Tangent and Normal lines
to a curve
Rate of Change
DIFFCAL / CALENG1
Dr. Susan A. Roces
1
Tangent and Normal Line to a Curve
y = f(x)
Q(x2, y2)
∆y
P(x1, y1)
Normal LINE
∆x
TANGENT LINE
2
The tangent line at point P is that line that
passes through this point with slope:
∆y
dy
mTL = lim
=
∆x
dx
∆x → 0
The slope of the tangent line is also
referred to as the slope of the curve at
point P.
3
Normal Line to a curve
The normal line to a curve at point P is
defined to be the line perpendicular to the
tangent line at that point.
If two lines are perpendicular, the slope of
one is the negative reciprocal of the slope
of the other line.
mNL
1
=−
mTL
4
Point – slope Form
Equation of the tangent line:
Given:
P( x1 , y1 )
and
mTL
y − y1 = mTL ( x − x1 )
5
Point – slope Form
Equation of the normal line:
Given:
P( x1 , y1 )
y − y1 =
and
1
−
mTL
mNL = −
1
mTL
( x − x1 )
6
Examples:
1. Find the equations of the tangent and
normal lines to the curve
y = 3x2 – 2x + 1 at P(1,2).
Solution:
y = 3x2 – 2x + 1
dy
= 6x – 2
dx
mTL @ P (1.2) = dy = 6(1) – 2 = 4
dx P (1.2)
7
Equation of the tangent line:
P(1,2), mTL = 4
y − y1 = mTL ( x − x1 )
y − 2 = 4 ( x − 1)
y −2 =4 x −4
y − 4x + 2 = 0
8
Equation of the normal line:
P(1,2),
mNL = − 1 = − 1
y − y1 =
−
1
mTL
4
mTL
( x − x1 )
1
y − 2 = − 4 ( x − 1)
4 y − 8 = − x +1
4y + x − 9 =0
9
2. Find the point of intersection/s of the
2
x
normal line to the curve y = − 4
x
where the slope of the curve is 2, with
the line x – y = 0.
x–y=0
?
Given:
x2 − 4
y=
x
10
Solution:
x2 − 4
y=
x
x ( 2 x ) − ( x 2 − 4)(1)
y '=
x2
2x2 − x2 + 4
y '=
x2
x 2 + 4 but: m = 2
y '=
TL
2
x
11
x2 + 4
y '=
=2
2
x
2
2
x + 4 = 2x
2
x =4
x = 2,−2
12
when : x = 2
( 2) 2 − 4 0
y=
= = 0,
2
2
P ( 2 ,0 )
when : x = − 2
( −2) 2 − 4 0
y=
= = 0, P (− 2 , 0 )
2
2
13
Equation of the normal line:
mNL = − 1
P(-2,0),
y − y1 =
−
1
mTL
2
( x − x1 )
1
y − 0 = − ( x + 2)
2
2y =− x − 2
2y + x =− 2
14
Equation of the normal line:
mNL = − 1
P(2,0),
y − y1 =
−
1
mTL
2
( x − x1 )
1
y − 0 = − ( x − 2)
2
2y =− x + 2
2y + x =2
15
Points of intersection:
2y + x =2
x=y
2y + y =2
3y = 2
y =2/3
x =2/3
P( 2 / 3, 2 / 3)
2y + x =− 2
x=y
2y + y =− 2
3y = − 2
y =−2/3
x =−2/3
P( − 2 / 3, − 2 / 3)
16
3. Find the equation/s of the tangent line
to the ellipse x2 + 2 y2 = 9 that is
perpendicular to the line 4x – y = 6.
Given:
?
?
4x – y = 6
y = mx + b
y = 4x - 6
Therefore:
mL = 4
mTL = −
1
4 17
Solution:
x2 + 2 y2 = 9
2 x + 4 y y’ = 0
(1)
4 y y’ = - 2 x
2x
x
1
y '= −
=−
=−
4y
2y
4
2y=4x
y=2x
(2)
18
Solve simultaneously equations (1) and (2):
x2 + 2 ( 4 x 2 ) = 9
9x2 =9
x2 =1
x = 1 , -1
when: x = 1, y = 2(1) = 2; P(1, 2)
when: x = -1, y = 2(-1) = - 2; P(-1, -2)
19
Equation of the tangent line:
P(1, 2),
mTL = − 1
4
y − y1 = mTL ( x − x1 )
1
y − 2 = − ( x − 1)
4
4 y − 8 = − x +1
4y + x − 9 =0
20
Equation of the tangent line:
P(- 1, - 2),
mTL = − 1
4
y − y1 = mTL ( x − x1 )
1
y + 2 = − ( x + 1)
4
4 y + 8 = − x −1
4y + x + 9 =0
21
Assignments
p. 129, Nos. 43-44
p. 144, Nos. 113-116
22
Rate of Change
Average Rate of Change is the ratio
∆y
which is the average rate of
∆x
change over the interval ∆x .
∆y y2 − y1 f ( x2 ) − f ( x1 )
=
=
∆x x2 − x1
∆x
23
If let ∆x → 0 , this ratio in general
approaches a limiting value, which is
defined as the rate of change of y
corresponding to the given value of x
or the INSTANTANEOUS RATE
OF CHANGE.
∆y
dy
= lim
dx
∆x
∆x → 0
24
Examples:
1. The supply for a certain brand of dress
is S = x4 - x3 + x2 – x, where x is the
price per dress. Find the rate of
change of the supply per peso change
in the price:
a). when the price is increased from
one peso to two pesos.
b). when the price is one peso.
25
Given: S = x4 - x3 + x2 – x
Req’d:
dS
∆S
b)
a)
∆x
Solution:
x1 =1, x2 = 2
S2 − S1
=
∆x x2 − x1
dx
x =1
a) ∆S
[(2)4 – (2)3+ (2)2 – (2)]
=
− [(1)4 – (1)3+ (1)2 – (1)]
2 −1
26
[16 − 8 + 4 − 2] − [1 − 1 + 1 − 1]
=
2 −1
10 − 0
= 10
=
1
b) dS
b)
dx
= 4 x3 - 3 x2 + 2 x – 1
= 4 (1)3 - 3 (1)2 + 2 (1) – 1
= 4 - 3 + 2 – 1=2
27
Examples:
2. Suppose a right circular cylinder has a
constant height of 10 m. If V in m3 is
the volume of the right circular
cylinder, and r is the radius of its base,
find the rate of change of the volume
with respect to r if r changes :
a). from 5 to 5.1.
b). from 5 to 5.4.
c). when r = 5.
28
V = π r 2 h = 10 π r 2
Given:
Req’d:
b) ∆V
a) ∆V
∆r
r1 = 5, r2 = 5.1
Solution:
V2 − V1
=
=
∆r r2 − r1
a) ∆V
∆r
c) dV
r1 = 5, r2 = 5.4
dr
r =5
10 π (5) 2
10π (5.1) 2
−
5.1 − 5
 (5.1) 2 − (5) 2 
= 10π 
=101π

0 .1


29
b) ∆V
∆r
=
10 π (5.4) 2
= 10π
c)
− 10π (5) 2
5.4 − 5
 (5.4) 2 − (5) 2 


0
.
4


= 104 π
2
V = 10 π r
dV
= 10 π ( 2 r )
dr
= 10 π ( 2) (5) = 100 π
30
3. Find the rate of change of the area of an
equilateral triangle with respect to the
side:
a) when the side is 4 feet;
b) when the side changes from 4 to 4.4
feet.
Given:
Req’d:
s
h
1
s
2
s a) dA b) ∆A
ds s = 4 ∆s
s1 = 4, s2 = 4.4
31
Solution:
1
A = 2 (base )( height )
1
A = 2 ( s )( h )
dA
then: h = f (s)
sin ce:
ds
a)
A = f (s)
32
2
h
1
s
2
1 

s = ( h) +  s 
 2 2
1 
2
2
h = (s) −  s 
2 
4 ( s ) 2 − ( s ) 2 3s 2
=
=
4
4
3s 2
s 3
=
=
4
2
2
s
h
2
h
2
33
1
A = ( s )( h )
2
1
(s ) 3
(
)
=
s
A 2
2
3 2
s
A=
4
a) dA = 2 3 ( s ) = 2 3 (4)
ds
4
4
=2 3
34
b)
3
A=
4
s
2
∆ A A2 − A1
=
s2 − s1
∆s
3  (4.4) 2 − (4) 2 
=


0.4
4 

3 (3.36)
=
= 2.1
1. 6
3
35
Assignments
p. 113, Nos. 67-69
p. 144, Nos. 106-110
36