12 Tangents and Gradients

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12: Tangents and Gradients

Christine Crisp

Teach A Level Maths

Vol. 1: AS Core Modules


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Gradients and Tangents

Module C1

AQA

Edexcel

OCR

MEI/OCR

Module C2

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We need to be ableto find these points

using algebra

e.g. Find the coordinates of the points on the curvewhere the gradient equals 4783

xxy

Gradient of curve= gradient of tangent= 4

Points with a Given Gradient


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Gradient is 4

dx

dy

e.g. Find the coordinates of the points on the curve

where the gradient is 4783 xxy

The gradient of the curve is given bySolution:

83 2

x

483

2x

dx

dyxxy 78

3

4dx

dy

123 2

x

Quadrat ic equation

w i th no l inear x -te rm

2x42

x

7)2(8)2(2 3

yx

7)2(8)2(2 3yx

1

15
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)1,2(

)15,2( x

x


The points on with gradient 4783 xxy


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SUMMARY

To find the point(s) on a curve with a givengradient: )(xfy

let equal the given gradientdx

dy

solve the resulting equation

find the gradient functiondx

dy


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Find the coordinates of the points on the curves withthe gradients given

where the gradient is -2342

xxy1.

where the gradient is 320213 23 xxxy2.

Ans: (-3, -6)

Ans: (-2, 2) and (4, -88)

( Watch out for the common factor in thequadratic equation )

Exercises


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Increasing and Decreasing Functions

An increasing function is one whosegradient is always greater than or equal tozero.

for all values of x0

dx

dy

A decreasing function has a gradient thatis always negative or zero.

for all values of x0dxdy


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43 2 xdxdy

e.g.1 Show that is an increasingfunction

xxy 43

xxy 43

Solution:

a positive number ( 3 ) a perfect

square ( which is positive or zero forall values of x, and

for all values of x0dx

dy

is the sum ofdx

dy

a positive number ( 4 )

so, is an increasing functionxxy 43


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962

xxdx

dy

Solution: xxxy 93 2331

e.g.2 Show that is an

increasing function.xxxy 93

23

31

To show that is never negative ( inspite of the negative term ), we need to

complete the square.

962

xx

22)3(96 xxx

xxxy 93 23

31

is an increasing function.


dy

Since a square is always greater than or equalto zero,


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xxy 43

The graphs of the increasing functions

and arexxxy 93 2331

xxy 43

xxxy 93 23

3

1

and


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Exercises

xxxy 52 2331 2. Show that is an increasingfunction and sketch its graph.

1. Show that is a decreasing function andsketch its graph.

3

xy

Solutions are on the next 2 slides.


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1. Show that is a decreasing function andsketch its graph.

Solutions

3

xy

23x

dx

dy

Solution: . This is the product of a

square which is always and a negative number,

0dxdy

0

so for all x. Hence is a

decreasing function.

3xy

3xy


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Solutions

xxxy 52 23

3

1

2. Show that is an increasing

function and sketch its graph.

542

xxdx

dySolution: .

Completing the square: 1)2( 2 xdx

dy

which is the sum of a square which is 0and a positive number. Hence yis an increasingfunction.

xxxy 52 23

31


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(-1, 3)x

123 23

xxxySolution:

263 2

xxdx

dy

At x= 1

2)1(6)1(3 2

m

c)1(53

cmxy cxy 5

c2

So, the equation of the tangent is 25xy

5

Gradient= -5

(-1, 3)on line:

The gradient of a curve at a point and the gradient ofthe tangent at that point are equal

The equation of a tangent

e.g. 1 Find the equation of the tangent at the point

(-1, 3)on the curve with equation123

23

xxxy


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An Alternative Notation

The notation for a function of xcan be usedinstead of y.

)(xf

When is used, instead of using for thegradient function, we write dx

dy)(xf

)(/

xf ( We read this as f

dashed x )

132)( 23

xxxxfe.g.

343)( 2/ xxxf

This notation is helpful if we need to substitute for x.

173)2(4)2(3)( 2/

f 2
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)1(432)( 23

xxxxf

3)2(4)2(3)2( 2/

fm

c)2(72

Solution: To use we need to know y atthe pointas well as xand m

cmxy

c12

So, the equation of the tangent is 127 xy

3812 7

4)2(3)2(2)2()2( 23 fy

24688

343)( 2/

xxxfFrom (1),

cmxy cxy

7

(2, 2)on the line

e.g. 2 Find the equation of the tangent where x= 2onthe curve with equation where)(xfy

432)( 23 xxxxf
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if the y-value at the point is not given,substitute the x -value into the equation of the

curve to findy

SUMMARY

To find the equation of the tangent at a point onthe curve :)(xfy

find the gradient function ))(( / xfordx

dy

substitute the x-value into

to find the gradient of the tangent, m

))(( /

xfor

dx

dy

substitute for y, mand xinto to find ccmxy


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Exercises

115 xyAns:

Ans:

Find the equation of the tangent to the curve

32 23

xxxy

1.

at the point (2, -1)

Find the equation of the tangent to the curve2.

at the point x= -1, where)(xfy

12 xy

6113)( 23 xxxxf

d d


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G d d


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The following slides contain repeats ofinformation on earlier slides, shown withoutcolour, so that they can be printed and

photocopied.

For most purposes the slides can be printedas Handouts with up to 6slides per sheet.


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SUMMARY

To find the point(s) on a curve with agiven gradient: )(xfy

let equal the given gradientdxdy

solve the resulting equation

find the gradient functiondx

dy


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Gradient is 4

dx

dy

e.g. Find the coordinates of the points on the curve

where the gradient is 4783 xxy

The gradient of the curve is given bySolution:

83 2

x

483 2

x

dx

dyxxy 78

3

4dxdy

123 2x

Quadrat ic equationw i th no l inear x -term

2x42

x7)2(8)2(2

3

yx

7)2(8)2(2 3yx

1

15


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Increasing and Decreasing Functions

An increasing function is one whosegradient is always greater than or equal tozero.

for all values of x0

dx

dy

A decreasing function has a gradient thatis always negative or zero.

for all values of x0dxdy


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962

xxdx

dy

Solution: xxxy 93 2331

e.g. Show that is an

increasing function.xxxy 93

23

31

To show that is never negative ( inspite of the negative term ), we need to

complete the square.

962

xx

22)3(96 xxx

xxxy 93 23

31

is an increasing function.


dy

Since a square is always greater than or equalto zero,


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if the y-value at the point is not given,substitute the x -value into the equation of thecurve to find y

To find the equation of the tangent at a point onthe curve :)(xfy

find the gradient function ))(( / xfordx

dy

substitute the x-value into

to find the gradient of the tangent, m

))(( /

xfor

dx

dy

substitute for y, mand xinto to find ccmxy

SUMMARY


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(-1, 3)x

123 23

xxxySolution:

263

2

xxdx

dy

At x= -1

2)1(6)1(3 2

m

c)1(53

cmxy cxy 5

c2


5

Gradient= -5

(-1, 3)on line:

e.g. 1 Find the equation of the tangent at the point

(-1, 3)on the curve with equation123

23

xxxy

The equation of a tangent


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)1(432)( 23

xxxxf

3)2(4)2(3)2( 2/

fm

c)2(72

Solution: To use we need to know y atthe pointas well as xand m

cmxy

c12


3812 7

4)2(3)2(2)2()2( 23 fy24688

343)( 2/

xxxfFrom (1),

cmxy cxy

7

(2, 2)on the line

e.g. 2 Find the equation of the tangent where x= 2onthe curve with equation where)(xfy

432)( 23 xxxxf

12 Tangents and Gradients

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