If y = f(x) be a given function, then the differential coefficient f' (x) or dy at the point P (x , y ) is (i) If the tangent at P (x1,y1) of the curve y = f(x) is parallel to the x- axis (or perpendicular to y- axis) then  = 0 i.e. its slope will be zero. FGdy J dx 1 1 m = H K = 0 the trigonometrical tangent of the angle  (say) which the positive direction of the tangent to the curve at P makes with the positive direction of x- axis Gdy J, therefore represents the slope of the tangent. Thus dx (x1,y1) The converse is also true. Hence the tangent at (x1,y1) is parallel to x- axis.  GJ = 0 (x1,y1) (ii) If the tangent at P (x , y ) of the curve y = 1 1 f (x) is parallel to y - axis (or perpendicular to x-axis) then  =  / 2 , and its slope will be infinity i.e. dy m = =  dx (x ,y ) The converse is also true. Hence the tangent at (x1, y1) is parallel to y- axis Fdy (x1,y1) Thus (i) The inclination of tangent with x- axis. dy (iii) If at any point P (x1, y1) of the curve y = f(x), the tangent makes equal angles with the axes, then at the point P,  =  / 4 or 3 / 4 , = tan–1 GHdxJK Hence at P, tan  = dy/dx =  1. The (ii) Slope of tangent = dy dx (iii) Slope of the normal = – dx/dy Ex.1 Find the following for the curve y2 = 4x at point (2,–2) (i) Inclination of the tangent (ii) Slope of the tangent (iii) Slope of the normal Sol. Differentiating the given equation of curve, we get dy/dx = 2/y = –1 at (2,–2) so at the given point. (i) Inclination of the tangent = tan–1(–1) = 135º (ii) Slope of the tangent = –1 (iii) Slope of the normal = 1 converse of the result is also true. thus at (x1,y1) the tangent line makes equal angles with the axes.  GJ =  1 (x1,y1) Ex.2 The equation of tangent to the curve y2 = 6x at (2, – 3). (A) x + y – 1 = 0 (B) x + y + 1 = 0 (C) x – y + 1 = 0 (D) x + y + 2 = 0 Sol. Differentiating equation of the curve with respect to x (a) Equation of tangent to the curve y = f(x) at A (x1,y1) is 2y dy = 6  dx FGdy J dx (2,3) = 3 = –1 3 y – y1 = FGdy J (x1,y1) (x–x1) Therefore equation of tangent is y + 3 = – (x – 2)  x + y + 1 = 0 Ans. [B] Ex.3 The equation of tangent at any of the curve x = at2, y = 2at is - (A) x = ty + at2 (B) ty + x + at2 = 0 (C) ty = x + at2 (D) ty = x + at3 2 a 1 Sol. dy/dx = (dy/dt)/(dx/dt) = 2 at = t  equation of the tangent at (x,y) point is (y – 2 at) = 1 (x – at2) t  ty = x + at2 Ans.[C] Ex.4 The equation of the tangent to the curve x2 (x – y) + a2 (x + y) = 0 at origin is- (A) x + y + 1 = 0 (B) x + y + 2 = 0 (C) x + y = 0 (D) 2x – y = 0 Sol. Differentiating equation of the curve w.r.t. x dy/dx = – y x (i) If tangent line is parallel to x - axis, then dy/dx = 0  y = 0 and x = a Thus the point is (a,0) (ii) If tangent is parallel to y – axis , then dy/dx =   x = 0 and y = a Thus the point is (0,a) (iii) If tangent line makes equal angles with both axis , then dy/dx =  1  y =

1. Total No.of questions in Tangent & Normal, Rolle's & LMVT,
Derivative as a rate measure are -
In Chapter Examples 21
Solved Examples 15
Total No. of questions..................................................36
TANGENT&NORMAL,ROLLE'S&LMVT,DERIVATIVEASARATEMEASURE

2. 1. GEOMETRICAL INTERPRETATION OF THE
DERIVATIVE
If y = f(x) be a given function, then the differential
coefficient f' (x) or
dy
dx
at the point P (x1, y1) is
the trigonometrical tangent of the angle  (say)
which the positive direction of the tangent to the
curve at P makes with the positive direction of x-
axis
dy
dx
F
H
GI
K
J, therefore represents the slope of the
tangent.
Thus
f' (x) =
dy
dx x y
F
H
GI
K
J
( , )
1 1
= tan 
Thus
(i) The inclination of tangent with x- axis.
= tan–1
dy
dx
F
H
GI
K
J
(ii) Slope of tangent =
dy
dx
(iii) Slope of the normal = – dx/dy
Ex.1 Find the following for the curve y2 = 4x at
point (2,–2)
(i) Inclination of the tangent
(ii) Slope of the tangent
(iii) Slope of the normal
Sol. Differentiating the given equation of curve,
we get dy/dx = 2/y = –1 at (2,–2)
so at the given point.
(i) Inclination of the tangent = tan–1(–1) = 135º
(ii) Slope of the tangent = –1
(iii) Slope of the normal = 1
2. EQUATION OF TANGENT
(a) Equation of tangent to the curve y = f(x) at
A (x1,y1) is
y – y1 =
dy
dx x y
F
H
GI
K
J
( , )
1 1
(x–x1)
(i) If the tangent at P (x1,y1) of the curve y =
f(x) is parallel to the x- axis (or perpendicular
to y- axis) then  = 0 i.e. its slope will be
zero.
m =
dy
dx x y
F
H
GI
K
J
( , )
1 1
= 0
The converse is also true. Hence the tangent
at (x1,y1) is parallel to x- axis.

dy
dx x y
F
H
GI
K
J
( , )
1 1
= 0
(ii) If the tangent at P (x1, y1) of the curve y =
f (x) is parallel to y - axis (or perpendicular
to x-axis) then  =  / 2 , and its slope will
be infinity i.e.
m =
dy
dx x y
F
H
GI
K
J
( , )
1 1
= 
The converse is also true. Hence the tangent
at (x1, y1) is parallel to y- axis

dy
dx x y
F
H
GI
K
J
( , )
1 1
= 
(iii) If at any point P (x1, y1) of the curve y = f(x),
the tangent makes equal angles with the
axes, then at the point P,  =  / 4 or 3 4
 / ,
Hence at P, tan  = dy/dx =  1. The
converse of the result is also true. thus at
(x1,y1) the tangent line makes equal angles
with the axes.

dy
dx x y
F
H
GI
K
J
( , )
1 1
=  1
Equation of tangent
Ex.2 The equation of tangent to the curve y2 = 6x
at (2, – 3).
(A) x + y – 1 = 0 (B) x + y + 1 = 0
(C) x – y + 1 = 0 (D) x + y + 2 = 0
Sol. Differentiating equation of the curve with
respect to x
2y
dy
dx
= 6 
dy
dx
F
H
GI
K
J 
( , )
2 3
=
3
3

= –1
Therefore equation of tangent is
y + 3 = – (x – 2)
 x + y + 1 = 0 Ans. [B]

3. Ex.3 The equation of tangent at any of the curve
x = at2, y = 2at is -
(A) x = ty + at2 (B) ty + x + at2 = 0
(C) ty = x + at2 (D) ty = x + at3
Sol. dy/dx = (dy/dt)/(dx/dt) =
2
2
a
at
=
1
t
 equation of the tangent at (x,y) point is
(y – 2 at) =
1
t
(x – at2)
 ty = x + at2 Ans.[C]
Ex.4 The equation of the tangent to the curve
x2 (x – y) + a2 (x + y) = 0 at origin is-
(A) x + y + 1 = 0 (B) x + y + 2 = 0
(C) x + y = 0 (D) 2x – y = 0
Sol. The given equation of the curve is
x3 – x2 y + a2 x + a2 y = 0
Differentiating it w.r.t. x
3x2 – 2xy – x2.
dy
dx
+ a2 + a2.
dy
dx
= 0
– x2.
dy
dx
+ a2 .
dy
dx = – 3x2 + 2xy – a2
Now at origin i.e. x = 0, y = 0
 a2 (1 + dy/dx) = 0
dy/dx = –1
 the equation of tangent is
y – 0 = – 1 (x – 0) y = – x  x + y = 0
Ans.[C]
Ex.5 The slope of the tangent to the hyperbola
2x2 – 3y2 = 6 at (3, 2) is-
(A) –1 (B) 1 (C) 0 (D) 2
Sol. Differentiating the given equation of the curve
4x– 6y. dy/dx = 0
 dy/dx = 2x/3y

dy
dx
F
H
GI
K
J
( , )
3 2
=
2
3
3
2
. = 1 Ans.[B]
Ex.6 The point of contact on the curve
x + y = a where-
(i) tangent line is parallel to x - axis
(ii) tangent line is parallel to y - axis
(iii) tangent line makes equal angles
with both the axis.
(A) (a,0), (0,a), (a/4, a/4)
(B) (0,a), (a, 0), (a/4, a/2)
(C) (0,a), (a/4, a/4), (a,a)
(D) (0,0), (a,0), (0,a)
Sol. Differentiating equation of the curve w.r.t. x
dy/dx = –
y
x
(i) If tangent line is parallel to x - axis, then
dy/dx = 0  y = 0 and x = a
Thus the point is (a,0)
(ii) If tangent is parallel to y – axis , then
dy/dx =   x = 0 and y = a
Thus the point is (0,a)
(iii) If tangent line makes equal angles with
both axis , then dy/dx =  1
 y = x = a/4
Thus the point is (a/4, a/4)Ans.[A]
Ex.7 If a tangent to the curve
x
a
2
2 +
y
b
2
2 = 1 is
parallel to x- axis, then its point of contact is-
(A) (a, 0) (B) (0,–b)
(C) (0,  b) (D) (  a , 0)
Sol. Differentiating given equation, we have
dy
dx
= –
b x
a y
2
2
If tangent is parallel to x– axis, then
dy/dx = 0  –
b x
a y
2
2
= 0  x = 0.
Thus from the given equation y =  b
 required point = (0,  b) Ans.[C]
3. LENGTH OF INTERCEPTS MADE ON AXES BY
THE TANGENT
Equation of tangent at any point (x1, y1) to the
curve y = f(x) is
y– y1 =
dy
dx x y
F
H
GI
K
J
1 1
,
b g
(x–x1) ....(1)
Equation of x- axis, y = 0 ....(2)
Equation of y – axis, x = 0 ....(3)
Solving (1) and (2), we get.
x = x1 –
y
dy
dx x y
1
1 1
F
H
GI
K
J
R
S
|
|
T
|
|
U
V
|
|
W
|
|
,
b g

4.  x – intercept = OA = x1 –
y
dy
dx x y
1
1 1
F
H
GI
K
J
R
S
|
|
T
|
|
U
V
|
|
W
|
|
,
b g
Similarly solving (1) and (3), we get
y– intercept
OB = y1 – x1
dy
dx x y
F
H
GI
K
J
1 1
,
b g
Length of intercepts made
on axes by the tangent
Ex.8 The length of intercepts on coordinate axes
made by tangent to the curve
x + y = a at the point P (x1, y1) are-
(A) ax ay
1 1
, (B) a a
,
(C) x y
1 1
, (D) None of these
Sol. x + y = a
So
dy
dx x y
F
H
GI
K
J
1 1
,
b g
= –
y
x
1
1
 x – intercept = x1 –
y
y
x
1
1
1

F
H
G
I
K
J
R
S
|
|
T
|
|
U
V
|
|
W
|
|
= x1 + x y
1 1 = x1 x y
1 1

e j
= a x1 ( x1 + y1 = a )
and y – intercept = y1 – x1

F
H
G
I
K
J
y
x
1
1
= y1 + x y
1 1 = a y1
Second Method : Equation of tangent at (x1, y1)
y– y1 = –
y
x
1
1
(x– x1)

y y
y
 1
1
+
x x
x
 1
1
= 0

x
x1
+
y
y1
= x1 + y1

x
ax1
+
y
ay1
= 1
Obviously x – intercept = ax1
and y – intercept = ay1 Ans. [A]
Ex.9 The length of intercepts on coordinate axes
made by tangent to the curve y = 2x2 + 3x –2
at the point (1,3) are-
(A) 4, –4/7 (B) –4/7, 4
(C) 4/7, –4 (D) 4/7, 4
Sol. The given equation of curve is
y = 2x2 + 3x – 2
 dy/dx = 4x + 3

dy
dx
F
H
GI
K
J1 3
,
b g
= 4.1 + 3 = 7
Now OA = 1–3. (1/7) = 4/7 and
OB = 3– 1.7 = –4
 required length of intercepts are 4/7, –4
Here negative sign shows that tangent cuts
the y- axis below the origin. Ans.[C]
4. LENGTH OF PERPENDICULAR FROM ORIGIN
TO THE TANGENT
The length of perpendicular from origin (0,0)
to the tangent drawn at the point (x1, y1) of
the curve y = f(x).
p =
y x
dy
dx
dy
dx
1 1
2
1

F
H
GI
K
J

F
H
GI
K
J
Explanation : The equation of tangent at point P
(x1, y1) of the given curve
y – y1 =
dy
dx P
F
H
GI
K
J(x–x1)
p = perpendicular from origin to tangent =

5. y x
dy
dx
dy
dx
1 1
2
1

F
H
GI
K
J

F
H
GI
K
J
Ex.10 The length of perpendicular from (0,0) to the
tangent drawn to the curve y2 = 4 (x+2) at
point (2,4) is-
(A)
1
2
(B)
3
5
(C)
6
5
(D) 1
Sol. Differentiating the given curve w.r.t x
2y.
dy
dx
= 4
at point (2,4) ,
dy
dx
= 1/2
 p =
y x
dy
dx
dy
dx
1 1
2
1


F
H
GI
K
J
 p =
4 2
1
2
1
1
4

F
H
GI
K
J

.
=
6
5
Ans.[C]
5. EQUATION OF NORMAL
The equation of normal at (x1, y1) to the curve
y = f(x) is
(y – y1) = –
1
1 1
dy
dx x y
F
H
GI
K
J ,
b g
(x – x1)
or (y–y1) .
dy
dx x y
F
H
GI
K
J
1 1
,
b g
+ (x – x1) = 0
5.1 Some facts about the normal
(i) The slope of the normal drawn at point
P (x1,y1) to the curve y = f(x) = –
dx
dy x y
F
H
GI
K
J
1 1
,
b g
(ii) If normal makes an angle of  with positive
direction of x– axis then
–
dx
dy
= tan  or
dy
dx
= – cot 
(iii) If normal is parallel to x– axis then
–
dx
dy
= 0 or
dy
dx
= 
(iv) If normal is parallel to y – axis then
–
dx
dy
F
H
GI
K
J=  or
dy
dx
= 0
(v) If normal is equally inclined from both the
axes or cuts equal intercept then
–
dx
dy
F
H
GI
K
J=  1 or
dy
dx
F
H
GI
K
J=  1
(vi) The length of perpendicular from origin to
normal is
P' =
x y
dy
dx
dy
dx
1 1
2
1

F
H
GI
K
J

F
H
GI
K
J
(vii) The length of intercept made by normal on
x- axis is = x1 + y1
dy
dx
F
H
GI
K
J
and length of intercept on y - axis is
= y1+ x1
dx
dy
F
H
GI
K
J
Equation of normal and related facts
Ex.11 The equation of normal at (1, 6) to the curve
y = 2x2 + 3x + 1 is -
(A) x + 7y – 43 = 0 (B) 7x + y – 43 = 0
(C) 7x + y = 0 (D) None of these
Sol. From the given curve
dy
dx
= 4x + 3

dy
dx
F
H
GI
K
J1 6
,
b g
= 4.1 + 3 = 7  –
dx
dy
F
H
GI
K
J= –
1
7
 The equation of normal is
y – 6 = –
1
7
(x – 1)
 x + 7 y – 43 = 0 Ans.[A]
Ex.12 The equation of normal to the curve
y = x + sin x cos x at x =  / 2 is -
(A) x =  / 2 (B) y =  / 2
(C) x + y =  / 2 (D) x – y =  / 2

6. Sol. At x =  / 2 , we get y =  / 2
So the point is (  / 2 ,  / 2 )
Now the equation of curve y = x + sin x. cos x

dy
dx
= 1 + cos2x – sin2 x
= 1 + cos 2 x

dy
dx
F
H
GI
K
J 
/ , /
2 2
b g= 1 – 1 = 0
 slope of normal = 
 The equation of normal
 (y –  / 2 ) = –
1
0
(x –

2
)  x =

2
Ans.[A]
Ex.13 The length of perpendicular from (0,0) to the
normal drawn at point (1,6) to the curve
y = 2x2 + 3x + 1 is-
(A) 1 (B) 1/2
(C) 43/ 50 (D) 1/6
Sol.  P' =
x y
dy
dx
dy
dx
1 1
2
1

F
H
GI
K
J

F
H
GI
K
J
Since dy/dx = 7
P' =
1 6 7
1 49


.
=
43
50
Ans.[C]
Ex.14 The abscissa of a point where the normal
drawn to the curve xy = (x+ c)2 makes an
equal intercept with coordinate axes is-
(A)  2 c (B) 
c
2
(C) 
3
2
c (D)  3 c
Sol. Differentiating the given curve w.r.t. x
x
dy
dx
+ y = 2(x + c)

dy
dx
=
2 ( )
x c y
x
 

2 1 1
1
( )
x c y
x
 
=  1
and x1y1 = (x1 + c)2
=
2 1
1
2
1
1
( )
x c
x c
x
x
 

b g
=  1
=
2 2 2
1
2
1 1
2
1
2
1
2
x c x x c x c
x
   
=  1
=
x c
x
1
2 2
1
2

=  1
 x1
2 – c2 =  x1
2
 x1
2 – c2 = x1
2
and 2x1
2 – c2 = 0
 2x1
2 = c2  x1 = 
c
2
Ans.[B]
Ex.15 The coordinate of a point to the curve
y = x log x where the normal is parallel to
line 2x – 2y = 3 is -
(A) e–2, – 2e–2 (B) – 2e–2, e–2
(C) e–2, e–2 (D) None of these
Sol. Differentiating the equation of line
2– 2dy/dx = 0
dy/dx = 1
 slope of tangent = – 1
 – 1 = 1 + log x, log x = – 2 x = e–2
 y = e–2 log e–2
y = –2 e–2
 point is (e–2 , – 2e–2 ) Ans.[A]
6. ANGLE OF INTERSECTION OF TWO CURVES
If two curves y = f1(x) and y = f2(x) intersect at
a point P, then the angle between their tangents
at P is defined as the angle between these two
curves at P. But slopes of tangents at P are
dy
dx
F
H
GI
K
J
1
and
dy
dx
F
H
GI
K
J
2
, so at P their angle of
intersection  is given by
tan  = 
dy dx dy dx
dy dx dy dx
/ /
/ /
b g b g
b gb g
1 2
1 2
1


The other angle of intersection will be ( 180º–  )
Note : If two curves intersect orthogonally i.e. at right
angle then  =

2
so the condition will be
dy
dx
F
H
GI
K
J
1
.
dy
dx
F
H
GI
K
J
2
= – 1

7. Angle of intersection of two curves
Ex.16 The angle of intersection between the curves
y =x and y2 = 4x at (4,4).
(A) tan–1
1
2
F
H
GI
K
J (B) tan–1
1
3
F
H
GI
K
J
(C)

4
(D)

2
Sol. Differentiating given equations, we have
dy
dx
F
H
GI
K
J
1
= 1 and
dy
dx
F
H
GI
K
J
2
= 2/y
 at (4,4)
dy
dx
F
H
GI
K
J
1
= 1
dy
dx
F
H
GI
K
J
2
= 2/4 = 1/2
Hence angle of intersection
= tan–1
1
1
2
1 1
1
2


F
H
GI
K
J = tan–1 (1/3). Ans. [B]
Ex.17 The condition that the curves
x
a
2
2 –
y
b
2
2 = 1
and xy = c2 mutually intersect orthogonally is -
(A) a2 + b2 = 0 (B) a2 = b2
(C) a2 – b2 = 0 (D)
a
b
2
2 = 1
Sol. The given curves are
x
a
2
2 –
y
b
2
2 = 1 ... (1)
and xy = c2 ... (2)
from (1),
dy
dx
F
H
GI
K
J
1
=
b x
a y
2
2 and
from (2) ,
dy
dx
F
H
GI
K
J
2
= – y/x
(1) and (2) intersect orthogonally if
dy
dx
F
H
GI
K
J
1
dy
dx
F
H
GI
K
J
2
= – 1

b x
a y
2
2
F
H
G
I
K
J 
F
H
G I
K
J
y
x
= – 1
 a2 = b2 which is the required condition .
Ans. [B]
7. LENGTH OF TANGENT, NORMAL, SUB TANGENT
& SUB NORMAL
Let tangent and normal to the curve y = f (x) at
a point P (x,y) meets the x– axis at points Q
and R respectively. Then PQ and PR are called
length of tangent and normal respectively, at point
P. Also if PM be the perpendicular from P on x-
axis, then QM and MR are called length of sub
tangent and subnormal respectively at P. So from
the diagram at P (x,y)
(i) length of tangent = PQ
= y cosec 
= y
1
2
 dy dx
dy dx
/
/
b g
b g
(ii) length of the normal = PR
= y sec 
=y 1
2
 dy dx
/
b g
(iii) length of sub tangent = QM
= y cot 
= y /
dy
dx
F
H
GI
K
J
(iv) length of sub normal = MR
= y tan 
= y
dy
dx
F
H
GI
K
J

8. Length of tangent, normal,
subtangent & subnormal
Ex.18 Find the length of tangent, subtangent,
normal and subnormal at the point
(2,4) of the curve y2 = 8x.
Sol. Differentiating the equation of the curve w.r.t.
x, we get
2y
dy
dx
= 8  dy/dx = 8/2y = 4/y

dy
dx
F
H
GI
K
J2 4
,
b g
= 4/4 = 1
Therefore at the point (2,4):
length of tangent =
4 1 1
1
2
 bg = 4 2
length of sub tangent = 4/1 = 4
length of normal = 4 1 1
2
 bg= 4 2
length of sub normal = 4.1 = 4
Ex.19 Find the length of tangent, sub-tangent,
normal and sub-normal at the point  of the
curve x = a (  + sin  ), and y = a(1–cos  ).
Sol. 
dy
dx
F
H
GI
K
J=
dy d
dx d
/
/


b g
b g=
a
a
sin
cos


1
b g

dy
dx
=
2 2 2
2 2
2
sin / cos /
cos /
 


dy
dx
= tan

2
 length of tangent
= a 1 1 2
2
2
 
cos tan /
tan /
 

b g
=
a sin / sec /
sin /
cos /
2
2 2
2
2
 


= a. sin  / 2
length of normal
= a (1–cos) 1 2
2
 tan /

= a.2 sin2
 / 2 . sec  / 2
length of sub tangent
=
a 1
2
 cos
tan /


b g=
2 2
2
2
2
a sin /
sin /
cos /



= a sin 
length of sub normal = a (1–cos  ) tan  /2
= 2a sin2
 /2 tan  /2
8. POINT OF INFLEXION
If at any point P, the curve is concave on one
side and convex on other side with respect to x-
axis, then the point P is called the point of
inflexion. Thus P is a point of inflexion if at P,
d y
dx
2
2 = 0 , but
d y
dx
3
3  0
Also point P is a point of inflexion if f" (x) = f"'(x)
= ... = f n–1(x) = 0 and fn (x)  0 for odd n.
Point of inflexion
Ex.20 Prove that origin for the curve y = x3 is a
point of inflexion.
Sol.  y = x3

dy
dx
= 3x2 ,
d y
dx
2
2
= 6x,
d y
dx
3
3 = 6
clearly at (0,0)
d y
dx
2
2
= 0 and
d y
dx
3
3
 0
 There is a point of inflexion at (0, 0).
Ex.21 Prove that every point of the curve y = b sin
x a
/
b gis a point of inflexion, where it meets
the x- axis.
Sol. For the points of intersection of curve and x–
axis, y = 0.

9.  sin x/a = 0  x = a. n 
Here dy/dx =
b
a
cos
x
a
d y
dx
2
2 = –
b
a2 sin
x
a
d y
dx
3
3 = –
b
a3
cos
x
a
Now at x = an  , we have
d y
dx
2
2 = –
b
a2
sin n  = 0
and
d y
dx
3
3 = –
b
a3
cos n   0
Therefore x = an is a point of inflexion of
the curve. This is the point where the curve
meets the x- axis.
9. ROLLE'S THEOREM
If a function f defined on the closed interval [a, b], is
(i) Continuous on [a, b],
(ii) Derivable on (a, b) and
(iii) f(a) = f(b), then there exists atleast one real
number c between a and b (a < c < b) such
that f'(c) = 0
Geometrical interpretation
Let the curve y = f(x), which is continuous on
[a, b] and derivable on (a, b), be drawn.
The theorem states that between two points with
equal ordinates on the graph of f, there exists
atleast one point where the tangent is parallel to
x-axis.
Algebraic interpretation
Between two zeros a and b of f (x) (i.e., between
two roots a and b of f(x) = 0) there exists atleast
one zero of f'(x).
10. Lagrange's mean value theorem
If a function f defined on the closed interval [a, b], is
(i) Continuous on [a, b] and
(ii) Derivable on (a, b), then there exists atleast
one real num ber c between a and b
(a < c < b) such that
f'(c) =
a
b
)
a
(
f
)
b
(
f


Geometrical interpretation
The theorem states that between two points A
and B on the graph of f there exists atleast one
point where the tangent is parallel to the chord
AB.
11. DERIVATIVE AS A RATE MEASURE
Let y = f(x) be a function of x. Let y be the
change in y corresponding to a small change x
in x. Then
x
y


represents the change in y due to
a unit change in x. In other words,
x
y


represents
the average rate of change of y w.r.t. x as x
changes from x to x + x.
As x  0, the limiting value of this average
rate of change of y with respect to x in the interval

10. [x, x + x] becomes the intantaneous rate of
change of y w.r.t. x.
Thus,
x
y
lim
0
x 



= instantaneous rate of change of
y w.r.t. x

dx
dy
= rate of change of y w.r.t. x 









 dx
dy
x
y
lim
0
x

The word " instantaneous" is often dropped.
Hence,
dx
dy
represents the value of change of y
w.r.t. x for a definite value of x.
Related rates :
Generally we come across with the problems in
which the rate of change of one of the quantities
involved is required corresponding to the given
rate of change of another quantity. For example,
suppose the rate of change of volume of a
spherical balloon is required when the rate of
change of its radius is given. In such type of
problems, we must find a relation connecting such
quantities and differentiate this relation w.r. to
time.

11. SOLVED EXAMPLES
Ex.1 The slope of tangent to the curve x = a sin3 t,
y = a cos3 t at t =

3
is-
(A) 3 (B) – 3
(C)
1
3
(D) –
1
3
Sol. dx/dt = 3a sin2 t cos t
and dy/dt = – 3a cos2 t sin t
 dy/dx =
dy dt
dx dt
/
/
= –
3
3
2
2
a t t
a t t
cos sin
sin cos
= – cot t

dy
dx t
F
H
GI
K
J  /3
= –
1
3
 slope of tangent = –
1
3
Ans.[D]
Ex.2 The equation of tangent to the curve y = sin x
at the point (  , 0) is -
(A) x + y = 0 (B) x + y = 
(C) x– y =  (D) x–y = 0
Sol. y = sin x 
dy
dx
= cos x

dy
dx
F
H
GI
K
J
( , )
 0
= – 1
Therefore equation of tangent at ( , )
 0 is
given by
y – 0 = – 1 (x –  )  x + y = 
Ans.[B]
Ex.3 The equation of tangent at the point 't' to the
curve x = a cos3 t, y = a sin3 t is -
(A) x sec t – y cosec t = a
(B) x sec t + y cosec t = a
(C) x cosec t – y sec t = a
(D) x cosec t + y sec t = a
Sol. dx/dt = –3a cos2 t sin t
dy/dt = 3a sin2 t cos t
 dy/dx =
3
3
2
2
a t t
a t t
sin cos
cos sin

= –
sin
cos
t
t
(slope of tangent at point 't')
Therefore the equation of tangent at the point
't' is written as
y – a sin3 t = –
sin
cos
t
t
(x–a cos3 t)

y
t
sin
– a sin2 t = –
x
t
cos
+ a cos2 t
 x sec t + y cosec t = a Ans.[B]
Ex.4 The equation of the normal to the curve
y = x (2–x) at the point (2, 0) is-
(A) x –2y = 2 (B) x – 2y + 2 = 0
(C) 2x + y = 4 (D) 2x + y – 4 = 0
Sol. Differentiating y2 = 2x – x2, we get
dy
dx
= 2– 2x 
dy
dx
F
H
GI
K
J
( , )
2 0
= – 2
 normal at (2,0) is
y – 0 =
1
2
(x–2)
 x – 2y = 2 Ans. [A]
Ex.5 The point of the curve 3x2 – y2 = 8 at which
the normal is parallel to the line x+ 3y = 4is-
(A) (2, –2) (B) (  2 ,  2 )
(C) (–2, 2) (D) (3, 2)
Sol. Here 3x2 – y2 = 8 ...(1)
 6x – 2y .
dy
dx
= 0 
dy
dx
=
3x
y
 slope of normal = – y/3x
slope of given line = –1/3
Since normal is parallel to the given line,
therefore – y/3x = – 1/3  y = x
Putting this value of y in the equation of the
curve, we get x =  2 and y =  2
So required point is (  2,  2) Ans.[B]
Ex.6 If tangent line drawn to the curve
x = at2, y = 2at is perpendicular to x– axis,
then point of contact is-
(A) (0,0) (B) (0,a)
(C) (a,0) (D) (a,a)
Sol.
dx
dt
= 2at and
dy
dt
= 2a 
dy
dx
=
2
2
a
at
=
1
t
[Slope of tangent at the point (at2, 2at)]
Since tangent line is perpendicular to x -
axis , then

dy
dx
=  
1
t
   t = 0
Therefore required point = (0, 0)
Ans.[A]
Ex.7 The equation of the tangent to the curve
y = be–x/a at the point where it meets y– axis is-
(A)
x
b
+
x
b
= 1 (B)
x
a
+
y
b
= 1
(C)
x
b
+
y
a
= 2 (D)
x
a
+
y
b
= 2

12. Sol. The point of intersection of the curve with
y- axis is (0,b).Differentiating y = be–x/a we
have
dy
dx
= –
b
a
e–x/a

dy
dx b
F
H
GI
K
J
( , )
0
= –
b
a
 the equation of the required tangent is
y – b = –
b
a
(x– 0) 
x
a
+
y
b
= 1
Ans.[B]
Ex.8 If parametric equation of the curve is
x = et cos t, y = et sin t, then the angle
made by the tangent with x– axis at
t =  / 4 is -
(A) 0 (B)  / 4 (C)  / 3 (D)  / 2
Sol. Here dy/dx =
dy dt
dx dt
/
/
=
e t e t
e t e t
t t
t t
cos sin
sin cos

 
=
sin cos
cos sin
t t
t t



dy
dx t
F
H
GI
K
J  /4
=
2
0
= 
Thus angle made by tangent with x- axis is  / 2.
Ans.[D]
Ex.9 For what values of a the curve y = 1– ax2
and y = x2 are perpendicular to each other ?
(A) 0 (B) 1/3 (C) –1/3 (D) 1
Sol. y = 1 – ax2
 dy/dx = –2ax
and y = x2
 dy/dx = 2x
The curves are perpendicular to each other when
(– 2ax) (2x) = – 1  4ax2 = 1 ...(1)
Again, solving the equation of curves, we get
1 – ax2 = x2
 ( 1+ a) x2 = 1
 x2 =
1
1 a

4
1
a
a

= 1  a = 1/3
Ans.[B]
Ex.10 The angle of intersection of the curves
y = 4 – x2 and y= x2 is -
(A)  / 2 (B) tan–1 (4/3)
(C) tan–1 (4 2 /7) (D) None of these
Sol. Solving the given equations, we get
x =  2 , y = 2
 point of intersection =  2 2
,
e j. Now
y = 4– x2
 dy/dx = – 2x
y = x2
 dy/dx = 2x
 at  2 2
,
e j
dy
dx
F
H
GI
K
J
1
=  2 2 ,
dy
dx
F
H
GI
K
J
2
==  2 2
So required angle
= tan–1
 

2 2 2 2
1 2 2 2 2
 
e j e j
= tan–1 4 2 7
/
e j Ans.[C]
Ex. 11 The length of tangent to the curve x = a (cos t
+ log tan t/2), y= a sin t at any point is -
(A) a (B) ax (C) ay (D) xy
Sol.
dy
dx
=
dy
dt
/
dx
dt
But
dx
dt
= a  
F
H
G I
K
J
sin
tan /
sec /
t
t
t
1
2
2
1
2
2
b g
= a  
F
H
G I
K
J
sin
sin
t
t
1
=
a t
t
cos
sin
2
and
dy
dt
= a cos t

dy
dx
=
( cos ) sin
cos
a t t
a t
2 = tan t
 Length of tangent
=
y
dy
dx
dy
dx
1
2

F
H
GI
K
J
F
H
GI
K
J
=
a t t
t
sin tan
tan
1 2

=
a t t
t
sin . sec
tan
= a Ans.[A]
Ex.12 The length of normal at any point to the curve
y =
1
2
a(ex/a + e–x/a) is-
(A) y/a (B) y2/a
(C) y2/a2 (D) Constant
Sol. y = a(ex/a + e–x/a) = a cosh
x
a
F
H
GI
K
J

dy
dx
= a.
1
a
sinh
x
a
F
H
GI
K
J= sinh
x
a
F
H
GI
K
J
 Length of normal = y 1
2

F
H
GI
K
J
dy
dx
= y 1 2

F
H
GI
K
J
sinh
x
a
= y cosh
x
a
F
H
GI
K
J=
y
a
2
Ans.[B]

13. Ex.13 If the length of the subnormal at any point of
the curve y = a1–n xn is constant, then n is
equal to-
(A) 2 (B) 1/2 (C) –1 (D) –1/2
Sol.
dy
dx
= na1–n xn–1, so at any point
Length of subnormal = y
dy
dx
F
H
GI
K
J
= (a1–n xn) (na1–n xn–1)
= na2 (1–n) x2n–1
It will be constant when 2n–1 = 0 i.e., when
n = 1/2. Ans.[B]
Ex.14 The slope of tangent at (2, –1) to the curve
x = t2 + 3t –8 and y = 2t2 –2t – 5 is-
(A) –6 (B) 0
(C) 6/7 (D) 22/7
Sol. x = t2 + 3t – 8 
dx
dt
= 2t + 3
y = 2t2 – 2t = 5 
dy
dt
= 4t – 2

dy
dx
=
4 2
2 3
t
t


Since (2,–1) is a point on the curve,
therefore 2 = t2 + 3t – 8  t = – 5, 2
and –1 = 2t2 – 2t – 5  t = 2, –1
Thus t = 2
Now slope of tangent at the given point
=
4 2 2
2 2 3
x
x


=
6
7
Ans.[C]
Ex.15 The point of inflexion for the curve y = (x–a)n,
where n is odd integer and n  3 is-
(A) (a, 0) (B) (0,a)
(C) (0,0) (D) None of these
Sol. Here
d y
dx
2
2 = n ( n–1) ( x– a)n–2
Now
d y
dx
2
2 = 0  x = a
Differentiating equation of the curve n times,
we get
d y
dx
n
n = n!
 at x = a,
d y
dx
n
n  0 and
d y
dx
n
n


1
1 = 0
Where n is odd.
Therefore (a, 0) is the point of inflexion.
Ans.[A]