Transcript Tangents and Gradients

Gradients and Tangents
e.g. Find the gradient of the line joining the
points with coordinates ( 1, 1 ) and ( 3, 7 )
Solution:
m
y 2  y1
x 2  x1

m
71
31

6
3
2
difference in the
y-values
( 3, 7 )x
7-1=6
( 1, 1 )x
3-1=2
difference in the
x-values
Gradients and Tangents
The gradient of a straight line is given by
the difference in the
y - values
m
the difference in the x  values
We use this idea to get the gradient at a point on a
curve
Gradients are important as they measure the rate
of change of one variable with another. For the
graphs in this section, the gradient measures how y
changes with x
This branch of Mathematics is called
Calculus
Gradients and Tangents
The Gradient at a point on a Curve
Definition: The gradient of a point on a curve equals
the gradient of the tangent at that point.
e.g.
y  x
2
12
(2, 4)x
3
Tangent at ( 2, 4)
The gradient of the tangent at (2, 4) is
m
So, the gradient of the curve at (2, 4) is 4
12
3
4
Gradients and Tangents
The gradient changes as we move along a curve
e.g.
y  x
m  6
2
Gradients and Tangents
y  x
2
m  4
Gradients and Tangents
y  x
2
m  2
Gradients and Tangents
The Rule
for Differentiation
y  x
2
m0
Gradients and Tangents
The Rule
for Differentiation
y  x
2
m2
Gradients and Tangents
y  x
2
m4
Gradients and Tangents
y  x
2
m6
Differentiation from first principles
400
360
320
P
280
f(x+h)
240
200
160
120
80
A
f(x)40
0
0
0.5
x1
1.5
2
2.5
f( x  h )  f( x )
Gradient of AP =
h
3
x3.5
+h
4
Differentiation from first principles
400
360
320
280
240
200
P
f(x+h)
160
120
80
A
f(x)40
0
0
0.5
x1
1.5
2
2.5
f( x  h )  f( x )
Gradient of AP =
h
x +3 h
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
P
120
f(x+h)
80
A
f(x)40
0
0
0.5
x1
1.5
2
2.5
x+
h
f( x  h )  f( x )
Gradient of AP =
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
P
80
f(x+h)
A
f(x)40
0
0
0.5
x1
1.5
2
x+
h
2.5
f( x  h )  f( x )
Gradient of AP =
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
80
f(x+h)
f(x)40
A
P
0
0
0.5
x1
x 1.5
+h
2
2.5
f( x  h )  f( x )
Gradient of AP =
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
80
A
40
f(x)
0
0
0.5
x1
1.5
2
2.5
3
L im it f( x  h )  f( x )
Gradient of tangent at A =
h 0
h
3.5
4
Differentiation from first principles
f(x) = x2
f( x  h )  f( x )
( x  h)  x
2

2
x  2 xh  h  x
2

h
h

(2 x  h ) h
h
 2x  h
h
f '( x ) 
L im it f( x  h )  f( x )
h 0
h
2
 2x
2

2 xh  h
h
2
Differentiation from first principles
f(x) = x3
f( x  h )  f( x )
( x  h)  x
3

3
x  3 x h  3 xh  h  x
3

2
h
h

h
(3 x  3 xh  h ) h
2
2
 3 x  3 xh  h
2
h
f '( x ) 
L im it f( x  h )  f( x )
h 0
h
2
 3x
2
3
3
3 x h  3 xh  h
2

2
h
2
Generally with
L im it f( x  h )  f( x )
h 0
h
h is written as δx
And f(x+ δx)-f(x) is written as δy
Limit
x  0
x
dy

y
dx
Generally
f ( x )  kx
y  kx
n
f ' ( x )  nkx
n 1
dy
n
 nkx
n 1
dx
f ( x )  kx
y  kx
n
f ' ( x )  nkx
n 1
dy
dx
n
 nkx
n 1
Gradients and Tangents
Points with a Given Gradient
e.g. Find the coordinates of the points on the curve
3
y  x  8 x  7 where the gradient equals 4
Gradient of curve
= gradient of tangent
= 4
We need to be able
to find these points
using algebra
Gradients and Tangents
Gradients and Tangents
Gradients and Tangents
Exercises
Find the coordinates of the points on the curves with
the gradients given
1.
y  x  4x  3
2
where the gradient is -2
Ans: (-3, -6)
3
2
2. y  x  3 x  21 x  20 where the gradient is 3
( Watch out for the common factor in the
quadratic equation )
Ans: (-2, 2) and (4, -88)
Gradients and Tangents
Increasing and Decreasing Functions
•
An increasing function is one whose
gradient is always greater than or equal to
zero.
dy
 0
dx
•
for all values of x
A decreasing function has a gradient that
is always negative or zero.
dy
dx
 0
for all values of x
Gradients and Tangents
e.g.1 Show that y  x 3  4 x is an increasing
function
dy
2
3
 3x  4
Solution: y  x  4 x 
dx
dy
is the sum of
dx • a positive number ( 3 )  a perfect
square ( which is positive or zero for
all values of x, and
• a positive number ( 4 )

dy
dx
 0 for all values of x
so, y  x 3  4 x is an increasing function
Gradients and Tangents
e.g.2 Show that y  1 x 3  3 x 2  9 x is an
3
increasing function.
Solution: y 
1
3
dy
dx
x  3x  9x 
3
2
dy
 x  6x  9
dx
 0
for all values of x
2
Gradients and Tangents
The graphs of the increasing functions
3
2
3
y  x  4 x and y  13 x  3 x  9 x
y  x  4x
3
and
y
1
3
x  3x  9x
3
2
are
Gradients and Tangents
Exercises
1. Show that y   x 3 is a decreasing function and
sketch its graph.
2. Show that y  1 x 3  2 x 2  5 x is an increasing
3
function and sketch its graph.
Solutions are on the next 2 slides.
Gradients and Tangents
Solutions
1. Show that y   x 3 is a decreasing function and
sketch its graph.
Solution:
dy
dx
  3 x . This is the product of a
2
square which is always  0 and a negative number,
dy
so
 0 for all x. Hence y   x 3 is a
dx
decreasing function.
y  x
3
Gradients and Tangents
Solutions
2. Show that
function
Solution:
y
dy
1
3
3
2
x  2 x  5 x is an increasing
 x  4x  5 .
dx
dy
2
Completing the square:
 ( x  2)  1
dx
which is the sum of a square which is  0
2
and a positive number. Hence y is an increasing
function.
y
1
3
x  2x  5x
3
2
Gradients and Tangents
The equation of a tangent
e.g. 1 Find the equation of the tangent at the point
(-1, 3) on the curve with equation
3
2
y  x  3x  2x  1
3
2
Solution:
y

x

3
x
 2 at
x  a1 point and the gradient of
The gradient of a curve
dy the tangent
Gradient
=equal
-5
2
at
that
point
are

 3x  6x  2
dx
 At x = 1
(-1, 3)
x
m  3(  1)  6(  1)  2
2
 5
y  mx  c 
y  5 x  c
(-1, 3) on line:  3   5 (  1 )  c

2c
So, the equation of the tangent is y   5 x  2