# Tangents and normals.

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Introduction

Tangents and normals

If you differentiate the equation of a curve, you will get a formula for the gradient of the curve. Before you learnt calculus, you would have found the gradient of a curve by drawing a tangent to the curve and measuring the gradient of this. This is because the gradient of a curve at a point is equal to the gradient of the tangent at that point.

Example:

Find the equation of the tangent to the curve y = x³ at the point (2, 8).

dy = 3x²

dx

Gradient of tangent when x = 2 is 3×2² = 12.

From the coordinate geometry section, the equation of the tangent is therefore:

y - 8 = 12(x - 2)

so y = 12x - 16

You may also be asked to find the gradient of the normal to the curve. The normal to the curve is the line perpendicular (at right angles) to the tangent to the curve at that point.

Remember, if two lines are perpendicular, the product of their gradients is -1.

So if the gradient of the tangent at the point (2, 8) of the curve y = x³ is 8, the gradient of the normal is -1/8, since -1/8 × 8 = -1.

Integration

Introduction

Integration is the reverse of differentiation.

If y = 2x + 3, dy/dx = 2

If y = 2x + 5, dy/dx = 2

If y = 2x, dy/dx = 2

So the integral of 2 can be 2x + 3, 2x + 5, 2x, etc.

Middle

Coordinate geometry

The distance between two points

The length of the line joining the points (x1, y1) and (x2, y2) is:

Example:

Find the distance between the points (5, 3) and (1, 4).

Distance = √ [(1 - 5)² + (4 - 3)²]

= √ [16 + 1] = Ö17

The midpoint of a line joining two points

The midpoint of the line joining the points (x1, y1) and (x2, y2) is:

[½(x1 + x2), ½(y1 + y2)]

Example:

Find the coordinates of the midpoint of the line joining (1, 2) and (3, 1).

[½(4), ½(3)] = (2, 1.5)

The gradient of a line joining two points

The gradient of the line joining the points (x1, y1) and (x2, y2) is:

y2 - y1

x2 - x1

Example:

Find the gradient of the line joining the points (5, 3) and (1, 4).

Gradient = 4 - 3 = 1 = -0.25

1 - 5 -4

The equation of a line using one point and the gradient

The equation of a line which has gradient m and which passes through the point (x1, y1) is:

y - y1 = m(x - x1)

Example:

Find the equation of the line with gradient 2 passing through (1, 4).

y - 4 = 2(x - 1)

y - 4 = 2x - 2

y = 2x + 2

Since m = y2 - y1

x2 - x1

The equation of a line passing through (x1, y1) and (x2, y2) is therefore:

y - y1 = y2 - y1

x - x1 x2 - x1

Sin, Cos, Tan

The sine, cosine and tangents of common angles:

30 | 45 | 60 | |

sin | 1/2 | 1/√2 | √3/2 |

cos | √3/2 | 1/√2 | 1/2 |

tan | 1/√3 | 1 | √3 |

These occur frequently and need to be remembered.

Quadrants and the 'cast' rule

Conclusion

So we can write arcsin(0.5) = 30º, 150º, 390º, ...

Solving Equations

Example:

Solve the equation sinø = 0.6428, for 0 < ø < 360º

therefore ø = arcsin(0.6428)

= 40º, 140º, 400º, ...

but the question asks for solutions between 0 and 360º, so the answer is 40º and 140º .

Graphing sinø, cosø and tanø

The following are graphs of sinø, cosø and tanø:

Points to note:

The graphs of sinø and cosø are periodic, with period of 360º (in other words the graphs repeat themselves every 360º .

The graph of cosø is the same as the graph of sinø, although it is shifted 90º to the right/ left. For this reason sinø = cos(90 - ø) or cosø = sin(90 - ø).

The graph of tanø has asymptotes. An asymptote is a line which the graph gets very close to, but does not touch. The red lines are asymptotes.

These graphs obey the usual laws of graph transformations.

Sine and cosine formulae

Sine and Cosine Formulae

sin x = sin (180 - x)

cos x = -cos (180 - x)

The Sine Rule

This works in any triangle:

a = b = c

sinA sinB sinC

alternatively, sinA = sinB = sinC

a b c

NOTE: the triangle is labelled as follows:

The Cosine Rule

c² = a² + b² - 2abcosC

which can also be written as:

a² = b² + c² - 2bccosA

This also works in any triangle.

The area of a triangle

The area of any triangle is ½ absinC (using the above notation)

This formula is useful if you don't know the height of a triangle (since you need to know the height for ½ base × height).

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

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