calculus 11.2

GradientLimits Derivatives of functionsFirst principles Composite functions

D I F F E R E N T I A L C A L C U L U S

Second derivatives Stationary points Graph sketchingTangents and NormalsHome

If I have seen further…..

…. it is because I have stood on the shoulders of giants.

ISaac Newton to Robert Hooke in 1675

ISAAC NEWTON 1643 -1727




1




Introduction

The concept of gradient is so important for a thorough understanding of differential calculus.

The two-point formula Gradients of secants

The graphs of some linear functions are less steep >

… and others have negative slopes >

The graphs of some linear functions are steep with a positive slope >

Gradient is a measure of this steepness or slope.

It is defined as the ratio of the rise to the run.

The gradient of the green function is 2.

Check the gradient of the red function is 1/3

And the blue line has a gradient of -1

Gradient =

rise

run

2




The given line passes through the points P and Q where:

P = ( 2, 3 )

Q = ( 1, 1 )

In the interval PQ:

rise = 3 - 1

run = 2 - 1

P

Q

X

Xrun

rise

Gradient = rise

run

=3 −12 −1

=21

= 2

Generally, the straight line passing through the two points; ,

Has a gradient given by:

( x

1, y

1)

( x

2, y

2)

Gradient = y2−y

1

x2−x

1

Introduction The two-point formula Gradients of secants

3




y = x2

Gradient of secant PQ =y

2−y

1

x2−x

1

=9 −13 −1

=4

Here is the graph of the function

And here is a secant PQ

where and .

y =x2

P = (1 ,1 )

Q = (3 , 9 )

Using the formula for the gradient of a line

through two points, we have:

Introduction The two-point formula Gradients of secants

4




Introduction Case study

The gradient of the graph of a linear function is easy to find; we can use the two-point formula as shown in the previous section. But how can we find the gradient at different points on a non-linear function, such as the one shown here?

Clearly the parabola gets steeper as the x-values increase…

… but how can we measure the actual gradient at any particular point on the curve?

>

DEFINITION The gradient at a point P on the

curve is defined as the gradient of the tangent to the curve at that point.

P1

P2

Gradient of the function at the point ( 3, 9)

Gradient of the function at the point ( 2, 4)

y =x2

Limiting processAn algebraic approach

5




y =x2

P (2 , 4)

Introduction Case study

By definition this means finding the gradient of the tangent to the curve at that point… A ( 4 , 16 )Our goal here is to find exactly the gradient of the function at the point y =x2

P (2, 4)

As a first approximation, consider the secant AP

Gradient AP =y2 −y1x2 −x1

=16−44−2

=122

= 6

As a second approximation, consider the secant BP

Gradient BP =y2 −y1x2 −x1

=9−43−2

=51

= 5

B (3, 9)

P

C

Gradient CP =y2 −y1x2 −x1

=6.25−42.5−2

=2.250.5

= 4.5

An algebraic approach Limiting process

For a third approximation we will need to zoom in and consider secant CP….Note now how close the tangent is to the curve

6




y =f(x)

Introduction An algebraic approachCase study

Our goal here is to find exactly the gradient of the function at the point

By definition this means finding the gradient of the tangent to the curve at that point…

As a first approximation, consider the secant A x+h, f (x+h)( )

P x , f (x )( )

y =x2 P (2, 4)

A P

Gradient AP =y2 −y1x2 −x1

=f (x +h)−f(x)

x+h−x

=f (x +h)−f(x)

h

h

x + h

Limiting process

Gradient at P =lim

h→ 0

f (x +h)−f(x)

h

7




y =f(x)


Using first principles find the derived function for

A x+h, f (x+h)( )

P x , f (x )( )

y =3x2

h

x + h

Limiting process

y = 3x2

dy

dx=lim

h→ 0

f (x +h)−f(x)

h

=limh→ 0

3(x+h)2 −3x2

h

=limh→ 0

3(x2 +2xh+h2 )−3x2

h

=limh→ 0

3x2 +6xh+3h2 −3x2

h

=limh→ 0

6xh+3h2

h

=limh→ 0

h(6x+3h)

h

=limh→ 0

6x+3h

dy

dx=6x

8





Using first principles find the derived function for y =5x2

Limiting process

y = 5x2

dy

dx=lim

h→ 0

f (x +h)−f(x)

h

=limh→ 0

5(x+h)2 −5x2

h

=limh→ 0

5(x2 +2xh+h2 )−5x2

h

=limh→ 0

5x2 +10xh+5h2 −5x2

h

=limh→ 0

10xh+5h2

h

=limh→ 0

h(10x+5h)

h

=limh→ 0

10x+5h

dy

dx=10x

8

y =4x3

y = 4x3

dy

dx=lim

h→ 0

f (x +h)−f(x)

h

=limh→ 0

4(x+h)3 −4x3

h

=limh→ 0

4(x3 +3x2h+3xh2 +h3 )−4x3

h

=limh→ 0

4x3 +12x2h+12xh2 + 4h3 −4x3

h

=limh→ 0

12x2h+12xh2 + 4h3

h

=limh→ 0

h(12x2 +12xh+ 4h2 )

h

=limh→ 0

12x2 +12xh+ 4h2

dy

dx=12x2

Using first principles find the derived function for SUMMARY

y = 5x2

dy

dx=10x

y = 4x3

dy

dx=12x2

y = 2x3

dy

dx=6x2

y = 10x4

dy

dx=40x3

y = axn

dy

dx= anxn−1




A tangent to a curve is a straight line touching the curve at a single point.

Introduction

9

Equation of a tangent Equation of a normal

This diagram shows the TANGENT to the curve at the point (1, -2)

y = x3 −2x−1

FIGURE 1

TANGENT

This diagram shows the NORMAL to the curve at the point (1, -2)

y = x3 −2x−1

FIGURE 2

NORMAL

FACT SHEET

• A TANGENT touches a curve at a single point

(x1, y

1)

• Its equation is given by y−y

1=m

1(x−x

1)

• The NORMAL at a point is perpendicular to the tangent at that point. (It’s at 90 degrees)

(x

1, y

1)

• It’s equation is given by y−y

1=m

2(x−x

1)

• It’s gradient, , is given by the gradient or derived function at the value

x

1

m

1

• It’s gradient, , is found using the gradient of the tangent and the fact that

m

1m

2=−1

m

2

A normal is a straight line, perpendicular to the tangent.

ACTIVITY 1




Equation of a tangent

dy

dx = 2x−3

dy

dx=2 ×2 −3

= 1

y −y1= m(x −x

1)

y −0 = 1(x −2 )

y = x−2

Find the equation of the tangent to the curve at the point (2, 0) y = x2 −3x+ 2

y = x2 −3x+ 2

Introduction

METHOD

10

2 Find the gradient of the function at x=2

3 Substitute the gradient, m = 1 and the coordinates of the point into the point/

gradient form of a straight line.

1 Differentiate to obtain the gradient function

Equation of a normal ACTIVITY 1




dy

dx = 2x−3

dy

dx=2 ×2 −3

= 1

y = x2 −3x+ 2

y −y1= m(x −x

1)

y −0 = −1(x −2 )

y = −x+2

Find the equation of the normal to the curve at the point (2, 0) y = x2 −3x+ 2

Introduction Equation of a normalEquation of a tangent

METHOD

2 Find the gradient of the function at x=2

4 Substitute the gradient, m = -1 and the coordinates of the point into the point/

gradient form of a straight line.

3 Find the gradient of the normal at x=2

m1m

2=−1

m2 = -1

11

1 Differentiate to obtain the gradient function

ACTIVITY 1




Introduction Equation of a normalEquation of a tangent ACTIVITY

QUESTION 1

f(x)= 2x3 −x+ 3

f '(x) = 6x2 −1

f '(0) = −1

y −y1= m(x −x

1)

y −3 = −1(x −0 )

y = −x+ 3

QUESTION 2

y = 2x2 −1

dy

dx = 4x

At x=−1 m = −4

y −y1= m(x −x

1)

y −1 = −4(x + 1)

y = −4x− 3

QUESTION 3

f(t)= 4 t

= 4 t 0.5

f '(t) = 2 t −0.5

f '(1) = 2

y −y1= m(x −x

1)

y −4 = 2(x + 1)

y = 2x+6

QUESTION 4

F(x)=12x

= 12 x−1

F'(x) = -12 x−2

f '( 4 ) = −0.75

y −y1= m(x −x

1)

y −3 = −0.75(x −4 )

y = −0.75x+7

QUESTION 5

v = 20 t−5 t 2

dvdt

= 20 −10 t

At t =0 m = 20

y −y1= m(x −x

1)

y −0 = 20(x −0 )

y = 20 x

QUESTION 6

f(x)= x2 +1( )4

f '(x) = 4 x2 +1( )3×2 x

f '(x) = 8 x x2 +1( )3

f '(1) = 64

y −y1= m(x −x

1)

y −16 = 64(x −1)

y = 64 x− 48



Second derivatives Stationary points Graph sketchingTangents and NormalsHome 2




x

MAXIMUMSTATIONARY POINT A

x

x x

1 x

2 x

3

x

1 x

2 x

3

x

1 x

2 x

3

y =f(x)

y = ′f (x)

y = ′′f (x)

x

1, f(x

1)( )

MINIMUMSTATIONARY POINT B

LOCATED AT:

′f (x

1)=0

FIRST DERIVATIVE IS ZERO

′′f (x

1) < 0

SECOND DERIVATIVE IS NEGATIVE

INFLEXION POINT C

0

0

0

x

3, f(x

3)( )

LOCATED AT:

′f (x

3)=0

FIRST DERIVATIVE IS ZERO

′′f (x

3) > 0

SECOND DERIVATIVE IS POSITIVE

x

2, f(x

2)( )

LOCATED AT:

′′f (x

1)=0

SECOND DERIVATIVE IS ZERO

A

C

B




This diagram shows the STATIONARY POINTS on the graph of the cubic function

y = x3 −2x−1

FIGURE 2

This diagram shows the STATIONARY POINT on the graph of the quadratic function

y = - x2 −2x+1

FIGURE 1 FACT SHEET

• Stationary points lie on the graphs of functions where the gradient is zero. (The tangents to the curve are horizontal at these points; the function is neither increasing nor decreasing.)

• The stationary point in figure 1 is called a maxima.

• Figure 2 shows a function having both a maxima and minima




0 = 2x−3

2x = 3

x = 1.5

When x =1.5

y = 1.52 −3 ×1.5 + 2

y = −0.25

y = x2 −3x+ 2

Stationary point (1.5,−0.25 )

Find the stationary point on the graph of the function y = x2 −3x+2

METHOD

1 Find the gradient function

by differentiating

2 Find the x-value for which the gradient is

zero by solving the equation

4 Write down the coordinates of the

stationary point

3 Find the y-value of the function at x =

1.5 by substitution into

the original function

y = x2 −3x+ 2

dy

dx = 2x−3




dy

dx = 3x2 +6x−45

3x2 +6x−45 =0

3( x2 + 2x−15 ) = 0

3( x+ 5 )( x−3 ) = 0

x1=−5 and x

2=3

y1=85 and x

2=−81

d 2y

dx2= 6x+6

Find the stationary points on the function and determine their nature y = x3 + 3x2 −45x

METHOD

2 Find the x-values for which the gradient function is zero.

4 Find the sign of the second derivative to determine the

nature of each stationary point.

3 Substitute these x-values into the original function to determine the stationary points.

1 Differentiate the given function to obtain the gradient function

At x

1=−5

d2ydx2

=−24 (negative)⇒ (−5, 3 ) is a maxima

At x

1=3

d2ydx2

=24 (positive)⇒ (3, −81) is a minima

f(x) = 3x2 +6x−45

3x2 +6x−45 =0

3( x2 + 2x−15 ) = 0

3( x+ 5 )( x−3 ) = 0

x1=−5 and x

2=3

y1=85 and x

2=−81

′′f (x) = 6x+6

′′f (−5 )=−24 (negative)⇒ (−5, 3 ) is a maxima

′′f ( 3 )=24 (positive)⇒ (3, −81) is a minima

calculus 11.2

Education

linear functions

concept of gradient

actual gradient

normals graph

negative slopes gradient

point p

particular point

nonlinear function