Slide 1 ___________________________________ 2.1 DERIVATIVES
A DERIVATIVE is a special type of limit that occurs when we want to find
___________________________________ A) The slope of a tangent line
B) A velocity
___________________________________ C) Any instantaneous rate of change
___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 2 ___________________________________ Example 1:
A). Find the slope of the graph of f(x) = x2 + 1 at the point (2,5)
B). Find the equation of the tangent line of the graph of y = x2 + 1
at the point (2, 5).
f(x) = x2+1
10
___________________________________ The difficulty is that we have only one point, whereas we need two
points to calculate slope.
8
6
P(2, 5)
___________________________________ 4
2
-10
-5
5
10
-2
___________________________________ -4
-6
___________________________________ ___________________________________ ___________________________________ Slide 3 7
6
P(2,5)
5
P(2, 5)
5
4
4
3
___________________________________ 3
2
2
1
Q(x, f(x))
-4
___________________________________ 7
6
-2
Q approaches P from the left
(x → 2 from the left)
2
4
6
8
-4
2
-1
Q(x, f(x))
1
-2
2
4
2
-1
6
8
___________________________________ 7
Q(x, f(x))
7
6
Q(x, f(x))
6
5
P(2, 5)
5
P(2, 5)
4
4
3
___________________________________ 3
2
2
1
Q approaches P from the right
(x → 2 from the right)
-4
-4
-2
2
2
4
1
-2
6
2
8
-1
2
4
6
___________________________________ ___________________________________ ___________________________________ Slide 4 ___________________________________ PQ,
To find the slope of the secant lines, for the function f(x) = x2 + 1, we have: mPQ =
5 - f(x)
2- x
=
(
)
5 - x 2 +1
=
2- x
4 - x 2 (2- x)(2+ x)
=
=
=2+x
2-x
(2 - x)
5 - x 2 -1
2- x
___________________________________ ___________________________________ A). To find the slope of the tangent line, we let x →2:
5 - f(x)
5 - (x 2 + 1)
m = lim
= lim
x →2 2 - x
x→2
2-x
= lim
x →2
___________________________________ 5 - x2 − 1
4 - x2
(2-x)(2+x) = lim (2+ x) = 2+ 2= 4
= lim
= lim
x →2
x →2 2 - x
x →2
2-x
2-x
___________________________________ ___________________________________ ___________________________________ Slide 5 ___________________________________ B) To find the equation of the tangent line, use the slope of the tangent line and the point – slope equation:
y – y1 = m(x – x1) ___________________________________ y – 5 = 4(x – 2) → y = 4x ‐ 3
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 6 ___________________________________ DEFINITION OF DERIVATIVE
y
y
tangent line to f(x) at P
tangent line to f(x) at P
Q(x, f(x))
}
N
Q(a+h, f(a+h)
f(x)‐f(a)
P(a, f(a))
P(a, f(a))
x‐a
a
}
N
___________________________________ f(a+h)‐f(a)
h
x
x
a
a+h
___________________________________ x
︵ ︶
︵ ︶︵ ︶
a
f
h h
+
a
f
0
m→
ih
l
=
a
'
f
a
f a
- x x
f
a
m→
ix
l
=
a
'
f
The derivative of a function f at a number a is the slope of the tangent line at a and is given by:
OR ︵ ︶
︵
___________________________________ ︶︵ ︶
___________________________________ ___________________________________ ___________________________________ Slide 7 ___________________________________ THE VELOCITY PROBLEM
Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds.
___________________________________ t
n
e
m e
e m
c i
a t
l
p
s
i
d
=
y
t
i
c
o
l
e
v
e
g
a
r
e
v
a
The distance fallen after t seconds is given by the position function s(t) = 4.9t2
=
f(a+h)-f(a)
h
___________________________________ instantaneous velocity at time t =a
v(a) = lim
h →0
___________________________________ f (a + h ) − f (a )
h
___________________________________ ___________________________________ ___________________________________ Slide 8 instantaneous velocity at time t =a
f (a + h ) − f (a )
v(a) = lim
h →0
h
___________________________________ s=f(t)=4.9t 2
f(a+h) = 4.9(a+h)2 = 4.9(a2 +2ah+h2 )= 4.9a2 + 9.8ah + 4.9h2
___________________________________ (4.9a2 + 9.8ah + 4.9h2 ) - (4.9a2 )
h
h(9.8a + 4.9h)
9.8ah + 4.9h 2
= 9.8a + 4.9(0) = 9.8a
= lim
= lim
h
→
0
h →0
h
h
After 5 seconds, v(5) = 9.8(5) = 49 m/s
v(a) = lim
h →0
___________________________________ B). When does the ball hit the ground?
the ball hits the ground when s = 450m. 450 = 4.9t 2 → t 2 =
450
→ t ≈ 9.6s
4.9
___________________________________ C). How fast is the ball traveling when it hits the ground?
note: a = 9.6s. V(9.6) = 9.8(9.6) = 94m/s
___________________________________ ___________________________________ ___________________________________ Slide 9 ___________________________________ Try these problems:
1 a). Use the definition of derivative to find the slope of the tangent line to the curve f(x) = at the point (4, 2).
x
1/4
b). Find the equation of the tangent line to the curve at the given point. y = (1/4)x + 1 ___________________________________ 2 If an arrow is shot upward on the moon with a velocity
of 58m/s, its height (in meters) after t seconds is given by H = 58t – 0.83t2
a). Find the velocity of the arrow after one second. 56.34 m/s
___________________________________ b). Find the velocity of the arrow when t = a. (58 – 1.66a) m/s
___________________________________ c). When will the arrow hit the moon? After 69.9 s
d). With what velocity will the arrow hit the moon? ‐58 m/s
___________________________________ ___________________________________ ___________________________________