 1$ , gradient $8$ needs $3 x^{2} - 4 = 8$ , so $x = \pm 2$ , giving the points $(2, 1)$ and $(- 2, 1)$ .

Part 6 of 12

Stationary points: finding them and telling maximum from minimum

A stationary point is a point where the gradient is zero, so the tangent there is horizontal. To find one, solve $\frac{d y}{d x} = 0$ for $x$ , then put each $x$ into the equation of the curve to get its $y$ .
This syllabus deals with two kinds: a maximum, where the curve rises to the point and then falls, and a minimum, where it falls and then rises. Points of inflexion are not included. You must say which kind each point is, and justify it with one of two tests.
First derivative test. Find the sign of $\frac{d y}{d x}$ just to the left and just to the right of the point. Positive, then zero, then negative is a maximum: uphill, flat, downhill. Negative, then zero, then positive is a minimum.
Second derivative test. Find the value of $\frac{d^{2} y}{d x^{2}}$ at the point.
$\frac{d^{2} y}{d x^{2}} < 0 \Rightarrow maximum \frac{d^{2} y}{d x^{2}} > 0 \Rightarrow minimum$
Why those signs: at a maximum the gradient goes from positive, through $0$ , to negative, so the gradient is decreasing. A decreasing gradient has a negative derivative, and the derivative of the gradient is $\frac{d^{2} y}{d x^{2}}$ . At a minimum the gradient is increasing, so $\frac{d^{2} y}{d x^{2}}$ is positive.
If $\frac{d^{2} y}{d x^{2}} = 0$ at the point, the second derivative test gives no conclusion. Use the first derivative test instead.
Example: $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ . Then $\frac{d y}{d x} = 6 x^{2} - 6 x - 12 = 6 (x - 2) (x + 1)$ , which is zero at $x = 2$ and $x = - 1$ . From the curve's equation, the points are $(2, - 15)$ and $(- 1, 12)$ .
graph diagram — not drawn yet
Axes for −2.5 ≤ x ≤ 3.5 and −16 ≤ y ≤ 14 with the curve y = 2x³ − 3x² − 12x + 5 drawn smoothly through (−2.5, −15), (−2, 1), (−1, 12), (0, 5), (1, −8), (2, −15), (3, −4) and (3.5, 12). Draw a short horizontal tangent segment at (−1, 12), labelled 'maximum (−1, 12)', and another at (2, −15), labelled 'minimum (2, −15)'. Near the maximum write the gradient signs '+', '0', '−' from left to right; near the minimum write '−', '0', '+'.
At the maximum the gradient changes from positive to negative; at the minimum from negative to positive.
Now the nature. $\frac{d^{2} y}{d x^{2}} = 12 x - 6$ . At $x = 2$ it is $18 > 0$ , so $(2, - 15)$ is a minimum. At $x = - 1$ it is $- 18 < 0$ , so $(- 1, 12)$ is a maximum.
The syllabus expects full justification. Write the value, its sign and the conclusion together, as in ' $\frac{d^{2} y}{d x^{2}} = 18 > 0$ , so minimum'. The single word 'minimum' does not show how you know.

Part 7 of 12

Maxima and minima in practical problems

A practical problem asks for the largest or smallest value of a quantity: an area, a volume, a cost. The calculus is exactly the stationary-point method. The new work is in setting up the formula.
Step 1. Write the quantity as a formula. If it contains two unknown lengths, you cannot differentiate it yet.
Step 2. Use the fixed fact in the question (the length of fencing, the volume) to write one unknown in terms of the other. Substitute, so the quantity depends on one variable only.
Step 3. Differentiate, set the derivative equal to $0$ , and solve. Step 4. Show it is the maximum or minimum the question wants. Step 5. Answer the actual question, which may ask for the area rather than the $x$ that gives it.
Example: a rectangular pen is built against a long wall, so fencing is needed on three sides only, and $60$ m of fencing is available. Let each of the two sides at right angles to the wall be $x$ m. The side parallel to the wall is then $(60 - 2 x)$ m, so the area is
$A = x (60 - 2 x) = 60 x - 2 x^{2}$
$\frac{d A}{d x} = 60 - 4 x$ , which is $0$ when $x = 15$ . Then $\frac{d^{2} A}{d x^{2}} = - 4 < 0$ , so this is a maximum. The largest area is $15 \times 30 = 450$ m².
Check the answer is possible: the pen exists only for $0 < x < 30$ , and $x = 15$ is inside that range.

Part 8 of 12

Connected rates of change and small increments

$\frac{d A}{d t}$ is the rate at which $A$ changes as time $t$ passes. A positive rate means $A$ is increasing; a negative rate means it is decreasing.
When two quantities are linked by a formula, their rates are linked by the chain rule. If $A$ depends on $r$ , and $r$ changes with time:
$\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t}$
Example: the radius of a circle increases at $0.5$ cm/s. Since $A = π r^{2}$ , $\frac{d A}{d r} = 2 π r$ . When $r = 4$ , $\frac{d A}{d t} = 8 π \times 0.5 = 4 π$ , so the area increases at $4 π$ cm²/s.
Use the units to check the order. (cm² per cm) × (cm per s) = cm² per s. The quantity you do not want, $r$ , drops out, just as the $d r$ appears to cancel in the fractions.
If you are given $\frac{d V}{d t}$ and want $\frac{d r}{d t}$ , divide instead. Air is pumped into a spherical balloon at $100$ cm³/s. The volume of a sphere, $V = \frac{4}{3} π r^{3}$ , is on the List of formulas, and $\frac{d V}{d r} = 4 π r^{2}$ . When $r = 5$ , the radius grows at $\frac{1}{π}$ cm/s.
$\frac{d r}{d t} = \frac{d V}{d t} \div \frac{d V}{d r} = \frac{100}{4 π {(5)}^{2}} = \frac{1}{π}$
Small increments use the idea from the first section in reverse. For a small change $δ x$ , the chord gradient $\frac{δ y}{δ x}$ is close to $\frac{d y}{d x}$ . Rearranged, that estimates the change in $y$ .
$δ y \approx \frac{d y}{d x} \times δ x$
Example: $y = x^{3}$ and $x$ increases from $2$ to $2.01$ . $\frac{d y}{d x} = 3 x^{2} = 12$ at $x = 2$ , and $δ x = 0.01$ , so $δ y \approx 12 \times 0.01 = 0.12$ . The approximate new value is $y \approx 8 + 0.12 = 8.12$ . The exact value is ${2.01}^{3} = 8.120601$ , so the estimate is very close.
Evaluate $\frac{d y}{d x}$ at the starting value of $x$ , and take $δ x$ as the new value minus the old one, so a decrease makes $δ x$ negative. If the change is a letter, such as $p$ , the answer is an expression: here $δ y \approx 12 p$ .

Part 9 of 12

Integration: differentiation in reverse

Integration undoes differentiation. If $\frac{d y}{d x} = 2 x$ , then $y$ could be $x^{2}$ , or $x^{2} + 3$ , or $x^{2} - 7$ : each has derivative $2 x$ , because a constant differentiates to $0$ . So the answer carries an arbitrary constant $c$ , which stands for every one of those curves.
$\int 2 x d x = x^{2} + c$
An indefinite integral (one without limits) must always include $+ c$ . For powers, reverse the power rule: add one to the power, then divide by the new power. This works for every rational $n$ except $- 1$ .
$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c (n \neq - 1)$
Prepare the expression first. Rewrite roots and fractions as powers. Split a fraction with a single term underneath: $\frac{x^{2} + 3}{x} = x + \frac{3}{x}$ . Expand brackets: there is no product rule for integration, so $\int x (x + 1) d x$ must become $\int (x^{2} + x) d x$ .
Example. The middle term: $- 4 x^{- 3}$ becomes $\frac{- 4 x^{- 2}}{- 2} = 2 x^{- 2}$ . The last: dividing by $\frac{3}{2}$ is multiplying by $\frac{2}{3}$ .
$\int (6 x^{2} - \frac{4}{x^{3}} + \sqrt{x}) d x = \int (6 x^{2} - 4 x^{- 3} + x^{\frac{1}{2}}) d x = 2 x^{3} + 2 x^{- 2} + \frac{2}{3} x^{\frac{3}{2}} + c$
The power rule fails for $x^{- 1}$ , because it would divide by zero. Instead use $\frac{d}{d x} \ln x = \frac{1}{x}$ in reverse.
$\int \frac{1}{x} d x = \ln x + c (x > 0)$
When the inside is linear, $a x + b$ , reverse the chain-rule shortcut: integrate as if the inside were plain $x$ , then divide by $a$ . Differentiating multiplied by $a$ , so integrating divides by it.
$\int {(a x + b)}^{n} d x = \frac{{(a x + b)}^{n + 1}}{a (n + 1)} + c (n \neq - 1) \int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c \int \frac{1}{a x + b} d x = \frac{1}{a} \ln (a x + b) + c$
The trigonometric ones, again in radians. The sign trap: cosine differentiates to minus sine, so sine integrates to minus cosine.
$\int \cos (a x + b) d x = \frac{1}{a} \sin (a x + b) + c \int \sin (a x + b) d x = - \frac{1}{a} \cos (a x + b) + c \int \sec^{2} (a x + b) d x = \frac{1}{a} \tan (a x + b) + c$
Check any integral by differentiating your answer: you must get back exactly what you started with. $\int {(4 x + 1)}^{3} d x = \frac{{(4 x + 1)}^{4}}{16} + c$ , and $\frac{d}{d x} [\frac{{(4 x + 1)}^{4}}{16}] = \frac{4 \times 4 {(4 x + 1)}^{3}}{16} = {(4 x + 1)}^{3}$ . The shortcut works only for a linear inside: ${(x^{2} + 1)}^{3}$ must be expanded before you integrate it.
To find $c$ , use a point on the curve. If $\frac{d y}{d x} = 3 x^{2} - 2$ and the curve passes through $(1, 4)$ , then $y = x^{3} - 2 x + c$ , and $4 = 1 - 2 + c$ gives $c = 5$ . So $y = x^{3} - 2 x + 5$ .
Reversing a derivative you already know can integrate functions that none of these rules cover. Given $\frac{d}{d x} (x \ln x) = \ln x + 1$ , integrating both sides gives $\int (\ln x + 1) d x = x \ln x + c$ . Since $\int 1 d x = x$ , it follows that $\int \ln x d x = x \ln x - x + c$ .

Part 10 of 12

Definite integrals and plane areas

A definite integral has limits. Integrate, write the result in square brackets, then substitute the top limit and subtract the value at the bottom limit.
$\int_{a}^{b} f (x) d x = [F (x)]_{a}^{b} = F (b) - F (a)$
Leave out $+ c$ : it would appear in both $F (b)$ and $F (a)$ and cancel. For example, $\int_{1}^{4} 3 \sqrt{x} d x = [2 x^{\frac{3}{2}}]_{1}^{4} = 2 (8) - 2 (1) = 14$ , using $4^{\frac{3}{2}} = {(\sqrt{4})}^{3} = 8$ .
For a curve above the $x$ -axis between $x = a$ and $x = b$ , the area between the curve and the axis is $\int_{a}^{b} y d x$ .
Where the curve is below the axis, $y$ is negative, so the integral comes out negative. The area is the size of that number; an area is never negative.
If the curve crosses the axis inside the interval, split the integral at the crossing point and add the two sizes. One integral straight across lets the part below the axis cancel some of the part above.
Example: $y = x^{2} - 3 x$ crosses the axis at $x = 0$ and $x = 3$ . Between $x = 0$ and $x = 4$ the two pieces are:
$\int_{0}^{3} (x^{2} - 3 x) d x = {[\frac{1}{3} x^{3} - \frac{3}{2} x^{2}]}_{0}^{3} = - \frac{9}{2} \int_{3}^{4} (x^{2} - 3 x) d x = \frac{11}{6}$
graph diagram — not drawn yet
Axes for −0.5 ≤ x ≤ 4.5 and −3 ≤ y ≤ 5 with the curve y = x² − 3x drawn through (0, 0), (1, −2), (1.5, −2.25), (2, −2), (3, 0) and (4, 4). Shade the region between the curve and the x-axis from x = 0 to x = 3 (below the axis) and label it '−9/2 from the integral, area 9/2'. Shade the region between the curve and the x-axis from x = 3 to x = 4 (above the axis) in a second shade and label it 'area 11/6'. Draw a dashed vertical line at x = 4.
The part below the axis gives a negative integral; its area is the size, $\frac{9}{2}$ .
Total area $= \frac{9}{2} + \frac{11}{6} = \frac{19}{3}$ . A single integral from $0$ to $4$ gives $- \frac{8}{3}$ , which is not the area of anything. So find where the curve crosses the axis before you integrate; a quick sketch shows you where to split.
For the region between two graphs, integrate (upper minus lower) between the $x$ -values where they meet. This works even where the region is below the axis, because only the difference in height matters.
$Area = \int_{a}^{b} (y_{upper} - y_{lower}) d x$
Example: the curve $y = 4 x - x^{2}$ and the line $y = x$ meet where $4 x - x^{2} = x$ , so $x^{2} - 3 x = 0$ , giving $x = 0$ and $x = 3$ . Test a value in between to see which is on top: at $x = 1$ the curve is at $3$ and the line at $1$ .
$\int_{0}^{3} (4 x - x^{2} - x) d x = {[\frac{3}{2} x^{2} - \frac{1}{3} x^{3}]}_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}$
graph diagram — not drawn yet
Axes for −0.5 ≤ x ≤ 4.5 and −1 ≤ y ≤ 5 with the curve y = 4x − x² drawn through (0, 0), (1, 3), (2, 4), (3, 3) and (4, 0), and the line y = x drawn from (−0.5, −0.5) to (4.5, 4.5). Mark the meeting points (0, 0) and (3, 3). Shade the region between the curve (above) and the line (below) for 0 ≤ x ≤ 3.
The shaded height at each $x$ is curve minus line, $3 x - x^{2}$ ; the limits are where the graphs meet.
Two curves work the same way. $y = 8 - x^{2}$ lies above $y = x^{2}$ between their meeting points, where $8 - x^{2} = x^{2}$ , so $x = \pm 2$ . The area between them is $\int_{- 2}^{2} (8 - 2 x^{2}) d x = {[8 x - \frac{2}{3} x^{3}]}_{- 2}^{2} = \frac{32}{3} - (- \frac{32}{3}) = \frac{64}{3}$ .
Any valid method is allowed. The area between a curve and a line can also be found as the area under the curve minus a triangle or trapezium under the line.

Part 11 of 12

Kinematics: displacement, velocity and acceleration

A particle moves along a straight line. Its displacement $s$ is its position measured from a fixed point $O$ , with one direction counted as positive, so $s$ can be negative. Its velocity $v$ is the rate of change of displacement, and its acceleration $a$ is the rate of change of velocity.
$v = \frac{d s}{d t} a = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$
Going the other way, integrate. Each integration brings a constant, found from a stated condition. 'Starts from rest' means $v = 0$ when $t = 0$ . 'Starts at $O$ ' or 'leaves $O$ ' means $s = 0$ when $t = 0$ .
$v = \int a d t s = \int v d t$
Constant acceleration is the special case where $a$ is a number. If $a = 4$ and the particle starts with velocity $3$ m/s, then $v = 4 t + c$ with $c = 3$ , so $v = 4 t + 3$ . Variable acceleration, such as $a = 6 t - 12$ , is handled in exactly the same way.
The particle is instantaneously at rest when $v = 0$ . If $v$ changes sign there, the particle turns round and moves back the other way.
Speed is the size of the velocity, ignoring direction: a velocity of $- 3$ m/s is a speed of $3$ m/s.
Distance travelled is the total length of path covered, and it never decreases. Displacement says only where the particle is now. For a particle that starts at $O$ and moves in the positive direction, the two are equal only until it turns round.
Example: $s = t^{3} - 6 t^{2} + 9 t$ for $0 \leq t \leq 4$ , with $s$ in metres and $t$ in seconds. Then $v = 3 t^{2} - 12 t + 9 = 3 (t - 1) (t - 3)$ and $a = 6 t - 12$ . The particle is at rest at $t = 1$ and at $t = 3$ .
Find $s$ at the start, at each rest time and at the end: $s (0) = 0$ , $s (1) = 4$ , $s (3) = 0$ , $s (4) = 4$ . It goes $4$ m out, $4$ m back to $O$ , then $4$ m out again. Its displacement at $t = 4$ is $4$ m, but the distance it has travelled is $4 + 4 + 4 = 12$ m.
To get distance from a velocity, integrate $v$ separately between the rest times and add the sizes of the answers, exactly as for an area that crosses the axis.

Part 12 of 12

Kinematics graphs: reading gradient and area

On a displacement–time graph, the gradient is the velocity. Where the graph has a maximum or minimum point, $v = 0$ and the particle is at rest. For the example, the graph rises to $(1, 4)$ , falls to touch the $t$ -axis at $(3, 0)$ , then rises to $(4, 4)$ .
A distance–time graph never goes down, because distance travelled never decreases. Where the displacement–time graph goes down, the distance–time graph keeps going up by the same amount. For the example, distance rises to $4$ at $t = 1$ , to $8$ at $t = 3$ , and to $12$ at $t = 4$ .
graph diagram — not drawn yet
Two graphs side by side, each with t from 0 to 4 on the horizontal axis. Left, labelled 'displacement–time': s = t³ − 6t² + 9t drawn through (0, 0), (0.5, 3.125), (1, 4), (2, 2), (3, 0), (3.5, 0.875) and (4, 4), with a maximum at (1, 4) and a minimum touching the t-axis at (3, 0); vertical axis s from 0 to 12. Right, labelled 'distance–time', on the same vertical scale: a curve through (0, 0), (0.5, 3.125), (1, 4), (2, 6), (3, 8), (3.5, 8.875) and (4, 12), always rising, with zero gradient for an instant at t = 1 and t = 3.
Where displacement falls from $4$ to $0$ , distance keeps rising from $4$ to $8$ .
On a velocity–time graph, the gradient is the acceleration, and the area between the graph and the $t$ -axis is the displacement, with area below the axis counting as negative. For the example, the area under $v$ from $t = 0$ to $t = 1$ is $\int_{0}^{1} v d t = s (1) - s (0) = 4$ .
A speed–time graph is the velocity–time graph with every part below the $t$ -axis reflected up above it, since speed is never negative. The area under a speed–time graph is the distance travelled.
graph diagram — not drawn yet
Two graphs side by side, each with t from 0 to 4 on the horizontal axis and the vertical axis from −4 to 10. Left, labelled 'velocity–time': the U-shaped parabola v = 3t² − 12t + 9 through (0, 9), (0.5, 3.75), (1, 0), (2, −3), (3, 0), (3.5, 3.75) and (4, 9), with the part between t = 1 and t = 3 below the axis shaded. Right, labelled 'speed–time': the same curve, but with the part between t = 1 and t = 3 reflected above the axis, so the graph touches 0 at t = 1 and t = 3 and has a small hump reaching 3 at t = 2.
The dip to $v = - 3$ becomes a hump to speed $3$ : the particle is moving backwards at $3$ m/s at $t = 2$ .
An acceleration–time graph shows $a$ against $t$ . Here $a = 6 t - 12$ is a straight line from $- 12$ at $t = 0$ to $12$ at $t = 4$ . It crosses the $t$ -axis at $t = 2$ , which is exactly where the velocity–time graph has its lowest point, since there its gradient is zero.
When you sketch any of these, label the values that matter: the values at the ends of the time interval, the times when $v = 0$ , and any maximum or minimum. A sketch shows the shape and the key points; it does not need plotted points in between.

Worked examples

Example 1. Find the equation of the normal to the curve $y = \frac{x + 3}{2 x - 1}$ at the point where $x = 1$ . Give your answer in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.

Show the working

Find the point first. At $x = 1$ , $y = \frac{1 + 3}{2 - 1} = 4$ , so the point is $(1, 4)$ .
It is a quotient, so write the parts down: $u = x + 3$ , $v = 2 x - 1$ , $\frac{d u}{d x} = 1$ , $\frac{d v}{d x} = 2$ .
Apply the quotient rule, bottom first.
$\frac{d y}{d x} = \frac{(2 x - 1) (1) - (x + 3) (2)}{{(2 x - 1)}^{2}} = \frac{- 7}{{(2 x - 1)}^{2}}$
Expand carefully: $2 x - 1 - 2 x - 6 = - 7$ . The minus sign in front of the second product reaches both of its terms.
At $x = 1$ the tangent's gradient is $\frac{- 7}{1^{2}} = - 7$ . The normal is perpendicular, so its gradient is $\frac{1}{7}$ , since $- 7 \times \frac{1}{7} = - 1$ .
Use the point $(1, 4)$ and multiply through by $7$ to clear the fraction.
$y - 4 = \frac{1}{7} (x - 1) \Rightarrow 7 y - 28 = x - 1 \Rightarrow x - 7 y + 27 = 0$
Check that the point fits: $1 - 7 (4) + 27 = 0$ .

Answer: $x - 7 y + 27 = 0$ .

Example 2. Find the coordinates of the stationary point on the curve $y = x e^{- 2 x}$ and determine its nature.

Show the working

It is a product: $u = x$ , $v = e^{- 2 x}$ , $\frac{d u}{d x} = 1$ and, by the chain rule, $\frac{d v}{d x} = - 2 e^{- 2 x}$ .
$\frac{d y}{d x} = x (- 2 e^{- 2 x}) + e^{- 2 x} (1) = e^{- 2 x} (1 - 2 x)$
Set it equal to zero. $e^{- 2 x}$ is always positive, so it is never zero, and the only solution comes from $1 - 2 x = 0$ , which gives $x = \frac{1}{2}$ . There is exactly one stationary point.
Find $y$ from the curve's equation: $y = \frac{1}{2} e^{- 1} = \frac{1}{2 e}$ .
For the nature, differentiate $e^{- 2 x} (1 - 2 x)$ with the product rule again.
$\frac{d^{2} y}{d x^{2}} = - 2 e^{- 2 x} (1 - 2 x) + e^{- 2 x} (- 2) = e^{- 2 x} (4 x - 4)$
At $x = \frac{1}{2}$ : $\frac{d^{2} y}{d x^{2}} = e^{- 1} (2 - 4) = - \frac{2}{e}$ , which is negative, so the point is a maximum.
The first derivative test agrees, and is quicker here. Since $e^{- 2 x} > 0$ , the sign of $\frac{d y}{d x}$ is the sign of $1 - 2 x$ : positive just before $x = \frac{1}{2}$ and negative just after.

Answer: $(\frac{1}{2}, \frac{1}{2 e})$ is a maximum, because $\frac{d^{2} y}{d x^{2}} = - \frac{2}{e} < 0$ there.

Example 3. A closed cylindrical tin has radius $r$ cm and height $h$ cm, and its volume is $128 π$ cm³. Show that its total surface area, $S$ cm², is given by $S = 2 π r^{2} + \frac{256 π}{r}$ , and find the minimum value of $S$ .

Show the working

Use the fixed volume to remove $h$ : $π r^{2} h = 128 π$ , so $h = \frac{128}{r^{2}}$ .
The surface is two circular ends plus the curved side, which is $2 π r \times h$ . Substitute for $h$ ; this is the Show that, so write every line.
$S = 2 π r^{2} + 2 π r \times \frac{128}{r^{2}} = 2 π r^{2} + \frac{256 π}{r}$
Write the fraction as a power, $256 π r^{- 1}$ , and differentiate. $π$ is just a constant.
$\frac{d S}{d r} = 4 π r - \frac{256 π}{r^{2}}$
Set it to zero and multiply by $r^{2}$ : $4 π r^{3} = 256 π$ , so $r^{3} = 64$ and $r = 4$ .
Justify the minimum: $\frac{d^{2} S}{d r^{2}} = 4 π + \frac{512 π}{r^{3}}$ , which at $r = 4$ is $4 π + 8 π = 12 π > 0$ . So this is a minimum.
Answer the question asked, which is the value of $S$ : $S = 2 π (16) + \frac{256 π}{4} = 32 π + 64 π = 96 π$ . The tin then has $h = \frac{128}{16} = 8$ , so its height equals its diameter.

Answer: The minimum surface area is $96 π$ cm², when $r = 4$ .

Example 4. The curve $y = 4 - e^{2 x}$ meets the $y$ -axis at $A$ and the $x$ -axis at $B$ . Find the exact area of the region bounded by the curve and the two axes.

Show the working

Find the limits. At $x = 0$ , $y = 4 - 1 = 3$ , so $A$ is $(0, 3)$ . At $B$ , $e^{2 x} = 4$ , so $2 x = \ln 4$ and $x = \frac{1}{2} \ln 4 = \ln 2$ .
For $0 \leq x < \ln 2$ , $e^{2 x} < 4$ , so the curve is above the axis and one integral gives the area.
Integrate term by term. $\int e^{2 x} d x = \frac{1}{2} e^{2 x}$ : divide by $2$ , the coefficient of $x$ .
$\int_{0}^{\ln 2} (4 - e^{2 x}) d x = {[4 x - \frac{1}{2} e^{2 x}]}_{0}^{\ln 2}$
Substitute. At the top limit, $e^{2 \ln 2} = e^{\ln 4} = 4$ . At the bottom, $e^{0} = 1$ .
$(4 \ln 2 - \frac{1}{2} (4)) - (0 - \frac{1}{2}) = 4 \ln 2 - 2 + \frac{1}{2} = 4 \ln 2 - \frac{3}{2}$
Sense check: $\ln 2 \approx 0.69$ , so the area is about $2.77 - 1.5 = 1.27$ . The region fits inside the rectangle $\ln 2$ wide and $3$ high, about $2.08$ , so the size is reasonable.

Answer: $4 \ln 2 - \frac{3}{2}$ square units.

Example 5. A particle moves in a straight line. At time $t$ seconds after leaving a fixed point $O$ , its velocity is $v$ m/s, where $v = 8 + 2 t - t^{2}$ . (a) Find the acceleration of the particle when it comes to rest. (b) Find the distance it travels in the first $6$ seconds. (c) Describe its speed–time graph for $0 \leq t \leq 6$ .

Show the working

(a) At rest means $v = 0$ : $t^{2} - 2 t - 8 = 0$ , so $(t - 4) (t + 2) = 0$ . Time cannot be negative, so $t = 4$ .
$a = \frac{d v}{d t} = 2 - 2 t$ , so at $t = 4$ , $a = - 6$ m/s². The negative sign means the velocity is decreasing.
(b) Integrate for displacement. The particle leaves $O$ , so $s = 0$ when $t = 0$ , which makes the constant $0$ .
$s = 8 t + t^{2} - \frac{1}{3} t^{3}$
It turns round at $t = 4$ , so find $s$ there and at the end: $s (4) = 32 + 16 - \frac{64}{3} = \frac{80}{3}$ and $s (6) = 48 + 36 - 72 = 12$ .
It goes $\frac{80}{3}$ m out, then comes back from $\frac{80}{3}$ to $12$ , a further $\frac{80}{3} - 12 = \frac{44}{3}$ m.
$distance = \frac{80}{3} + \frac{44}{3} = \frac{124}{3}$
Compare: its displacement at $t = 6$ is only $12$ m. Giving $s (6)$ as the distance would leave out the journey back.
(c) The velocity graph is an upside-down parabola: $v = 8$ at $t = 0$ , a maximum of $9$ at $t = 1$ (where $a = 0$ ), $v = 0$ at $t = 4$ and $v = - 16$ at $t = 6$ . For speed, reflect the part after $t = 4$ above the axis.

Answer: (a) $- 6$ m/s² (b) $\frac{124}{3}$ m, about $41.3$ m (c) The speed rises from $8$ to $9$ at $t = 1$ , falls to $0$ at $t = 4$ , then rises again to $16$ at $t = 6$ ; it is never negative.

Where people go wrong

Mistake: Differentiating a product term by term: $\frac{d}{d x} (x^{2} \sin x) = 2 x \cos x$ .

Instead: Use the product rule: $\frac{d}{d x} (x^{2} \sin x) = x^{2} \cos x + 2 x \sin x$ .

Differentiation shares out over $+$ and $-$ , not over $\times$ . In a product both factors change at once, and the product rule counts both changes.

Mistake: Multiplying by $a$ instead of dividing when integrating a function of $a x + b$ , as in $\int e^{3 x} d x = 3 e^{3 x} + c$ .

Instead: $\int e^{3 x} d x = \frac{1}{3} e^{3 x} + c$ .

Differentiating multiplies by $a$ , so integrating, which reverses it, divides by $a$ . Differentiate your answer to check: $\frac{d}{d x} (\frac{1}{3} e^{3 x}) = e^{3 x}$ .

Mistake: Writing $\int \sin x d x = \cos x + c$ .

Instead: $\int \sin x d x = - \cos x + c$ .

$\frac{d}{d x} (\cos x) = - \sin x$ , so it is $- \cos x$ whose derivative is $\sin x$ .

Mistake: Finding the area of a region that crosses the $x$ -axis with one integral across the whole interval.

Instead: Split at each point where the curve crosses the axis, integrate each part separately and add the sizes.

The integral over the part below the axis is negative, so a single integral subtracts that area instead of adding it.

Mistake: Giving the final displacement as the distance travelled when the particle has turned round.

Instead: Find when $v = 0$ , work out $s$ at those times and at the ends, and add the length of each stage.

Distance counts the journey back as well. Displacement only says where the particle finished.

Check yourself

For $y = x^{3}$ , calculate $\frac{δ y}{δ x}$ for the chord from $x = 2$ to $x = 2.1$ . Then find $\frac{d y}{d x}$ at $x = 2$ and explain how the two are related.
Answer
$δ y = {2.1}^{3} - 8 = 9.261 - 8 = 1.261$ , so $\frac{δ y}{δ x} = \frac{1.261}{0.1} = 12.61$ . $\frac{d y}{d x} = 3 x^{2} = 12$ at $x = 2$ . The chord gradient is close to $12$ and tends to $12$ as $δ x \to 0$ : $\frac{d y}{d x}$ is the limit of the chord gradients.
For $y = x^{4} - 6 \sqrt{x}$ , find $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ .
Answer
$y = x^{4} - 6 x^{\frac{1}{2}}$ , so $\frac{d y}{d x} = 4 x^{3} - 3 x^{- \frac{1}{2}}$ and $\frac{d^{2} y}{d x^{2}} = 12 x^{2} + \frac{3}{2} x^{- \frac{3}{2}}$ .
Differentiate (a) $5 e^{2 x - 1}$ (b) $\ln (x^{2} + 4)$ (c) $\cos 3 x$ (d) $\sqrt{1 - 4 x}$ .
Answer
(a) $10 e^{2 x - 1}$ (b) $\frac{2 x}{x^{2} + 4}$ (c) $- 3 \sin 3 x$ (d) $\frac{1}{2} {(1 - 4 x)}^{- \frac{1}{2}} \times (- 4) = - \frac{2}{\sqrt{1 - 4 x}}$ .
Differentiate (a) $x^{3} \ln x$ (b) $\frac{e^{2 x}}{x + 1}$ , factorising each answer.
Answer
(a) $x^{3} \cdot \frac{1}{x} + 3 x^{2} \ln x = x^{2} (1 + 3 \ln x)$ (b) $\frac{(x + 1) (2 e^{2 x}) - e^{2 x} (1)}{{(x + 1)}^{2}} = \frac{e^{2 x} (2 x + 1)}{{(x + 1)}^{2}}$ .
Find the equation of the normal to $y = e^{2 x}$ at the point where $x = 0$ .
Answer
The point is $(0, 1)$ . $\frac{d y}{d x} = 2 e^{2 x} = 2$ at $x = 0$ , so the normal has gradient $- \frac{1}{2}$ : $y = 1 - \frac{1}{2} x$ , or $x + 2 y = 2$ .
Find the stationary point of $y = x + \frac{4}{x}$ for $x > 0$ , and use the second derivative to determine its nature.
Answer
$\frac{d y}{d x} = 1 - 4 x^{- 2} = 0$ gives $x^{2} = 4$ , so $x = 2$ (since $x > 0$ ) and $y = 2 + 2 = 4$ . $\frac{d^{2} y}{d x^{2}} = 8 x^{- 3} = 1 > 0$ at $x = 2$ , so $(2, 4)$ is a minimum.
(a) For $y = x^{2} + 3 x$ , find an approximate expression for the change in $y$ when $x$ increases from $2$ to $2 + h$ , where $h$ is small. (b) The radius $r$ cm of a circle is decreasing at $0.2$ cm/s. Find the rate of change of its area when $r = 5$ .
Answer
(a) $\frac{d y}{d x} = 2 x + 3 = 7$ at $x = 2$ , so $δ y \approx 7 h$ . (b) $\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t} = 2 π (5) \times (- 0.2) = - 2 π$ , so the area is decreasing at $2 π$ cm²/s.
Find (a) $\int {(2 x - 3)}^{4} d x$ (b) $\int \sec^{2} 3 x d x$ (c) $\int \frac{6}{3 x + 2} d x$ , and evaluate (d) $\int_{0}^{\frac{π}{6}} \cos 2 x d x$ .
Answer
(a) $\frac{{(2 x - 3)}^{5}}{10} + c$ (b) $\frac{1}{3} \tan 3 x + c$ (c) $2 \ln (3 x + 2) + c$ (d) ${[\frac{1}{2} \sin 2 x]}_{0}^{\frac{π}{6}} = \frac{1}{2} \sin \frac{π}{3} = \frac{\sqrt{3}}{4}$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A curve has equation $y = \frac{2 x}{x^{2} + 1}$ .

(a) Show that $\frac{d y}{d x} = \frac{2 (1 - x^{2})}{{(x^{2} + 1)}^{2}}$ . 3 marks No calculator

Mark scheme

M1 quotient rule applied with $u = 2 x$ and $v = x^{2} + 1$ : $\frac{(x^{2} + 1) (2) - 2 x (2 x)}{{(x^{2} + 1)}^{2}}$ oe, or product rule on $2 x {(x^{2} + 1)}^{- 1}$
A1 numerator expanded correctly to $2 x^{2} + 2 - 4 x^{2}$ oe
A1 given answer $\frac{2 (1 - x^{2})}{{(x^{2} + 1)}^{2}}$ reached with no error in the working

(b) Find the coordinates of the stationary points of the curve and determine their nature. 4 marks No calculator

Mark scheme

M1 sets $\frac{d y}{d x} = 0$ and solves $1 - x^{2} = 0$
A1 $(1, 1)$ and $(- 1, - 1)$
M1 a valid method for the nature of at least one point: the sign of $\frac{d y}{d x}$ on each side, or the sign of $\frac{d^{2} y}{d x^{2}}$
A1 $(1, 1)$ maximum and $(- 1, - 1)$ minimum, fully justified, e.g. the denominator is always positive, so $\frac{d y}{d x}$ has the sign of $1 - x^{2}$ : positive then negative at $x = 1$ , negative then positive at $x = - 1$ ; or $\frac{d^{2} y}{d x^{2}} = - 1 < 0$ at $x = 1$ and $1 > 0$ at $x = - 1$

Question 2.

(a) Find $\frac{d}{d x} (x \sin 2 x)$ . 2 marks No calculator

Mark scheme

M1 product rule used, with $\frac{d}{d x} (\sin 2 x) = 2 \cos 2 x$
A1 $\sin 2 x + 2 x \cos 2 x$ oe

(b) Hence find $\int x \cos 2 x d x$ . 3 marks No calculator

Mark scheme

M1 uses the reverse of (a): $\int (\sin 2 x + 2 x \cos 2 x) d x = x \sin 2 x$ , rearranged to $2 \int x \cos 2 x d x = x \sin 2 x - \int \sin 2 x d x$ oe
B1 $\int \sin 2 x d x = - \frac{1}{2} \cos 2 x$
A1 $\frac{1}{2} x \sin 2 x + \frac{1}{4} \cos 2 x + c$ oe, including the arbitrary constant

Question 3. A particle $P$ moves in a straight line so that, $t$ seconds after passing through a fixed point $O$ , its velocity is $v$ m/s, where $v = 2 t^{2} - 10 t + 8$ .

(a) Find the acceleration of $P$ when $t = 3$ . 2 marks

Mark scheme

M1 differentiates $v$ : $a = 4 t - 10$
A1 $2$ m/s²

(b) Find the total distance travelled by $P$ in the first $4$ seconds. 4 marks

Mark scheme

M1 integrates $v$ : $s = \frac{2}{3} t^{3} - 5 t^{2} + 8 t$ [ $+ c$ , with $c = 0$ ]
M1 solves $v = 0$ and uses $t = 1$ to split the motion into $0 \leq t \leq 1$ and $1 \leq t \leq 4$
A1 $s (1) = \frac{11}{3}$ and $s (4) = - \frac{16}{3}$ , or stage lengths $\frac{11}{3}$ and $9$
A1 $\frac{38}{3}$ m oe, awrt $12.7$ m

(c) Sketch the speed–time graph of $P$ for $0 \leq t \leq 5$ . 2 marks

Mark scheme

B1 correct shape: falling from $t = 0$ to $0$ at $t = 1$ , a hump above the axis between $t = 1$ and $t = 4$ , rising after $t = 4$ , with no part below the $t$ -axis
B1 $8$ at $t = 0$ and at $t = 5$ , zeros at $t = 1$ and $t = 4$ , and the hump's height $4.5$ at $t = 2.5$ shown

How this is examined

Setting up a practical quantity in one variable, then finding its maximum or minimum: a Show that an area, volume or cost equals a printed expression in one variable, followed by a Find of its greatest or least value For the Show that: the fixed fact (such as the volume) used to write the second variable in terms of the first, and each substitution line written until the printed expression appears. Then: the derivative set equal to $0$ and solved, the nature justified by the sign of the second derivative or a sign change of the first, and the quantity the question asks for (the area, not only the $x$ that gives it).
The area enclosed by a curve and a line, or by a curve that crosses the axis: a Find the area of a shaded region, where the limits come from solving the two equations together The meeting points found by solving the equations simultaneously, a correct integral of (upper minus lower) or of each separate piece, the limits substituted and subtracted, and a correct final area (exact when the question asks for it). A piece below the axis gives a negative integral; its area is the size of that number.
Distance travelled versus displacement for a particle that turns round: a multi-part kinematics question: differentiate $v$ for $a$ , integrate $v$ for $s$ , then a Find of the distance travelled and a Sketch of a speed–time or velocity–time graph $v = 0$ solved to find when the particle turns, $s$ found by integrating $v$ with the constant from the starting condition, and the distance given as the sum of the stage lengths. On the sketch: the correct shape, never below the axis for speed, with the end values and the times of rest labelled.
Finding stationary points and proving which is a maximum and which a minimum: a Find the coordinates of the stationary points, then a Determine their nature $\frac{d y}{d x}$ correct and set equal to $0$ , every $x$ with its $y$ taken from the curve's equation, then for each point either the value of $\frac{d^{2} y}{d x^{2}}$ with its sign and the conclusion, or the sign of $\frac{d y}{d x}$ on each side. A conclusion with no sign stated is not a justification.
Differentiating a product, quotient or composite, then using the gradient for a tangent or normal: a Show that $\frac{d y}{d x}$ equals a printed factorised form, then a Find of the equation of the tangent or normal at a given point The rule written with its parts visible, a correct unsimplified derivative, and algebra that reaches the printed form exactly. For the line: the gradient from $\frac{d y}{d x}$ at the point, $- \frac{1}{m}$ for a normal, and $y - y_{1} = m (x - x_{1})$ through the point, in the form the question asks for.
Linking rates with the chain rule, and estimating small changes: a short Find of a rate, which may use a formula from the List of formulas such as the volume of a sphere, or a Find of the approximate change in $y$ when $x$ changes by $p$ The chain rule written in terms of the quantities, such as $\frac{d r}{d t} = \frac{d V}{d t} \div \frac{d V}{d r}$ , the derivative evaluated at the given value, and the rate with its units. For small changes: $δ y \approx \frac{d y}{d x} \times δ x$ , with $\frac{d y}{d x}$ taken at the starting value.

Words to know

Derived function (derivative): The function $f^{'} (x)$ , or $\frac{d y}{d x}$ , that gives the gradient of $y = f (x)$ at every value of $x$ .
Second derivative: $f^{''} (x) = \frac{d^{2} y}{d x^{2}}$ , the derivative of the derivative; it measures how fast the gradient changes.
δx: A small change in $x$ , read 'delta $x$ '. $δ x \to 0$ means the change is made smaller and smaller towards zero.
Chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ , for differentiating a function of a function.
Tangent: The straight line that touches a curve at a point and has the same gradient as the curve there.
Normal: The straight line through a point on a curve perpendicular to the tangent there; its gradient is $- \frac{1}{m}$ .
Stationary point: A point where $\frac{d y}{d x} = 0$ . In this syllabus, a maximum or a minimum.
Indefinite integral: An integral without limits; the answer includes an arbitrary constant $c$ .
Definite integral: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , a number found by substituting limits.
Displacement: Position measured from a fixed point $O$ along the line, positive in one direction and negative in the other.
Velocity: Rate of change of displacement, $v = \frac{d s}{d t}$ ; its sign gives the direction of motion.
Acceleration: Rate of change of velocity, $a = \frac{d v}{d t}$ .
Speed: The size of the velocity, ignoring direction; never negative.
Distance travelled: The total length of path covered, counting every change of direction; it never decreases.

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