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Differentiation Differentiation

Differentiation - PowerPoint Presentation

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Differentiation - PPT Presentation

Introduction In this chapter you will learn how to differentiate equations that are given parametrically You will also learn to differentiate implicit functions which arent necessarily written in the form y ID: 360154

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Slide1

DifferentiationSlide2

IntroductionIn this chapter you will learn how to differentiate equations that are given parametricallyYou will also learn to differentiate implicit functions, which aren’t necessarily written in the form ‘y = …’These will be combined with calculating

normals and tangents as you have so far

We will also look at creating differential equations from information we are givenSlide3

Teachings for Exercise 4ASlide4

Differentiation4A

You can find the gradient of a curve given in parametric coordinates

When a curve is given in parametric equations:

Differentiate y and x with respect to the parameter t

We can then use the following calculation, a variation of the chain rule

 

 

 

 

 

 

 

 

 

 

Rewrite as a multiplication

Cancel

dts

Written differentlySlide5

Differentiation4A

You can find the gradient of a curve given in parametric coordinates

Find the gradient at the point P where t = 2, on the curve given by parametric equations:

 

 

 

 

 

 

 

Differentiate

Differentiate

 

 

 

 

 

 

 

 

 

Write in terms of t using

dy

/

dt

and

dx

/

dt

 This is the gradient function for the curve, but in terms of t

Sub in t = 2

Work outSlide6

Differentiation4A

You can find the gradient of a curve given in parametric coordinates

Find the equation of the normal at the point P where

θ = π

/6, on the curve with parametric equations:

 

 

 

We need to find the gradient at point P by differentiating and substituting

θ

in

We then need to find the gradient of the normal

We also need the coordinates of x and y at point P

Then we can use the formula y – y

1

= m(x – x

1

) to obtain the equation of the line

 

 

 

 

Differentiate

Differentiate

 

 

 

 

Sub in

dx

/

d

θ

and

dy

/

d

θ

Sub

θ

=

π

/

6

in to obtain the gradient at P

Work out the fraction

 

Gradient of the tangent at P:

 

Gradient of the normal at P:Slide7

Differentiation4A

You can find the gradient of a curve given in parametric coordinates

Find the equation of the normal at the point P where

θ = π

/6, on the curve with parametric equations:

 

 

 

We need to find the gradient at point P by differentiating and substituting

θ

in

We then need to find the gradient of the normal

We also need the coordinates of x and y at point P

Then we can use the formula y – y

1

= m(x – x

1

) to obtain the equation of the line

 

Gradient of the normal at P:

 

 

 

 

Sub in P =

π

/

6

Sub in P =

π

/

6

 

 

Work out

Work out

The coordinates of P are:

 Slide8

Differentiation4A

You can find the gradient of a curve given in parametric coordinates

Find the equation of the normal at the point P where

θ = π

/6, on the curve with parametric equations:

 

 

 

We need to find the gradient at point P by differentiating and substituting

θ

in

We then need to find the gradient of the normal

We also need the coordinates of x and y at point P

Then we can use the formula y – y

1

= m(x – x

1

) to obtain the equation of the line

 

Gradient of the normal at P:

The coordinates of P are:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sub in y

1

, m and x

1

Multiply by 2

Multiply by 5

(You could multiply by 10 in one step if confident)

Multiply the bracket out

Add 25√3

Divide by 2Slide9

Teachings for Exercise 4BSlide10

Differentiation4B

You can differentiate equations which are implicit, such as x

2 + y2

= 8x, or cos(x + y) =

siny

Implicit effectively means all the terms are mixed up, not necessarily in the form ‘y = …’

 This technique is useful as some equations are difficult to arrange into this form…

 

 

Differentiate the y term as you would for an x term

Write

dy

/

dx

after differentiating the y term

This is written differently

 The reason is that as the equation is not written as ‘y =…’, y is not a function of x

 

For example:

 

 

 

 

 

This is what happens when you differentiate an equation which starts ‘y =…’Slide11

Differentiation4B

You can differentiate equations which are implicit, such as x

2 + y2

= 8x, or cos(x + y) =

siny

 

 

Find

dy

/

dx

in terms of x and y for the following equation:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Differentiate each part one at a time

Isolate the terms with

dy

/

dx

in

Factorise by taking out

dy

/

dx

Divide by (3y

2

+ 3)

You now have a formula for the gradient, but in terms of x AND y, not just x

 

It would be very difficult to rearrange this into the form ‘y = …’Slide12

Differentiation4B

You can differentiate equations which are implicit, such as x

2 + y2

= 8x, or cos(x + y) =

siny

 

 

Find the value of

dy

/

dx

at the point (1,1) where:

 

 

 

 

 

 

 

 

 

 

 

Differentiate each one at a time – remember the product rule for the first term!!!!

 

 

 

Add 5, subtract 4y

2

You can substitute x = 1 and y = 1 now (or rearrange first)

 Do not replace the terms in

dy

/

dx

Divide by 8

 

 

 

Keep your workings tidy – you may need to use the product rule on multiple terms as well as factorise – show everything you’re doing!Slide13

Differentiation4B

You can differentiate equations which are implicit, such as x

2 + y2

= 8x, or cos(x + y) =

siny

 

 

Find the value of

dy

/

dx

at the point (1,1) where:

 

 

 

 

 

 

 

 

 

 

Give your answer in terms of e.

 

 

 

 

 

 

 

 

 

 

Differentiate (again watch out for the product rule!

Sub in x = 1 and y = 1

ln1 = 0 since e

0

= 1, cancelling the term out

Subtract

dy

/

dx

Factorise

Divide by (e

2

– 1)Slide14

Teachings for Exercise 4CSlide15

Differentiation4CYou can differentiate the general power function

ax where a is a constant

Differentiate

 

w

here a is a constant

 

 

 

 

 

 

Take natural logs

Use the power law

Differentiate each side

ln

a is a constant so can be thought of as just a number

This is important – as a is a constant it can be treated as a number

Multiply both sides by y

y

=

a

x

as from the first lineSlide16

Differentiation4CYou can differentiate the general power function

ax where a is a constant

Differentiate

 

 

 

 

Differentiate

 

 

Differentiate

Differentiate

 

where r is a constantSlide17

Teachings for Exercise 4DSlide18

DifferentiationYou can relate one rate of change to another. This is useful when a situation involves more than two variables.

4D

Given that the area of a circle A, is related to its radius r by the formula A =

πr

2, and the rate of change of its radius in cm is given by dr

/

dt

= 5, find

dA

/

dt

when r = 3

You are told a formula linking A and r

You are told the radius is increasing by 5 at that moment in time

You are asked to find how much the area is increasing when the radius is 3

This is quite logical – if the radius is increasing over time, the area must also be increasing, but at a different rate…

 

 

 

We are trying to work out

dA

/

dt

We have been told

dr

/

dt

in the question

You need to find a derivative that will give you the dA

and cancel the dr out

 

 

 

 

 

 

Differentiate

 

 

 

Replace

dr

/

dt

and

dA

/

dr

r

= 3Slide19

DifferentiationYou can relate one rate of change to another. This is useful when a situation involves more than two variables.

4D

The volume of a hemisphere is related to its radius by the formula:

 

The total surface area is given by the formula:

 

Find the rate of increase of surface area when the rate of increase of volume:

 

 

 

 

 

You need to use the information to set up a chain of derivatives that will leave

dS

/

dt

Rate of change of surface area over time

We are told

dV

/

dt

in the question so we will use it

We have a formula linking V and r so can work out

dV

/

dr

We have a formula linking S and r so can work out

dS

/

dr

This isn’t all correct though, as multiplying these will not leave

dS

/

dt

However, if we flip the middle derivative, the sequence will work!

 

 Slide20

DifferentiationYou can relate one rate of change to another. This is useful when a situation involves more than two variables.

4D

The volume of a hemisphere is related to its radius by the formula:

 

The total surface area is given by the formula:

 

Find the rate of increase of surface area when the rate of increase of volume:

 

 

 

 

You need to use the information to set up a chain of derivatives that will leave

dS

/

dt

 

 

 

 

 

 

 

 

 

 

Differentiate

Differentiate

Flip to obtain the derivative we want

 

 

 

 

Sub in values

You can write as one fraction

SimplifySlide21

Teachings for Exercise 4ESlide22

DifferentiationYou can set up differential equations from information given in context

4E

Differential equations can arise anywhere when variables change relative to one another

In Mechanics, speed may change over time due to acceleration

The rate of growth of bacteria may change when temperature is changed

As taxes change, the amount of money people spend will change

Setting these up involve the idea of proportion, a topic you met at GCSE…

Radioactive particles decay at a rate proportional to the number of particles remaining. Write an equation for the rate of change of particles…

Let N be the number of particles and t be time

 

 

The rate of change of the number of particles

The number of particles remaining multiplied by the proportional constant, k.

Negative

because the number of particles is decreasingSlide23

DifferentiationYou can set up differential equations from information given in context

4E

A population is growing at a rate proportional to its size at a given time. Write an equation for the rate of growth of the population

Let P be the population and t be time

 

 

The rate of change of the population

The population multiplied by the proportional constant k

 Leave positive since the population is increasingSlide24

DifferentiationYou can set up differential equations from information given in context

4E

Newton’s law of cooling states that the rate of loss of temperature is proportional to the excess temperature the body has over its surroundings. Write an equation for this law…

Let the temperature of the body be

θ

degrees and time be t

The ‘excess’ temperature will be the difference between the objects temperature and its surroundings –

ie

) One subtract the other

(

θ

θ

0

) where

θ

0 is the temperature of the surroundings

 

 

The rate of change of the objects temperature

The difference between temperatures (

θ

-

θ

0) multiplied by the proportional constant k

 Negative since it is a loss of temperatureSlide25

DifferentiationYou can set up differential equations from information given in context

4E

The head of a snowman of radius r cm loses volume at a rate proportional to its surface area. Assuming the head is spherical, and that the volume V =

4

/3πr

3

and the Surface Area A = 4

π

r

2

, write down a differential equation for the change of radius of the snowman’s head…

 

 

The first sentence tells us

The rate of change of volume

Is the Surface Area multiplied by the proportional constant k (negative as the volume is falling)

 

 

 

We want to know the rate of change of the radius r

We know

dV

/

dt

We need a derivative that will leave

dr

/

dt

when cancelled

 

 

Differentiate

 

Flip overSlide26

DifferentiationYou can set up differential equations from information given in context

4E

The head of a snowman of radius r cm loses volume at a rate proportional to its surface area. Assuming the head is spherical, and that the volume V =

4

/3πr

3

and the Surface Area A = 4

π

r

2

, write down a differential equation for the change of radius of the snowman’s head…

 

 

The first sentence tells us

The rate of change of volume

Is the Surface Area multiplied by the proportional constant k (negative as the volume is falling)

 

 

 

 

 

Differentiate

 

Flip over

 

 

 

 

 

 

 

 

 

 

Fill in what we know

We know A from the question

SimplifySlide27

SummaryYou have learnt how to differentiate equations given parametricallyYou can also differentiate implicit equationsYou can differentiate the general power function

ax

You can also set up differential equations based on information you are given