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