Tangents, Normals and Stationary Points (Further)

Find tangents, normals and stationary points of curves with e, trig, ln, products and quotients, justify maximum or minimum, and solve practical optimisation problems.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 57 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Tangents and normals on any curve

You already found tangents and normals to polynomial curves. The method does not change for $e^{x}$ , trig, products or quotients (and ln for 0606 and 4037). Only the differentiating is new, and you now have every rule you need.

Formula

y - y_{1} = m (x - x_{1})

$(x_{1}, y_{1})$ — the point on the curve. Find $y_{1}$ by putting $x_{1}$ into the equation of the curve.
$m$ — the gradient: the value of $\frac{d y}{d x}$ at $x_{1}$ for a tangent.

Formula

m_{normal} = - \frac{1}{m _{tangent}}

$m_{tangent}$ — the gradient of the tangent. Flip it and change its sign, because the normal is at right angles to the tangent.

Rule

Neither formula sheet gives these. Learn them.

Worked example

1
Find the equation of the tangent to $y = e^{2 x} + 3 x$ at the point where $x = 0$ .
2
$y = e^{0} + 0 = 1$ — Find the point: (0, 1).
3
$\frac{d y}{d x} = 2 e^{2 x} + 3$ — Differentiate.
4
$m = 2 e^{0} + 3 = 5$ — Gradient at x = 0.
5
$y - 1 = 5 (x - 0) \Rightarrow y = 5 x + 1$ — Put the point and the gradient into the line formula.

Worked example

1
Find the equation of the normal to $y = 3 cos 2 x$ at the point where $x = \frac{π}{4}$ .
2
$y = 3 cos \frac{π}{2} = 0$ — Find the point: $(\frac{π}{4}, 0)$ .
3
$\frac{d y}{d x} = - 6 sin 2 x, m = - 6 sin \frac{π}{2} = - 6$ — Gradient of the tangent at the point.
4
$m_{normal} = \frac{1}{6}$ — Flip −6 and change the sign.
5
$y - 0 = \frac{1}{6} (x - \frac{π}{4})$ — Line through the point with the normal's gradient.
6
$24 y = 4 x - π$ — Multiply by 24 to clear the fractions. Exact, with π left in.

Common mistake

Putting the expression for $\frac{d y}{d x}$ into the line. The gradient must be a number: substitute $x_{1}$ first.

Key idea

Point from the curve, gradient from dy/dx, then the line formula. For a normal, flip and change the sign first.

Where tangents and normals meet the axes

Many questions go one step further: where does the line cross an axis, or what area does it cut off? Put $y = 0$ to find where it crosses the x-axis, and $x = 0$ for the y-axis.

Worked example

1
The curve $y = \frac{8}{x} + x$ has a point P where $x = 2$ . The tangent and the normal at P meet the x-axis at A and B. Find the area of triangle ABP.
2
$y = 4 + 2 = 6$ — P is (2, 6).
3
$\frac{d y}{d x} = - \frac{8}{x ^{2}} + 1, m = - 2 + 1 = - 1$ — Write $\frac{8}{x}$ as $8 x^{- 1}$ to differentiate. Then put x = 2.
4
$Tangent: y - 6 = - (x - 2) \Rightarrow y = 8 - x$ — Gradient −1 through P.
5
$Normal: y - 6 = 1 (x - 2) \Rightarrow y = x + 4$ — Flip −1 and change the sign: the normal's gradient is 1.
6
$y = 0 : A = (8, 0), B = (- 4, 0)$ — Where each line meets the x-axis.
7
$Area = \frac{1}{2} \times 12 \times 6 = 36$ — The base AB is 8 − (−4) = 12. The height is the y of P, 6.

Common mistake

Measuring the height of the triangle along the slanted lines. The height is the straight-up distance from P to the x-axis: its y-coordinate.

Key idea

A sketch with the point, the two lines and the axes stops most errors. Draw one, even a rough one.

Stationary points of any curve

At a stationary point the gradient is 0. So solve $\frac{d y}{d x} = 0$ . With products and quotients, factorise the derivative first, then look at each factor.

Steps

1
- Differentiate to get $\frac{d y}{d x}$ .
- Factorise it. Take out any $e^{x}$ -type factor.
- Set each factor that can be zero equal to 0. A quotient is 0 only when its top is 0.
- Put each x back into the equation of the curve to find y.

Worked example

1
Find the stationary point of $y = (x - 2) e^{x}$ .
2
$\frac{d y}{d x} = (x - 2) e^{x} + e^{x} (1)$ — Product rule: $u = x - 2$ , $v = e^{x}$ .
3
$\frac{d y}{d x} = e^{x} (x - 1)$ — Take out the common factor $e^{x}$ .
4
$e^{x} > 0 always, so x - 1 = 0$ — e to any power is positive, so it can never be 0.
5
$x = 1, y = (1 - 2) e^{1} = - e$ — The stationary point is (1, −e). Leave e exact.

Worked example

1
Find the stationary points of $y = \frac{x ^{2}}{x + 1}$ .
2
$\frac{d y}{d x} = \frac{( x + 1 ) ( 2 x ) - x ^{2} ( 1 )}{( x + 1 ) ^{2}} = \frac{x ( x + 2 )}{( x + 1 ) ^{2}}$ — Quotient rule, then factorise the top.
3
$x (x + 2) = 0 \Rightarrow x = 0 or x = - 2$ — A fraction is 0 when its top is 0.
4
$(0, 0) and (- 2, - 4)$ — Find y: $\frac{0}{1} = 0$ and $\frac{4}{- 1} = - 4$ .

Common mistake

Writing $e^{x} = 0$ and trying to solve it. It has no solution. Say “ $e^{x} \neq = 0$ ” and move on.

Rule

Curves with ln are for 0606 and 4037 only; 4PM1 does not differentiate ln. With ln, keep only x-values where the inside of the ln is positive. 0606 and 4037 also leave out points of inflexion.

Key idea

Factorise first. Then each factor tells you either a stationary point or “never zero”.

Maximum or minimum? Justify it

Finding the point is not enough: exams expect you to show whether it is a maximum or a minimum. There are two tests. Use either, unless the question names one.

Maximum: $\frac{d ^{2} y}{d x ^{2}} < 0$ . Or $\frac{d y}{d x}$ goes + then 0 then −: uphill, flat, downhill.

Minimum: $\frac{d ^{2} y}{d x ^{2}} > 0$ . Or $\frac{d y}{d x}$ goes − then 0 then +: downhill, flat, uphill.

Worked example

1
Second derivative test: classify (1, −e) on $y = (x - 2) e^{x}$ , where $\frac{d y}{d x} = e^{x} (x - 1)$ .
2
$\frac{d ^{2} y}{d x ^{2}} = e^{x} (x - 1) + e^{x} (1) = x e^{x}$ — Product rule again, then tidy.
3
$x = 1 : \frac{d ^{2} y}{d x ^{2}} = e > 0$ — Substitute and state the sign.
4
$so (1, - e) is a minimum$ — Write the conclusion in words.

Worked example

1
First derivative test: classify (−2, −4) on $y = \frac{x ^{2}}{x + 1}$ , where $\frac{d y}{d x} = \frac{x ( x + 2 )}{( x + 1 ) ^{2}}$ .
2
$x = - 2.5 : \frac{( - 2.5 ) ( - 0.5 )}{2.25} > 0$ — Just left of −2. Only the sign matters.
3
$x = - 1.5 : \frac{( - 1.5 ) ( 0.5 )}{0.25} < 0$ — Just right of −2. Do not go as far as −1, where the bottom is 0.
4
$+, 0, - so (- 2, - 4) is a maximum$ — Uphill, flat, downhill.

Common mistake

Stopping at “ $\frac{d ^{2} y}{d x ^{2}} = e$ ”. You must also say it is positive, and so a minimum. A full justification has the value, the sign and the conclusion.

Common mistake

If $\frac{d ^{2} y}{d x ^{2}} = 0$ , the second derivative test tells you nothing. Use the first derivative test instead.

Key idea

Value, sign, conclusion: three parts for the justification mark.

Practical maximum and minimum problems

Real problems ask for the least material, the lowest cost or the biggest volume. The hard part is usually the first step: writing the quantity using one letter only.

Steps

1
- Write a formula for the quantity you want to make biggest or smallest.
- If it has two letters, use the other fact in the question to replace one.
- Differentiate, set equal to 0, and solve.
- Justify maximum or minimum.
- Answer the question that was asked: a length, a value, or both.

Worked example

1
A closed cylinder has volume $128 π cm^{3}$ . Find the radius that makes its surface area as small as possible, and the smallest area.
2
$π r^{2} h = 128 π \Rightarrow h = \frac{128}{r ^{2}}$ — The volume fact lets you replace h.
3
$S = 2 π r^{2} + 2 π r h = 2 π r^{2} + \frac{256 π}{r}$ — Two circles plus the curved side. Now S has one letter.
4
$\frac{d S}{d r} = 4 π r - \frac{256 π}{r ^{2}}$ — Write $\frac{256 π}{r}$ as $256 π r^{- 1}$ to differentiate.
5
$4 π r = \frac{256 π}{r ^{2}} \Rightarrow r^{3} = 64 \Rightarrow r = 4$ — Set equal to 0 and solve. No calculator needed.
6
$\frac{d ^{2} S}{d r ^{2}} = 4 π + \frac{512 π}{r ^{3}} = 12 π > 0$ — At r = 4 this is positive, so it is a minimum.
7
$S = 2 π (16) + \frac{256 π}{4} = 96 π cm^{2}$ — Put r = 4 back into S. Then h = 8 cm.

Common mistake

Differentiating S while it still has both r and h. Replace h first, so that S is a function of r only.

Common mistake

Giving r = 4 when the question asks for the smallest area. Read the last line of the question again.

Key idea

In the 1630s Pierre de Fermat found maximum and minimum values with a method very like setting the gradient to zero. That was decades before Newton and Leibniz published calculus.

Which method? Read the question

The command words tell you what to do.

You see	You do
“Tangent at x = a”	Point from y, gradient from $\frac{d y}{d x}$ , then the line formula
“Normal at x = a”	The same, but flip the gradient and change its sign
“Meets the axes” or “area of triangle”	Put y = 0 or x = 0 in each line, then sketch
“Stationary point” or “turning point”	Solve $\frac{d y}{d x} = 0$ after factorising
“Determine the nature” or “justify”	Second derivative sign, or the gradient either side
“Use the first derivative test”	Sign of $\frac{d y}{d x}$ just left and right of the point
A shape with a fixed volume, area or length	Replace one letter, differentiate, solve, justify, answer

Common mistake

Leaving the nature unjustified. “It is a minimum” with no test earns no mark for the nature.

Key idea

Same steps as for polynomials. The new rules only change how you differentiate.

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