1.07PureStretch

Tangents and Normals

Q: What if the tangent is horizontal?

Then m = 0 and the tangent is y = y_1. The normal is vertical, x = x_1, since - 10 is undefined.

Q: Do I have to eliminate the parameter for parametric curves?

No. Use dy dx = dy/ dt dx/ dt and work with t throughout.

Q: Why does y^2 differentiate to 2y, dy/ dx?

y is itself a function of x, so y^2 is a composite and the chain rule supplies the extra factor.

A $\textbf{tangent}$ touches a curve at a point and has the same gradient as the curve there. The $\textbf{normal}$ passes through the same point at right angles to the tangent. Finding either is a three-step routine — gradient, point, line — but exam questions wrap that routine inside curves defined parametrically or implicitly, and ask what the lines do next.

40 min Differentiation

The big picture

This lesson is where differentiation meets coordinate geometry. The derivative supplies the gradient; the coordinates of the point come from the curve; and $y - y_{1} = m(x - x_{1})$ from the straight-lines topic turns the two into an equation. Normals use the perpendicular gradient rule $m_{1}m_{2} = -1$ . The same routine extends to curves where $y$ is not given as a simple function of $x$ : parametric curves, where the chain rule gives $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$ , and implicit curves such as $x^{2} + xy + y^{2} = 7$ , which you differentiate term by term. Typical follow-up questions — where does the normal meet the curve again, what area does the tangent enclose with the axes — test whether you can use the line once you have it.

What you'll be able to do

Find equations of tangents and normals to $y = \mathrm{f}(x)$
Find points on a curve where the tangent has a given gradient
Differentiate parametric equations and find tangents and normals to parametric curves
Differentiate implicitly and find tangents to implicitly defined curves

The routine

Step 1: differentiate and substitute the $x$ -coordinate to get the gradient of the tangent, $m$ .

Step 2: make sure you have both coordinates of the point — find $y$ from the curve if needed.

Step 3: use $y - y_{1} = m(x - x_{1})$ . For the normal, use the gradient $-\dfrac{1}{m}$ instead.

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

Line through a point with gradient m.

m_{\text{normal}} = -\dfrac{1}{m_{\text{tangent}}}

Perpendicular gradient.

Example 1Find the equations of the tangent and normal to

y = x^{3} - 4x + 1

at the point where

x = 2

1Point:

y = 8 - 8 + 1 = 1

, so

(2, 1)

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - 4

, which is

8

x = 2

3Tangent:

y - 1 = 8(x - 2)

, so

y = 8x - 15

4Normal gradient

-\tfrac18

y - 1 = -\tfrac18(x - 2)

, i.e.

x + 8y - 10 = 0

AnswerTangent

y = 8x - 15

; normal

x + 8y - 10 = 0

Tip — Questions often ask for the answer "in the form $ax + by + c = 0$ where $a$ , $b$ , $c$ are integers". Clear fractions at the end, not in the middle.

Working backwards from a gradient

If a question gives the gradient and asks for the point, set the derivative equal to that gradient and solve for $x$ .

Parallel tangents have equal gradients; a tangent parallel to the $x$ -axis has gradient zero; a tangent perpendicular to a given line has gradient $-\tfrac{1}{m}$ of that line.

Example 1Find the points on

y = x^{3} - 3x^{2} + 2

where the tangent is parallel to

y = 9x

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - 6x = 9

, so

x^{2} - 2x - 3 = 0

(x - 3)(x + 1) = 0

x = 3

x = -1

3At

x = 3

y = 27 - 27 + 2 = 2

. At

x = -1

y = -1 - 3 + 2 = -2

Answer

(3, 2)

and

(-1, -2)

Parametric curves

If $x$ and $y$ are both given in terms of a parameter $t$ , the chain rule gives the gradient without eliminating $t$ .

Find the parameter value at the point, then the coordinates and gradient all come from that $t$ .

\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}

Parametric differentiation.

Example 1A curve has

x = t^{2}

y = 2t

. Find the equation of the normal at the point where

t = 3

1Point:

(9, 6)

\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{2t} = \dfrac{1}{t}

, which is

\tfrac13

t = 3

3Normal gradient

-3

y - 6 = -3(x - 9)

Answer

y = -3x + 33

Why this works

Parametric gradients often come out simpler in terms of $t$ than the Cartesian equation would give. Here the curve is $y^{2} = 4x$ , but $\tfrac{1}{t}$ is quicker to use than differentiating $y = 2\sqrt{x}$ .

Implicit differentiation

When $x$ and $y$ are mixed in one equation, differentiate every term with respect to $x$ , treating $y$ as a function of $x$ . By the chain rule, $\dfrac{\mathrm{d}}{\mathrm{d}x}y^{n} = ny^{n-1}\dfrac{\mathrm{d}y}{\mathrm{d}x}$ .

Products like $xy$ need the product rule: $\dfrac{\mathrm{d}}{\mathrm{d}x}(xy) = x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y$ .

Then collect the $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ terms on one side and factorise.

Example 1Find the gradient of

x^{2} + xy + y^{2} = 7

at the point

(1, 2)

1Differentiate:

2x + \left(x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y\right) + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0

2Substitute

(1, 2)

2 + \dfrac{\mathrm{d}y}{\mathrm{d}x} + 2 + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0

5\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4

Answer

\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\tfrac45

Tip — Substitute the point before rearranging when you only need a numerical gradient. It avoids a messy general expression.

Think like an examiner

Find both coordinates of the point before writing the line.
Use $-\tfrac{1}{m}$ for the normal, and show it.
Give the line in the form requested, with integer coefficients if specified.
In implicit differentiation, attach $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to every $y$ term and use the product rule on $xy$ .

Common misconceptions

Using the derivative expression, not its value, as the gradient.

Substitute the

x

-coordinate to get a number.

Using gradient

-m

for the normal.

The normal gradient is

-\tfrac{1}{m}

Differentiating

y^{2}

2y

in implicit differentiation.

2y\dfrac{\mathrm{d}y}{\mathrm{d}x}

Computing

\dfrac{\mathrm{d}x/\mathrm{d}t}{\mathrm{d}y/\mathrm{d}t}

for a parametric gradient.

\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}

—

y

on top.

Tangents and normals

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

Equation of a line.

m_{1}m_{2} = -1

Perpendicular lines.

\tfrac{\mathrm{d}y}{\mathrm{d}x} = \tfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}

Parametric.

\tfrac{\mathrm{d}}{\mathrm{d}x}y^{n} = ny^{n-1}\tfrac{\mathrm{d}y}{\mathrm{d}x}

Implicit.

Stretch yourself

The normal to the curve $y = x^{2}$ at the point $P(1, 1)$ meets the curve again at $Q$ . Find the coordinates of $Q$ and the area of the triangle formed by the tangent at $P$ , the normal at $P$ and the $y$ -axis.

Hint — Find the normal, solve it simultaneously with $y = x^{2}$ . For the area, find where the tangent and normal cross the $y$ -axis — the triangle has its base on that axis.

Reveal solution

SolutionAt

P

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x = 2

, so the tangent is

y = 2x - 1

and the normal has gradient

-\tfrac12

y = -\tfrac12 x + \tfrac32

. Meeting the curve:

x^{2} = -\tfrac12 x + \tfrac32

, so

2x^{2} + x - 3 = 0

, giving

(2x + 3)(x - 1) = 0

. The new root is

x = -\tfrac32

, so

Q = \left(-\tfrac32, \tfrac94\right)

. The tangent meets the

y

-axis at

(0, -1)

and the normal at

\left(0, \tfrac32\right)

, a base of length

\tfrac52

. The third vertex

P

is at horizontal distance

1

from the axis, so the area is

\tfrac12 \times \tfrac52 \times 1 = \tfrac54

\blacksquare

Questions students ask

What is the difference between a tangent and a normal?

Both pass through the same point on the curve. The tangent has the curve’s gradient there; the normal is perpendicular to the tangent.

What if the tangent is horizontal?

Then

m = 0

and the tangent is

y = y_{1}

. The normal is vertical,

x = x_{1}

, since

-\tfrac{1}{0}

is undefined.

Do I have to eliminate the parameter for parametric curves?

No. Use

\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}

and work with

t

throughout.

Why does

y^{2}

differentiate to

2y\,\mathrm{d}y/\mathrm{d}x

y

is itself a function of

x

, so

y^{2}

is a composite and the chain rule supplies the extra factor.

Key takeaways

Tangent: gradient from the derivative at the point, then $y - y_{1} = m(x - x_{1})$ .
Normal: same point, gradient $-\tfrac{1}{m}$ .
Given a gradient, solve $\mathrm{f}^{\prime}(x) = m$ to find the point.
Parametric: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$ .
Implicit: differentiate every term, attaching $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to $y$ terms.