Tangents and Normals
MODULE - V
Calculus
24
Notes
TANGENTS AND NORMALS
In the previous lesson, you have learnt that the slope of a line is the tangent of the angle which
this line makes with the positive direction of x-axis. It is denoted by the letter 'm'. Thus, if θ is
the angle which a line makes with the positive direction of x-axis, then m is given by tan θ .
You have also learnt that the slope m of a line, passing through two points ( x1, y 1 ) and ( x 2 , y 2 ) is
given by m =
y 2 − y1
x 2 − x1
In this lesson, we shall find the equations of tangents and normals to different curves, using the
knowledge of differential calculus. We shall also study about Rolle's Theorem and Mean Value
Theorems and their applications.
OBJECTIVES
After studying this lesson, you will be able to :
define tangent and normal to a curve (graph of a function) at a point;
•
find equations of tangents and normals to a curve under given conditions;
•
state Rolle's Theorem and Lagrange's Mean Value Theorem; and
•
test the validity of the above theorems and apply them to solve problems.
•
EXPECTED BACKGROUND KNOWLEDGE
•
Knowledge of coordinate Geometry
•
Concept of tangent and normal to a curve
•
Concept of diferential coefficient of various functions
•
Geometrical meaning of derivative of a function at a point
MATHEMATICS
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24.1 SLOPE OF TANGENT AND NORMAL
Let y = f ( x ) be a continuous curve and let
P ( x1, y1 ) be a point on it then the slope PT at
P ( x1, y1 ) is given by
Notes
 dy 
  at ( x1, y1 )
 dx 
....(i)
and (i) is equal to tan θ
We know that a normal to a curve is a line
perpendicular to the tangent at the point of contact
We know that
α=
π
+θ
2
Fig. 24.1
(From Fig. 24.1)
π

tan α = tan  + θ  = − cot θ
2

⇒
=−
∴ Slope of normal
=−
1
tan θ
1
−1
 dx 
=
at ( x1, y 1 ) or −   at
dy
m 

 dy 


dx


( x1, y 1)
Note
1.
The tangent to a curve at any point will be parallel to x-axis if θ = 0 , i.e, the derivative
at the point will be zero.
i.e.
2.
 dx 
  at ( x1, y1 ) = 0
 dy 
The tangent at a point to the curve y = f ( x ) will be parallel to y-axis if
dy
= 0 at that
dx
point.
Let us consider some examples :
Example 24.1 Find the slope of tangent and normal to the curve
x 2 + x 3 + 3xy + y 2 = 5 at (1, 1)
Solution : The equation of the curve is
x 2 + x 3 + 3xy + y 2 = 5
.....(i)
Differentialing (i),w.r.t. x, we get
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Calculus
.....(ii)
dy
 dy

2x + 3x + 3  x
+ y.1 + 2y
=0
dx
 dx

2
Substituting x = 1, y = 1 , in (ii), we get
dy
 dy 
2 × 1 + 3 × 1 + 3  + 1 + 2
=0
dx
 dx 
5
or
dy
= −8 ⇒
dx
Notes
dy
8
=−
dx
5
∴ The slope of tangent to the curve at (1, 1) is −
∴
The slope of normal to the curve at (1, 1) is
8
5
5
8
Example 24.2 Show that the tangents to the curve y =
1 5
3x + 2x 3 −3x 


6
at the points x = ±3 are parallel.
3x5 + 2x 3 − 3x
Solution : The equation of the curve is y =
6
Differentiating (i) w.r.t. x, we get
(
4
2
dy 15x + 6x − 3
=
dx
6
.....(i)
)
15 3 4 + 6 ( 3) 2 − 3
 dy  at =  ( )

 dx  x =3
6
 
=
1
[15 × 9 × 9 + 54 − 3]
6
=
3
[405 + 17] = 211
6
1
 dy 
  atx = − 3 =
6
 dx 
∴
15 (− 3)4 + 6 (− 3)2 − 3

 = 2111
The tangents to the curve at x = ±3 are parallel as the slopes at x = ±3 are equal.
Example 24.3 The slope of the curve 6y3 = px 2 + q at (2, −2 ) is
1
.
6
Find the values of p and q.
Solution : The equation of the curve is
6y3 = px 2 + q
MATHEMATICS
.....(i)
301
Tangents and Normals
MODULE - V Differentiating (i) w.r.t. x, we get
Calculus
2 dy
18y
dx
= 2px
.....(ii)
Putting x = 2, y = −2, we get
18 ( − 2)
Notes
2
dy
=2p ⋅2 = 4p
dx
dy p
=
dx 18
∴
It is given equal to
1
6
1 p
=
⇒ p=3
6 18
∴
∴ The equation of curve becomes
6y3 = 3x 2 + q
Also, the point (2, –2) lies on the curve
∴
6 ( − 2) =3 ( 2 )
⇒
− 48 − 12 = q or
3
2
+q
q = −60
∴ The value of p = 3, q = −60
CHECK YOUR PROGRESS 24.1
1.
Find the slopes of tangents and normals to each of the curves at the given points :
(i) y = x 3 − 2x at x = 2
(ii) x 2 + 3y + y 2 = 5 at ( 1, 1 )
(iii) x = a (θ − sin θ) , y = a (1 − cos θ) at θ =
π
2
2.
Find the values of p and q if the slope of the tangent to the curve xy + px + qy = 2 at
(1, 1) is 2.
3.
Find the points on the curve x 2 + y 2 = 18 at which the tangents are parallel to the line
x+y = 3.
4.
At what points on the curve y = x 2 − 4x + 5 is the tangent perpendiculat to the line
2y + x − 7 =0 .
24.2 EQUATIONS OF TANGENT AND NORMAL TO A CURVE
We know that the equation of a line passing through a point ( x1, y 1 ) and with slope m is
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y − y1 =m ( x −x1 )
As discussed in the section before, the slope of tangent to the curve y = f (x) at ( x1, y 1 ) is given
 dy 
by   at ( x1, y 1 ) and that of normal is
 dx 
 dx 
−
 at ( x1, y1 )
 dy 
Notes
∴ Equation of tangent to the curve y = f (x) at the point ( x1, y 1 ) is
 dy 
y − y1=  
 dx  ( x
1
MODULE - V
Calculus
,y
1)
[x
−x1 ]
And, the equation of normal to the curve y = f(x) at the point ( x1, y 1 ) is


 −1 
y − y1 = 

 dy 
 dx ( x
 x −x 

1
1 , y1
)
Note
(i)
 dy 
The equation of tangent to a curve is parallel to x-axis if  
= 0 . In that case
 dx ( x ,y )
1 1
the equation of tangent is y = y1 .
(ii) In case  dy 
→ ∞, the tangent at
 dx ( x ,y )
1 1
( x1 , y1 ) is parallel to
y − axis and its
equation is x = x1
Let us take some examples and illustrate
Example 24.4 Find the equation of the tangent and normal to the circle x 2 + y2 = 25 at the
point (4, 3)
Solution : The equation of circle is
x 2 + y2 = 25
....(i)
Differentialing (1), w.r.t. x, we get
2x + 2y
⇒
∴
MATHEMATICS
dy
=0
dx
dy − x
=
dx
y
4
 dy 
 dx  at = − 3
  ( 4,3)
303
Tangents and Normals
MODULE - V ∴ Equation of tangent to the circle at (4, 3) is
Calculus
4
y −3 = −
or
Notes
3
(x − 4)
4 ( x − 4 ) + 3 ( y − 3) = 0
=
Also, slope of the normal
or,
4x + 3y = 25
−1
3
=
 dy 
4
 
 dx ( 4,3 )
∴ Equation of the normal to the circle at (4,3) is
y −3 =
3
( x −4 )
4
or
4y − 12 = 3x − 12
⇒
3x = 4y
∴ Equation of the tangent to the circle at (4,3) is 4x+3y = 25
Equation of the normal to the circle at (4,3) is 3x = 4y
Example 24.5 Find the equation of the tangent and normal to the curve 16x 2 + 9y2 = 144 at
the point ( x1, y 1 ) where y1 > 0 and x1 = 2
Solution : The equation of curve is
16x 2 + 9y2 = 144
.....(i)
Differentiating (i), w.r.t. x we get
32x + 18y
dy
=0
dx
dy
16x
=−
dx
9y
or
As x1 = 2 and ( x1, y 1 ) lies on the curve
∴
⇒
As
16 ( 2 ) + 9 ( y2 ) = 144
2
y2 =
80
4
5
⇒ y=±
9
3
y1 > 0 ⇒ y =
4
5
3
 4

5  is
∴ Equation of the tangent to the curve at  2,
 3

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Tangents and Normals
 16x 
4
y−
5 = −

3
 9y a t 2, 4


5

3 
[x
−
2]
4
16 2 × 3
5 =−
⋅
( x −2 )
3
9 4 5
or
y−
or
4
3 5y −
5 ⋅ 3 5 + 8 (x − 2 ) = 0
3
MODULE - V
Calculus
or y −
3 5y − 20 + 8x − 16 = 0
4
8
5 +
( x −2 ) =0
3
3 5
Notes
or 3 5y + 8x = 36
 4

5  is
Also, equation of the normal to the curve at  2,
 3

y−
4
 9y 
5 =

3
 16x a t 2, 4


5
3 
[ x − 2]
y−
4
9 2 5
5 =
×
( x −2 )
3
16
3
y−
4
3 5
5=
(x − 2 )
3
8
3 × 8(y) − 32 5 = 9 5 (x − 2 )
24y − 32 5 =9 5 x −18 5
or
9 5x − 24y + 14 5 = 0
x2 y2
Example 24.6 Find the points on the curve
−
= 1 at which the tangents are parallel
9 16
to x-axis.
Solution : The equation of the curve is
x2 y2
−
=1
9 16
....(i)
Differentiating (i) w.r.t. x we get
2x 2y dy
−
⋅ =0
9 16 dx
or
For tangent to be parallel to x-axis,
MATHEMATICS
dy 16x
=
dx 9y
dy
=0
dx
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Tangents and Normals
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Calculus
16x
=0
9y
⇒
⇒
x=0
Putting x = 0 in (i), we get y 2 = −16 y = ± 4i
This implies that there are no real points at which the tangent to
Notes
x2 y2
−
= 1 is parallel to x9 16
axis.
Example 24.7
y=
Find the equation of all lines having slope – 4 that are tangents to the curve
1
x −1
y=
Solution :
1
x −1
.....(i)
dy
1
=−
dx
(x −1) 2
∴
It is given equal to – 4
∴
⇒
−1
(x −1) 2
( x − 1)2 =
When
1
4
x = 1±
⇒
Substituting x =
= −4
1
3 1
⇒x= ,
2
2 2
1
in (i), we get
2
y=
1
1
=
= −2
1
1
−1 −
2
2
x=
3
, y =2
2
3  1

∴ The points are  , 2  ,  , − 2 
2
2

 

∴ The equations of tangents are
306
(a)
3

y − 2 = −4  x − 
2

⇒
y − 2 = − 4x + 6
(b)
1

y + 2 = −4  x − 
2

⇒
y + 2 = − 4x + 2
4x + y = 8
or
or
4x + y = 0
MATHEMATICS
Tangents and Normals
Example 24.8 Find the equation of the normal to the curve y = x 3 at (2 , 8 )
y = x3
Solution :
⇒
dy
= 3x 2
dx
 dy 
= 12
 dx 
  atx = 2
∴
MODULE - V
Calculus
Notes
1
12
Equation
of
the
normal
is
∴
∴ Slope of the normal = −
y −8 = −
or
12(y − 8) + (x − 2) = 0
1
(x − 2)
12
or
x + 12y = 98
CHECK YOUR PROGRESS 24.2
1.
Find the equation of the tangent and normal at the indicated points :
(i) y = x 4 − 6x3 + 13x 2 − 10x + 5 at (0, 5)
(ii) y = x 2 at (1, 1)
(iii) y = x3 − 3x + 2 at the point whose x–coordinate is 3
2.
2
2
Find the equation of the targent to the ellipse x + y = 1 at ( x1, y 1 )
a2 b2
3.
Find the equation of the tangent to the hyperbola
x2
a2
4.
−
y2
b2
=1
at ( x0, y 0 )
Find the equation of normals to the curve
y = x 3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0
5.
Prove that the curves x = y2 and xy = k cut at right angles if 8k 2 = 1
24.3 ROLLE'S THEOREM
Let us now study an important theorem which reveals that between two points a and b on the
graph of y = f ( x) with equal ordinates f (a) and f (b), there exists at least one point c such that
the tangent at [c,f(c)] is parallel to x-axis. (see Fig. 24.2).
MATHEMATICS
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Notes
Fig. 24.2
24.3.1 Mathematical formulation of Rolle's Theorem
Let f be a real function defined in the closed interval [a, b] such that
(i)
f is continuous in the closed interval [ a, b ]
(ii)
f is differentiable in the open inteval ( a, b )
(iii)
f (a) = f (b)
Then there is at least one point c in the open inteval (a, b) such that f '(c) = 0
Remarks
(i)
The remarks "at least one point" says that there can be more than one point c ∈ (a , b) such
that f ' (c ) = 0 .
(ii) The condition of continuity of f on [a, b] is essential and can not be relaxed
(iii) The condition of differentiability of f on ( a,b ) is also essential and can not be relaxed.
For example f(x) =| x | , x ∈ [ − 1,1] is continuous on [−1,1] and differentiable on (–1, 1) and
Rolle's Theorem is valid for this
Let us take some examples
Example 24.9
Verify Rolle's for the function
f(x) = x(x −1)(x − 2),x ∈ [0,2]
Solution :
f(x) = x(x −1)(x − 2)
= x 3 − 3x 2 + 2x
308
(i)
f (x) is a polynomial function and hence continuous in [0, 2]
(ii)
f (x) is differentiable on (0,2)
(iii)
Also f (0) = 0 and f (2) = 0
MATHEMATICS
Tangents and Normals
MODULE - V
Calculus
f(0) = f(2)
∴
∴ All the conditions of Rolle's theorem are satisfied.
f '(x) = 3x 2 − 6x + 2
Also,
∴ f '(c) = 0 gives 3c 2 − 6c + 2 = 0
⇒
c = 1±
⇒
c=
6 ± 36 − 24
6
Notes
1
3
We see that both the values of c lie in (0, 2)
Example 24.10 Discuss the applicability of Rolle's Theorem for
f(x) = s i n x − sin2x , x ∈ [0, π ]
(i)
....(i)
is a sine function. It is continuous and differentiable on (0, π)
Again, we have, f (0) = 0 and f ( π) = 0
f ( π) = f (0 ) = 0
⇒
∴ All the conditions of Rolle's theorem are satisfied
f '(c) = 2  2cos 2 c −1 − cosc = 0


Now
or
4 c o s 2 c − cosc − 2 = 0
∴
cosc =
=
As
1 ± 1 + 32
8
1 ± 33
⋅
8
33 < 6
7
= 0.875
8
which shows that c lies between 0 and π
cosc <
∴
CHECK YOUR PROGRESS 24.3
Verify Rolle's Theorem for each of the following functions :
x3 5x 2
−
+2x , x ∈ [0,3]
3
3
(i)
f(x) =
(iii)
 π
f(x) = s i n x + c o s x −1 on  0, 
 2
MATHEMATICS
(ii)
f(x) = x 2 −1 on [–1, 1]
(iv)
f(x) = x 2 −1 ( x −2 ) on [ 1,2]
−
(
)
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Notes
24.4 LANGRANGE'S MEAN VALUE THEOREM
This theorem improves the result of Rolle's Theorem saying that it is not necessary that tangent
may be parallel to x-axis. This theorem says that the tangent is parallel to the line joining the end
points of the curve. In other words, this theorem says that there always exists a point on the
graph, where the tangent is parallel to the line joining the end-points of the graph.
24.4.1 Mathematical Formulation of the Theorem
Let f be a real valued function defined on the closed interval [a, b] such that
(a) f is continuous on [a, b], and
(b) f is differentiable in (a, b)
(c) f (b) ≠ f (a)
then there exists a point c in the open interval (a, b) such that
f '(c) =
f(b) − f(a)
b −a
Remarks
When f (b) = f (a), f '(c) = 0 and the theorem reduces to Rolle's Theorem
Let us consider some examples
Example 24.11 Verify Langrange's Mean value theorem for
f(x) = (x − 3)(x −6)(x −
9) on [3, 5]
f(x) = (x − 3)(x −6)(x −
9)
Solution :
= (x − 3)(x 2 − 15x + 54)
f(x) = x 3 −18x 2 +99x −162
or
(i)
...(i)
is a polynomial function and hence continuous and differentiable in the given interval
Here, f(3) = 0, f(5) = (2)(–1)( −4) = 8
f(3) ≠ f(5)
∴
∴ All the conditions of Mean value Theorem are satisfied
f(5) − f(3) 8 −0
=
=4
5 −3
2
∴
f '(c) =
Now
f '(x) = 3x 2 − 36x + 99
∴
∴
3c 2 − 36c + 99 = 4 or 3c 2 − 36c + 95 = 0
c=
36 ± 1296 − 1140 36 ± 12.5
=
6
6
= 8.08 or 3.9
c = 3.9 ∈ (3,5)
∴ Langranges mean value theorem is verified
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Example 24.12 Find a point on the parabola y = (x − 4 )2 where the tangent is parallel to the
MODULE - V
Calculus
chord joining (4 , 0) and (5,1)
Solution : Slope of the tangent to the given curve at any point is given by(f '(x) ) at that point.
Notes
f '(x) = 2 ( x − 4 )
Slope of the chord joining (4 , 0) and (5,1) is

y2 − y1 
Q m = x − x 

2
1
1− 0
=1
5− 4
∴ According to mean value theorem
2 ( x − 4 ) =1
x=
⇒
( x − 4)
or
=
1
2
9
2
which lies between 4 and 5
Now
y = (x − 4 )
When
x=
2
2
9
9

, y =  −4 
2
2

1
=
4
9 1
∴ The required point is  , 
 2 4
CHECK YOUR PROGRESS 24.4
1.
2.
Check the applicability of Mean Value Theorem for each of the following functions :
(i)
f(x) = 3x 2 − 4 on [2,3]
(ii)
f(x) = l o g x on [1,2]
(iii)
1
f(x) = x + on [1,3]
x
(iv)
f(x) = x 3 − 2x 2 −x +
3 on [0,1]
Find a point on the parabola y = ( x + 3)2 , where the tangent is parallel to the chord
joining (3, 0 ) and (− 4,1)
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LET US SUM UP
•
The equation of tangent at ( x1, y 1 ) to the curve y = f(x) is given by
y − y1 =[ f '(x)]at
Notes
•
( x1, y1 ) {x
The equation of normal at ( x1, y 1 ) to the curve y = f(x) is given by
 −1 
y − y1 =

 f ' ( x ) ( x
•
−x1}
1 ,y1
)
(x
−x1 )
The equation of tangent to a curve y = f(x) at ( x1, y 1 ) and parallel to x-axis is
given by y = y1 and parallel to y-axis is given by x = x1
•
Rolle's Theorem states : If f (x) is a function which is
(i)
continuous in the closed interval [a,b]
(ii)
differentiable in the open interval (a,b)
(iii)
f(a) = f(b)
then there exists a point c in (a,b) such that f '(c) = 0
•
Lagranges Mean Value Theorem states that if f (x) is a function which is
(i)
continuous in the closed interval [a,b]
(ii)
differentiable in the open interval (a,b)
(iii)
f(b) ≠ f(a)
then there exists a point c in (a,b) such that
f '(c) =
f(b) − f(a)
b −a
SUPPORTIVE WEB SITES
l
l
http://www.wikipedia.org
http://mathworld.wolfram.com
TERMINAL EXERCISE
1.
312
Find the slopes of tangents and normals to each of the following curves at the indicated
points :
MATHEMATICS
Tangents and Normals
2.
(i)
y = x at x=9
(iii)
x = a( θ − sin θ) , y = a(1 +cos θ) at θ =
(iv)
y = 2x 2 + c o s x at x = 0
(ii)
MODULE - V
Calculus
y = x 3 + x at x = 2
π
2
(v) xy = 6 at (1,6)
Find the equations of tangent and normal to the curve
Notes
π
3
x = a c o s 3 θ , y = asin θ at θ = 4
x2 y2
−
= 1 at which the tangents are parallel to y-axis.
9 16
3.
Find the point on the curve
4.
Find the equation of the tangents to the curve
y = x 2 −2x +5 , (i) which is parallel to the line 2x + y + 7 = 0 (ii) which is perpendicular
to the line 5(y − 3x) = 12
5.
Show that the tangents to the curve y = 7x 3 + 11 at the points x = 2 and x = − 2 are
parallel
6.
Find the equation of normal at the point
(am2 ,am 3 ) to the curve ay2 = x3
7.
8.
Verify Rolle's Theorem for each of the following functions:
(
)
(i)
f ( x ) = x 2 −1 ( x − 2 ) on [− 1, 2]
(iii)
f (x ) =
(ii)
f ( x) =
x ( x − 2)
on [0 , 2]
x −1
 3
8x 2
− 2x , x ∈ 0, 
 4
3
If Rolle's theorem holds for f ( x ) = x 3 + bx 2 + ax, x ∈ [1, 3] with c = 2 +
1
, find the
3
values of a and b.
9.
Verify Mean Value Theorem for each of the following functions.
(i)
(iii)
10.
f ( x ) = ax 2 + bx 2 + cx + d on [0,1]
(ii)
f (x ) =
1
on [− 1, 4]
4x + 1
2
y = ( x + 3) on [− 4,3]
Find a point on the parabola f ( x ) = ( x − 3)2 , where the tangent is parallel to the chord
joining the points (3, 0 ) and (4,1)
MATHEMATICS
313
Tangents and Normals
MODULE - V
Calculus
ANSWERS
CHECK YOUR PROGRESS 24.1
1
10
1.
(i) 10, −
2.
p = 5, q = −4
Notes
(ii) −
2 5
,
5 2
(iii) 1, − 1
3. ( 3, 3 ), ( − 3, −3 )
4. ( 3, 2 )
CHECK YOUR PROGRESS 24.2
1.
2.
4.
Tangent
Normal
(i) y + 10x = 5
x − 10y + 50 = 0
(ii) 2x − y =1
x + 2y − 3 = 0
(iii) 24x − y = 52
x + 24y = 483
xx1
a2
+
yy1
b2
=1
xx 0
3.
a2
−
yy0
b2
=1
x + 14y − 254 = 0, x + 14y + 86 = 0
CHECK YOUR PROGRESS 24.3
(i) c =
5± 7
(ii) c = 0
3
(iii) c =
π
4
(iv) c =
2± 7
3
1
3
 43 1 
,
2.  −

 14 196 
CHECK YOUR PROGRESS 24.4
1. (i) c = 2.5 (ii) c = 1/log 2e (iii) c = 3
(iv) c =
TERMINAL EXRCISE
314
1
, −6
6
(ii) 13, −
1
13
1.
(i)
2.
2 2 ( x + y ) = a; x + y = 0
4.
(i) 2x + y − 5 = 0
6.
2x + 3my − am2 2 + 3m2 = 0
7.
c=
9.
(i) c =
(iii) 1, − 1
(iv) 0, not defined
(v) − 6,
1
6
3. (3, 0 ) , (− 3, 0 )
(ii) 12x + 36y = 155
(
)
2± 7
3
(ii) At no real point (iii) c =
3
8
8. a = 11 , b = − 6
1
2
(ii) Not applicable (iii) c = −
1
2
7 1
10.  , 
 2 4
MATHEMATICS