Transcript C4.4 Differentiation

A2-Level Maths:
Core 4
for Edexcel
C4.4 Differentiation
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Implicit differentiation
Contents
Implicit differentiation
Parametric differentiation
Differentiating exponential functions
Exponential growth and decay
Connected rates of change
Examination-style questions
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Implicit differentiation
When the equation of a curve is given in the form y = f(x) then
the variable y is said to be given explicitly in terms of the
variable x.
For example:
x
3
y = 2 x + 1 , y = 2 sin x ,
and y =
e +1
e 1
x
are explicit functions.
When neither x nor y is given explicitly in terms of the other
then the curve is said to be defined implicitly.
For example:
2
3
y + x = 5 , x sin y = 1,
and
1
x
2
+
1
y
2
= xy
are implicit functions.
It is not always easy, or even possible, to rearrange an implicit
function into an explicit form.
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Implicit differentiation
For this reason we need to develop a technique to differentiate
such functions in implicit form.
Differentiate y 2 + x 3 = 5 with respect to x.
Differentiating every term in the equation with respect to x
gives:
d
2
(y )+
dx
d
dx
3
(x ) =
d
(5 )
dx
where ddx is taken as an operator meaning ‘differentiate with
respect to x’.
The term in x and the constant can be differentiated directly to
give:
d
2
2
(y )+ 3x = 0
dx
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Implicit differentiation
To differentiate the term in y with respect to x we have to use
the chain rule.
d
2
(y ) =
dx
d
2
( y )×
dy
dy
= 2y
dx
dy
dx
This would normally be done in a single step so:
d
(y )+
dx
becomes
d
2
3
(x ) =
dx
2y
dy
d
(5 )
dx
2
+ 3x = 0
dx
We can now divide through by 2y and rearrange to find
dy
dx
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=
3x
dy
dx
:
2
2y
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Implicit differentiation
Differentiate x 3  2 y 2 + 7 y  5 x = 8 with respect to x.
Differentiating term by term with respect to x gives:
3x2
dydx
dy
= 0)
 4y
+ =7 2(3 t+52 )(6
dxdt
dx
Now rearrange to collect the terms in
7
dy
dx
dy
 4y
dy
dy
dx
together:
= 5  3x
2
(7  4 y ) = 5  3 x
2
5  3x
2
dx
dx
dy
dx
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=
7  4y
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Implicit differentiation
In some cases, we might also need to use the product rule. For
example:
Differentiate x 2 y = sin x + 3 y with respect to x.
Differentiating term by term with respect to x:
d
2
x y=
dx
d
dx
si n x +
d
3y
dx
The first term is treated as a product to give:
x
2
d
y+ y
dx
Using
d
dx
( uv ) = u
dv
dx
+v
2
x = cos x + 3
dx
dy
dx
du
dx
x
2
dy
dx
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d
+ 2 xy = co s x + 3
dy
dx
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Implicit differentiation
Rearrange to collect the terms in
x
2
dy
3
dx
dy
dy
dy
dx
together:
= co s x  2 xy
dx
( x  3 ) = co s x  2 xy
2
dx
dy
dx
=
co s x  2 xy
x 3
2
Once we have differentiated a curve that has been defined
implicitly, we can find the equation of the tangent or the normal
to the curve at a given point.
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Implicit differentiation
Find the equation of the tangent to the curve x2 + y2 – xy = 7
at the point (2, –1).
Differentiating with respect to x gives:
2x + 2 y
dy
x
dx
dy
 y=0
dx
Rearranging to collect the terms in
2y
dy
dx
dy
x
dy
dy
dx
together:
= y  2x
dx
(2 y  x ) = y  2 x
dx
dy
dx
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=
y  2x
2y  x
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Implicit differentiation
At the point (2, –1), x = 2 and y = –1:
dy
=
dx
=
1  4
2  2
5
4
=
5
4
The gradient of the tangent at (2, –1) is therefore
5
4
.
We can find the equation of the tangent using the coordinates:
y +1=
5
4
( x  2)
Using y – y1 = m(x – x1)
4 y + 4 = 5( x  2 )
4 y  5 x + 14 = 0
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Implicit differentiation
Find the equation of the normal to the curve 2x2 – (y + 3)2 = 1
at the point (5, 4).
Differentiating with respect to x gives:
4 x  2( y + 3 )
dy
=0
dx
2( y + 3 )
dy
= 4x
dx
( y + 3)
dy
= 2x
dx
dy
dx
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=
2x
y+3
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Implicit differentiation
We need the gradient at the point (5, 4):
dy
=
2(5 )
4+3
dx
=
10
7
The gradient of the normal at (5, 4) is therefore – 170 .
The equation of the normal is given by:
y4=–
7
10
( x  5)
Using y – y1 = m(x – x1)
10 y – 40 = – 7 x + 35
1 0 y + 7 x – 75 = 0
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Parametric differentiation
Contents
Implicit differentiation
Parametric differentiation
Differentiating exponential functions
Exponential growth and decay
Connected rates of change
Examination-style questions
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Parametric differentiation
Suppose we want to find dy for a curve that has been defined
dx
parametrically.
2
y = 3t
For example:
x = 2t + 5
We can differentiate both of these equations with respect to the
parameter t to give:
dx
and
=2
dt
dy
dx
dy
= 6t
dt
can be found using the chain rule:
dy
dx
=
dy
dt
×
dt
=
dx
=
dy
dx
dt
dt
6t
2
= 3t
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Parametric differentiation
In general, then, to differentiate a pair of parametric equations
we can use the chain rule in the form:
dy
dx
Find
dy
,
dx
=
dy
dx
dt
dt
in terms of t, for the curve defined by the parametric
equations x = cos 2t and y = sin t.
Differentiating each equation with respect to t gives:
dx
=  2 sin 2 t
dt
dy
= co s t
dt
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Parametric differentiation
dy
=
dx
=
dy
dx
dt
dt
co s t
 2 sin 2 t
This can be simplified further using the double angle formula
for sin 2t:
dy
dx
=
co s t
 2(2 sin t co s t )
=
1
4 sin t
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=  41 cosec t
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Parametric differentiation
A curve is defined by the parametric equations x = (3t + 2)2
and y = 2t3 + 9.
Find dd yx in terms of t and hence find the coordinates of the
Differentiating with respect to x gives
points where the gradient of the curve is –1.
dx
= 2(3 t + 2 )(3 ) = 6(3 t + 2)
dt
dy
= 6t
2
dt
dy
=
dx
dy
dx
dt
dt
6t
=
2
6(3 t + 2 )
=
t
2
3t + 2
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Parametric differentiation
The gradient of the curve is –1 when:
t
2
3t + 2
= 1
t = 3t  2
2
2
t + 3t + 2 = 0
( t + 1)( t + 2) = 0
So the gradient of the curve is –1 when t = –1 and when t = –2.
When t = –1:
x = (–3 + 2)2 = 1
When t = –2:
x = (–6 + 2)2 = 16 and y = 2(–2)3 + 9 = –7
and y = 2(–1)3 + 9 = 7
 The gradient of the curve is –1 at (1, 7) and (16, –7).
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Differentiating exponential functions
Contents
Implicit differentiation
Parametric differentiation
Differentiating exponential functions
Exponential growth and decay
Connected rates of change
Examination-style questions
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Differentiating exponential functions
Suppose we want to differentiate a general exponential
function of the form
y = ax
where a is a constant.
Take the natural log of both sides:
x
ln y = ln ( a )
ln y = x ln a
Now differentiating with respect to x gives:
1 dy
= ln a
y dx
dy
= y ln a
dx
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Differentiating exponential functions
Since y = ax we can substitute ax for y. So in general:
x
If y = a th e n
dy
x
= a ln a
dx
For example:
d
(3 ) = 3 ln 3  1 .0 9 8 6(3 )
x
x
x
dx
d
(8 ) = 8 ln 8  2 .0 7 9 4(8 )
x
x
x
dx
( 61 )  = ( 61 )

dx
d
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x
x
ln
1
6

  1 .7 9 1 8 ( 61 )
x

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Exponential growth and decay
Contents
Implicit differentiation
Parametric differentiation
Differentiating exponential functions
Exponential growth and decay
Connected rates of change
Examination-style questions
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Exponential growth
Exponential growth occurs when a quantity increases at a
rate that is proportional to its size.
In other words, when a quantity grows exponentially, the larger
it becomes, the quicker it grows.
Examples of quantities that grow exponentially include:
the number of micro-organisms in a culture dish,
investments with a fixed compound interest rate,
population growth.
For example, a bacteria colony starts with 100 bacteria and
doubles every minute.
The number of bacteria can be shown in a table.
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Exponential growth
Time (mins)
0
1
2
3
…
t
Number of
bacteria
100
200
400
800
…
100 × 2t
So if N is the number of bacteria after time t we have:
N = 100 × 2t
More generally, if b is the number of bacteria when t = 0,
N = b2t
In practice it is more usual to express exponential growth in
terms of the constant e.
We can write the number 2 in terms of e as eln 2, and so the
formula becomes
N = bet ln 2
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or
N = be 0.693t
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Exponential growth
In general, exponential growth can be modelled by the function
f(t) = Aekt
where t is time,
A is the original quantity (the quantity when t = 0),
f(t) is the quantity after time t and
k is a positive constant (the growth rate)
An exponential growth
curve has the following
basic shape:
f(t)
y = Aekt
A
0
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t
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Exponential decay
Exponential decay occurs when a quantity decreases at a
rate that is proportional to its size.
In other words, when a quantity decays exponentially, the
smaller it becomes, the more slowly it decays.
Examples of quantities that decay exponentially include:
the number of atoms in a radioactive isotope,
the value of a car as it depreciates,
the rate at which an object cools (when the external
temperature is constant).
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Exponential decay
In general, exponential decay can be modelled by the function
f(t) = Ae–kt
where t is time,
A is the original quantity (the quantity when t = 0),
f(t) is the quantity after time t, and
k is a positive constant (the decay rate).
An exponential decay
curve has the following
basic shape.
f(t)
A
0
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y = Ae–kt
t
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Exponential decay
Forensic scientists can predict a recently murdered victim’s
time of death from the temperature of the body.
They do this by applying Newton’s law of cooling:
d = ae–kt
where d is the temperature difference (between the body and
its surroundings), a and k are constants and t is the time that
has passed since the body started to cool. For example,
A body is discovered at 8.30 pm. The body’s temperature is
recorded as 30°C and room temperature as 20°C. One hour
later, the temperature of the body is 29°C.
Assuming that the room temperature remains constant
throughout, estimate the time of death.
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Exponential decay
Let t = 0 at 8.30 pm.
d = 30 – 20 = 10
And using d = ae–kt:
10 = ae–k(0) = ae0 = a
When t = 1, one hour later:
d = 29 – 20 = 9
Using d = ae–kt:
9 = ae–k(1) = ae–k
But a = 10 so,
9 = 10e–k
e–k = 0.9
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Exponential decay
We can solve e–k = 0.9 by taking natural logs on both sides:
ln (e–k ) = ln 0.9
–k = ln 0.9
–k = –0.105 (to 3 s.f.)
k = 0.105
Substitute these values of k and a into d = ae–kt:
d = 10e–0.105t
If we assume that the murder victim had a normal body
temperature of 36.9 °C when she died and that the room
temperature was 20°C, then at the time of death:
d = 36.9 – 20 = 16.9
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Exponential decay
Substituting d = 16.9 into d = 10e–0.105t gives
16.9 = 10e–0.105t
Solving this equation for t will give the time since the victim died.
e–0.105t = 1.69
–0.105t = ln 1.69
t=
ln 1 .6 9
 0 .1 0 5
t = –5.00 (to 3 s.f.)
t = 0 at 8.30 pm, so t = –5 at 3.30 pm.
 The victim died at about 3.30 pm.
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Connected rates of change
Contents
Implicit differentiation
Parametric differentiation
Differentiating exponential functions
Exponential growth and decay
Connected rates of change
Examination-style questions
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Rates of change
When we talk about rates of change we are usually talking
about the rate at which a variable changes with respect to time.
Suppose, for example, that a spherical balloon is slowly being
inflated.
There are several changes we could measure, such as:
The rate at which the radius r changes,
dr
dt
The rate at which the surface area A changes,
The rate at which the volume V changes,
dA
dt
dV
dt
We can connect these rates of change using the chain rule.
For example, the rate of change of the surface area is
connected to the rate of change of the radius by
dA
dt
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=
dA
dr
×
dr
dt
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Rates of change
We can find
dA
dr
by differentiating the formula for the surface
area of a sphere with respect to the radius.
A=
4πr2
dA
so
= 8 r
dr
So, the rate of change of the surface area is connected to the
rate of change of the radius by
dA
dt
= 8 r ×
dr
dt
If the surface area of the balloon is increasing at a rate of
15 cm2 s–1, find the rate at which the radius of the balloon
is increasing at the moment when the radius is 4 cm.
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Connected rates of change
We are given that
Using
dA
dt
= 8 r ×
dA
dt
dr
dt
= 15 and we want to find
dr
dt
when r = 4.
we have:
1 5 = 3 2 ×
dr
dt
dr
dt
=
15
3 2
= 0.149
So, the radius is increasing at a rate of 0.149 cm s–1 (to 3 s.f.)
at the moment when the radius of the balloon is 4 cm.
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Connected rates of change
If the air in the fully-inflated balloon is released at a rate of
30 cm3 s–1, find the rate at which the surface area is
decreasing at the moment when the radius is 5 cm.
We are given that
dV
dt
= –30 and we want to find
dA
dt
when r = 5.
Using the chain rule, the rate of change of the volume is
connected to the rate of change of the radius by:
dV
dt
V =
4
3
r 
3
dV
= 4 r
=
dV
×
dr
dr
dt
2
dr
So
dV
dt
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= 4 r ×
2
dr
dt
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Connected rates of change
Here
dV
dt
= –30 and r = 5:
3 0 = 1 0 0  ×
dr
dt
dr
30
3
=
=
dt
1 0 0
1 0
We can now find
dA
dt
using
dA
dt
dA
= 8 r ×
dt
dr
dt
4
= 4 0 ×
with r = 5:
3
1 0
= –12
So, the surface area is decreasing at a rate of 12 cm2 s–1 at
the moment when the radius of the balloon is 5 cm.
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Connected rates of change
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Connected rates of change
Contents
Implicit differentiation
Parametric differentiation
Differentiating exponential functions
Exponential growth and decay
Connected rates of change
Examination-style questions
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Examination-style question 1
The diagram shows the curve defined by the parametric
equations x = 2t2 and y = t2 + 2t.
y
P
O
Q
x
a) Find the gradient of the curve at the point P where t = 1.
b) Find the equation of the normal at the point P.
c) The normal to the curve at the point P cuts the curve again
at the point Q. Find the coordinates of the point Q.
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Examination-style question 1
dx
a)
= 4t
dt
dy
= 2t + 2
dt
Using
dy
=
dx
dy
dx
dt
dt
dy
gives:
=
2t + 2
4t
dx
when t = 1:
dy
dx
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=
2
=
t +1
2t
=1
2
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Examination-style question 1
b) The gradient of the normal to the curve at the point P is –1.
Also when t = 1, x = 2
y=3
If the gradient of the normal is –1 and it passes through the
point (2, 3) then its equation is given by:
y  3 = ( x  2 )
y+x5=0
c) Substituting x = 2t2 and y = t2 + 2t into the equation for the
normal gives:
t + 2t + 2t  5 = 0
2
2
3t + 2t  5 = 0
2
( t  1)(3 t + 5 ) = 0
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Examination-style question 1
The normal therefore cuts the curve when t = 1 and when t = 35 .
When t =
5
3
: x = 2( 35 ) 2
=5
5
9
y = ( 35 ) + 2( 35 )
2
=
5
9
The coordinates of the point Q are therefore (5 59 ,  59 ).
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Examination-style question 2
The mass, m grams, of a sample of radioactive iodine decays
according to the formula
m = 15 e
 0.083 t
where t is the number of days after it is first observed.
a) What is the initial mass of the sample?
b) Sketch the graph of m against t.
c) What is the mass of the sample after 2 days?
d) Calculate the time it takes for the sample to halve its
original mass.
e) Find the rate at which the sample is decaying when t = 5.
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Examination-style question 2
a) When t = 0,
m = 15e0
= 15
So, the initial mass of the sample is 15 g.
b) The graph of m = 15e–0.083t
will be an exponential decay
curve passing through the
point (0, 15).
m
15
t
c) When t = 2,
m = 15e–0.083 × 2
= 12.71 (to 2 d.p.)
So, the mass of the sample after 2 days is 12.71 g (to 2 d.p.).
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Examination-style question 2
d) The initial mass is 15 g, so we need to find the time it takes
for the sample to reach 7.5 g.
That is, when 7.5 = 15e–0.083t
0.5 = e–0.083t
Take logs:
ln 0.5 = ln e–0.083t
ln 0.5 = –0.083t
t=
ln 5
 0 .0 8 3
= 8.35 (to 2 d.p.)
The sample takes 8.35 days (to 2 d.p.) to halve its mass.
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Examination-style question 2
d) The rate at which the sample is decaying is given by
dm
.
dt
m = 15 e
dm
 0.083 t
=  0 .0 8 3 × 1 5 e
 0 .0 8 3 t
dt
When t = 5
dm
=  1.245 e
 0.083 t
=  1 .2 4 5 e
 0 .0 8 3 × 5
dt
=  0 .8 2 (to 2 d .p .)
Therefore, when t = 5, the sample is decaying at a rate of 0.82
grams per day (to 2 d.p.).
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Examination-style question 3
The height of a conical container is twice its radius, r cm, as
shown in the diagram.
r
2r
Liquid is poured into the container
at a rate of 5 litres per minute.
If x cm is the depth of the liquid at
time t minutes, write an expression
for the rate at which the depth is
increasing when x = 2 cm.
(The volume of a cone of radius r
and height h is given by 31  r 2 h ).
The volume of the liquid in the container at time t is given by
v=
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1
3
r x
2
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Examination-style question 3
Since the radius of the container is half its depth this can be
written in term of x as
2
 x
v=   x
3 2
1
 v=

dv
x
3
12
=
x
2
4
dx
The liquid enters the container at a rate of 5 litres per minute
so
dv
=5
dt
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Examination-style question 3
Using the chain rule
dv
dv
=
dt
×
dx
dx
dt
So
5=
x
2
×
4

dx
=
dt
dx
dt
20
x
2
At the instant when x = 2 cm the rate at which the liquid is
entering the container is
dx
dt
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=
20
4
=
5

cm /m in
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