Rate of Change and Derivative Calculus Lesson

1. 3.1 The Limit
Definition of the
Derivative
September 25, 2015

2. Objectives
O I can find the derivative of a function
using the limit definition of a derivative
O I can evaluate the slope of a curve (the
derivative) at a specific point on the curve
O I can write the equation of a line tangent
to a curve at a certain point

3. Agenda
O Discussion of Unit 2 Limits Exam (5
minutes)
O Lesson Warm-Up (20 minutes)
O Notes on the Limit Definition of a
Derivative with built-in Guided Practice
(39 minutes)
O In-class Practice Time (20 minutes)
O Exit Ticket (10 minutes)

4. Lesson Warm-Up (10 min.)
1. Find the slope of the line that connects the two
points P( 4, 5 ) and Q ( -2, 3 )
2. Write the equation of the line PQ.
3. For the function f(x) = x2 – 5, evaluate
1. f(4) = ?
2. f(h) = ?
3. f(x+ h) = ?

5. Lesson Warm-Up (10 min.)
1. Find the slope of the line that connects the two
points P( 4, 5 ) and Q ( -2, 3 )
Slope formula
m =
y2 - y1
x2 - x1
=
3-5
(-2)- 4
=
-2
-6
=
1
3

6. Lesson Warm-Up (10 min.)
1. Find the slope of the line that connects the two
points P( 4, 5 ) and Q ( -2, 3 )
2. Write the equation of the line PQ.
y- y1 = m(x - x1)
y-5= 1
3
(x -3)
y-5= 1
3
x -1 y = 1
3
x + 4

7. Lesson Warm-Up (10 min.)
1. Find the slope of the line that connects the two
points P( 4, 5 ) and Q ( -2, 3 )
2. Write the equation of the line PQ.
3. For the function f(x) = x2 – 5, evaluate
1. f(4) = ?
2. f(h) = ?
3. f(x+ h) = ?
f(4) = 42 – 5 = 16 – 5 = 11
f(h) = h2 – 5
f(x+h) = (x +h)2 – 5
= (x + h) (x + h) – 5
= x2 + xh + xh + h2 – 5
= x2 + 2xh + h2 – 5

8. Rate of Change
Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = 2t + 1
Where is the object at
the 1st second?
t = 1 second
s(1) = 2(1) + 1 = 3 meters

9. Rate of Change
Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = 2t + 1
Where is the object at
the 2nd second?
t = 2 seconds
s(1) = 2(2) + 1 =5 meters

10. Rate of Change
Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = 2t + 1
What is the rate of
change ?
m =
s(t2 )- s(t1)
t2 -t1

11. Rate of Change
Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = 2t + 1
What is the rate of
change ?
m =
5-3
2-1
= 2

12. Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = t2
What is the AVERAGE rate
of change between
t = 1 and t = 2 seconds?
m =
s(2)- s(1)
2-1
m =
4-1
2-1
= 3Secant line

13. Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = t2
What is the
INSTANTANEOUS rate of
change between at exactly
the FIRST second?
tangent line at t = 1

14. Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = t2
What is the
INSTANTANEOUS rate of
change between at exactly
the FIRST second?
m =
s(1.1)- s(1)
1.1-1

15. Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = t2
What is the
INSTANTANEOUS rate of
change between at exactly
the FIRST second?
m =
s(1.1)- s(1)
1.1-1
m =
s(1.01)- s(1)
1.01-1

16. Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = t2
What is the
INSTANTANEOUS rate of
change between at exactly
the FIRST second?
m =
s(1.1)- s(1)
1.1-1
m =
s(1.01)- s(1)
1.01-1
m =
s(1.001)- s(1)
1.001-1

17. Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = t2
What is the
INSTANTANEOUS rate of
change between at exactly
the FIRST second?
m =
s(1.1)- s(1)
1.1-1
m =
s(1.01)- s(1)
1.01-1
m =
s(1.001)- s(1)
1.001-1

18. Consider: An object is moving and its position s(t) is
measured in meters and depends on t in seconds
s(t) = t2
What is the
INSTANTANEOUS rate of
change between at exactly
the FIRST second?
m =
s(1.1)- s(1)
1.1-1
m =
s(1.01)- s(1)
1.01-1
m =
s(1.001)- s(1)
1.001-1

19. The Derivative
The derivative of f(x) at x = a,
f '(a) = limh®0
f (a+h)- f (a)
h
f '(a) = limx®a
f (x)- f (a)
x -a

20. Finding the Derivative
Example 1: Write the equation of the line that is
tangent to the curve y = x2 at the point (1, 1).

21. Finding the Derivative
Example 1: Write the equation of the line that is
tangent to the curve y = x2 at the point (1, 1).
Step 1: Find the derivative (= slope of the curve)
at the point (1, 1)
f '(a) = limx®a
f (x)- f (a)
x -a

22. Finding the Derivative
Step 1: Find the derivative (= slope of the curve)
at the point (1, 1)
f '(a) = limx®a
f (x)- f (a)
x -a
f '(1) = limx®1
f (x)- f (1)
x -1
f '(1) = limx®1
x2
- f (1)
x -1

23. Finding the Derivative
f '(a) = limx®a
f (x)- f (a)
x -a
f '(1) = limx®1
f (x)- f (1)
x -1
f '(1) = limx®1
x2
- f (1)
x -1
f '(1) = limx®1
x2
-1
x -1

24. Finding the Derivative
f '(1) = limx®1
x2
-1
x -1
= limx®1
(x -1)(x +1)
x -1
= limx®1(x+1)
= limx®1(x+1)=1+1= 2

25. Writing the Equation
Derivative = slope of the curve = slope of the
tangent
Slope = 2 m/s
Point = (1, 1)

26. Writing the Equation
Slope = 2 m/s
Point = (1, 1)
y- y1 = m(x - x1)
y-1= 2(x -1)
y-1= 2x-2
y = 2x -1

27. Finding the Derivative
Example 2: Write the equation of the line that is
tangent to the curve y = x3 + x when x = 0.
f '(a) = limx®a
f (x)- f (a)
x -a
f '(0) = limx®0
f (x)- f (0)
x -0
f '(0) = limx®0
x3
+ x -0
x -0

28. Finding the Derivative
f '(0) = limx®0
x3
+ x -0
x -0
= limx®0
x3
+ x
x
=
x(x2
+1)
x
= x2
+1= 02
+1=1

29. Writing the Equation
Slope = 1
Point = (0, 0)
y- y1 = m(x - x1)
y-0 =1(x -0)
y = x

30. Guided Practice Problems
1. Write the equation of the line tangent to
the curve f(t) = t – 2t2 at a = 3.
2. f(x) = 4 – x2 at a = -1
3. at a = 3
4. at a = -2
f (t)= t2
+1
f (x)=
1
x +3

31. Homework Assignment
Write the equation of the tangent line of the
following curves at the given points.
1. f(x) = 2x2 + 10x , a = 3
2. f(x) = 8x3 , a = 1
3. , a = 1
4. , a = 0
f (x)= x +4
f (x)=
1
x2
+1

32. Exit Ticket
1. Compute the derivative and
write the equation of the tangent
line at a = -1 for the following
function: f(x) = 3x2 + 4x + 2
2. In full sentences, explain the
relationship how a secant line is
different from a tangent line and
how average velocity is different
from instantaneous velocity.