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0412/IIT.14/CR/Bk.4/Ch.20/Pg.108

Applications of Derivatives

1. Tangents and Normals

Tangents Let y = f (x) be a continuous and differentiable curve

and P (x1, y1) is any point on it. Then,

1 1(x , y )

dy

dx

= tan = slope of tangent at P

Thus, the equation of the tangent at (x1, y1) would be

(y y1) =

1 1

1(x , y )

dy(x x )

dx

Normals Similarly, the equation of the normal at (x1, y1) would be

(y y1) =

1 1

1

(x , y )

1(x x )

dy

dx

for

1 1(x , y )

dy0

dx

and x = x1 for

1 1(x , y )

dy0

dx

Length of Tangent, Normal, Subnormal and Subtangent.

Let P (x, y) be any point on y = f (x). Let the tangent drawnat P meets the xaxis at M and normal drawn at P meets the

xaxis at N. If PL is perpendicular from P on the xaxis,

then

PM = length of tangent

PN = length of normal

ML = length of sub-tangent

LN = length of subnormal

Let PML =

We have, tan =

dy

dx and PL = y

then, PM = y cosec =

2dx

y 1dy

Length of tangent = PM = | y |

2dx

1dy

Now, NPL = PML =

PN =y

cos

= y sec = y

2dy

1

dx



Notes on Applications of Derivatives (109)

0412/IIT.14/CR/Bk.4/Ch.20/Pg.109

length of normal = PN =

2dy

| y | 1dx

ML =PL

tan =

y

(dy/dx) =

y dx

dy

length of subtangent = ML = dxydy

Similarly, length of subnormal = LN =dy

ydx

Parametric form

dy dy dx y

dx dt dt x

If the equations of the curve be

x = f(t), y = (t), then

Tangent is y (t) = y [x f ( t)]x

Normal is y (t) = x

[x f ( t)]y

Tangent is || to xaxis if y = 0.

Tangent is to xaxis if x = 0.

Partial Differentiation

Y y = x

y

f (X x)

f

or x y(X x)f (Y y)f 0 ……(1)

or x y(a,b) (a,b)

(x a) f (y b) f = 0 at the point (a, b).

Normal will bex y

X x Y y

f f

……(2)

orx y

(a,b) (a,b)

x a y b

f f

at the point (a, b)

Tangent is || to xaxis if f x = 0.it is to xaxis if f y = 0.

Condition for tangent to be parallel or perpendicular to x axis :

If a tangent is parallel to xaxis or normal is perpendicular to xaxis then m = 0 so that dy/dx = 0.

If the tangent is perpendicular to xaxis or normal is parallel to xaxis then m = .

dy/dx = or its reciprocal dx/dy = 0



(110) Vidyalankar : IIT Maths

0412/IIT.14/CR/Bk.4/Ch.20/Pg.110

Angle of intersection of two curves :

Angle of intersection of two curves is the angle between the tangents to the two curves at their

common point of intersection. Hence, if be the acute angle between the tangents, then,

tan = 1 21 2

m m

1 m m

where, m1 = value ofdy

dx

at the intersection point for 1st curve,

m2 = value ofdy

dx

at the intersection point for 2nd curve

If the angle of intersection is 90, i.e., m1 m2 = 1, then the intersection of two curves is called

orthogonal intersection. Hence two curves are said to cut orthogonally if their angle of intersection

is a right angle.

i.e., = 90, tan 90 = .2. Monotonicity of functions

Monotonicity means monotonous (increasing or decreasing) behaviour of functions. Consider a

function y = f (x) with D as its domain. Let D1 D, then,

i) Strictly Increasing (or Increasing) Function :

f (x) is said to be increasing in D1 if for every x1, x2 D1, x2 > x1

f (x2) > f (x1)

In the graph

f (x) =h 0

f (x h) f (x)Lim

h

=h 0

( )Lim

( )

( for (x + h) > x, f (x + h) > f (x))

f (x) > 0

ii) Monotonically increasing (or non decreasing) function

f(x) is said to be nondecreasing in D1 if for every

x1, x2 D1, x2 > x1 f (x2) f (x1). It means that the valueof

f (x) would never decrease with an increase in the value of x.

Obviously, from the graph we can say that f '(x) 0.

iii) Strictly decreasing (or decreasing) function : f (x) is said to be decreasing in D1 if for every

x1, x2 D1, x2 > x1 f (x2) < f (x1). It means that the

value of f(x) would decrease with an increase in the value

of x.




0412/IIT.14/CR/Bk.4/Ch.20/Pg.111

From the graph,

f (x) =h 0

f (x h) f (x)Lim

h

=h 0

( )Lim

( )

( for x + h > x, f (x + h) < f (x))

f (x) < 0

iv) Monotonically decreasing (or non increasing) function

f(x) is said to be nonincreasing in D1 if for every

x1, x2 D1, x2 > x1 f (x2) f (x1). It means that the value of

f (x) would not increase with an increase in the value of x.

For nonincreasing functions, f (x) 0

Summary

A function y = f (x) continuous in [a, b] and differentiable in (a, b)

is said to be increasing, nondecreasing, decreasing and nonincreasing in the interval [a, b] if for

every x1, x2 [a, b] andx2 > x1, we have f (x2) > f (x1), f (x2) f (x1), f (x2) < f (x1) and f (x2) f (x1) respectively.

Remarks :

1) If f (0) = 0 and f (x) 0, x R then f (x) 0, x ( , 0) and f (x) 0,

x (0, ).

2) If f (0) = 0 and f (x) 0, x R, then f (x) 0, x ( , 0) and f (x) 0,

x (0, ).

3) The point at which f (x) = 0 or doesn't exist are called critical points in the given interval.

4) The points in domain at which f (x) = 0 are called stationary points.

3. Maxima and Minima

(1) Maxima and Min ima for Continuous Functions

a) Basic Definition of Local Maxima and Minima

i) A function y = f (x) is said to have a local (or relative) maxima at x = a if f (a) is the greatest

of all the values of f (x) in the interval (a h, a + h) for small positive h, i.e., f(a) > f (a h)

and f (a) > f (a + h)

ii) A function y = f (x) is said to have a local (or relative) minima at x = b if f (b) is the least of

all the values of f (x) in the interval (b h, b + h) for small positive h, i.e., f(b) < f (b h)

and f (b) < f (b + h).




0412/IIT.14/CR/Bk.4/Ch.20/Pg.112

b) Test for Local Maxima / Minima

i) Test for local maxima / minima at x = a if f (x) is differentiable at x = a.

If f (x) is differentiable at x = a and

x a

dy0

dx

, then we have the following two

methods to check whether f (x) has a local minima or local maxima or neither at x = a.

Method 1 :

Calculate dy/dx = 0 and solve for x and say x = a, b, c etc.

Put values of x slightly less than a in dy/dx and values of x slightly greater than a.

Ifdy

dxchanges sign from +ve to ve, then Maximum at x = a.

Ifdy

dx changes sign from ve to +ve then Minimum. at x = a.

In case there is no change of sign, then neither a maximum nor a minimum.

Method 2 :

Calculate dy/dx = 0 and solve for x. Suppose one root of dy/dx = 0 is at x = a.

If2

2

d y

dx = ve for x = a , then Max. at x = a

If2

2

d y

dx

= +ve for x = a , then Min. at x = a.

If2

2

d y

dx= 0 at x = a , then find

3

3

d y.

dx If

3

3

d y

dx 0 at x = a, then neither maximum nor minimum

at x = a. If3

3

d y

dx = 0 at x = a, then find

4

4

d y

dx. If

4

4

d y

dx > 0, i.e., +ve at x = a, then y is minimum at

x = a and if4

4

d y

dx < 0 i.e., ve at x = a, then y is maximum at x = a and so on.

ii) Test for local maxima / minima at x = a if f (x) is not differentiable at x = a.

Case 1 : If f (a h) and f (a + h) both exist and are nonzero, then f (x) has a localmaxima if f (a h) > 0 and f (a + h) < 0 and f (x) has a local minima if,

f (a h) < 0 and f (a + h) > 0.

Case 2 : If f (a h) and f (a + h) both exist and one of them is zero, we should decide the

existence of local maxima / minima at x = a from the basic definition of local

maxima / minima.

(2)Maxima and M in ima for Discontinuous Functions

If f (x) is not continuous at x = a, then also we should decide the existence of local maxima /

minima or neither of them at x = a from the basic definition of local maxima / minima. It is

always advisable to draw the graph of the function because the graph would give us the clear

picture of the existence of local maxima / minima at x = a.




0412/IIT.14/CR/Bk.4/Ch.20/Pg.113

(3) Global Maxima and M in ima (or Absolute Maximum / M in imum value)

Let y = f (x) be a given function in an interval [a, b]. Then the global maxima is the largestvalue of the function in that interval. It can be attained either at the critical points of f (x) within

[a, b] or at the endpoint of the interval. Similarly, global minima is the least value of the function

in that interval.

Thus, we can say that if C1, C2, …… , Cn are the critical points in the interval [a, b], then the

greatest value of the function in [a, b] is given by,

G = max { f (a), f (C1), f (C2), …… , f (b) } and the lowest value of the function in [a, b] is given by

L = min { f (a), f (C1), f (C2), …… , f (b) }

Note :

Greatest and least value in open interval (a, b) : Let C1, C2, …… , Cn be the critical point in the

interval (a, b).

Let, G = max { f (C1), f (C2), …… , f(Cn) } andL = min { f (C1), f (C2), …… , f (Cn) }

Now, if the limiting values at the end pointsx bLim f (x)

are greater than G or less than L, then f (x)

would not have local maximum / minimum in (a , b).

But, if G >

x a(and x b )

Lim f (x)

and L <

x a(and x b )

Lim f (x)

, then G and L would respectively be the global

maximum and global minimum of f (x) in (a, b).

4. Rolle's Theorem

Let y = f(x) be a function defined on [a, b]

and if it is

1) continuous in [a, b]

2) differentiable in (a, b) and

3) f (a) = f (b),

then there exist atleast one point in the interval

(a, b) (say c (a , b)) such thati.e., f (c) = 0.




0412/IIT.14/CR/Bk.4/Ch.20/Pg.114

Geometrical Interpretation :

Rolle’s theorem says that the curve which is differentiable has atleast one horizontal tangent

between any two points where it crosses the x-axis.

5. Lagrange's Mean Value Theorem

Let y = f(x) be a function defined on [a, b] and it is

(i) continuous in [a, b](ii) differentiable in (a, b), then there exists atleast

one point c (say) in (a, b) [i.e. c (a, b)] such

that,

f ( b) f ( a)f (c)

b a

.

Geometrical Interpretation :

The Mean Value Theorem says that somewhere between A and B the curve has atleast one tangent

parallel to chord AB.

Remember

In usual notations,

Area of a square = x2, its perimeter = 4x, where side is x.

Area of rectangle = xy, perimeter = 2 (x + y), where sides are x and y.

Area of a trapezium =

1

2 (sum of parallel sides) perpendicular distance between them.

Area of a circle = 2r , perimeter = 2r, where radius is r.

Volume of a sphere = 34r

3 and its surface = 4r

2.

Volume of right cone = 21r h

3 .

Total surface = r (r + ), where is the slant height. Where as its curved surface is r only.

Volume of a cylinder = r 2h.

Total surface = 2r(r + h). Its curved surface is 2rh.

Volume of a cuboid = xyz and surface = 2 (xy + yz + zx).

50413 applicationofderivatives cn iitadvanced mathsfinalchecked

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