application of derivatives class 12 isc

8/10/2019 APPLICATION OF DERIVATIVES CLASS 12 ISC

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Mathematics



Session

Applications of Derivatives - 1



Session Objectives

Rate of Change of Quantities

Slope and Equation of Tangent

Slope and Equation of Normal

Angle Between Two Curves

Increasing and Decreasing Functions

Use of Derivative




represents the rate

of change of y with respect to x at x = x0

0 0x=x

dyor f' x

dx

Let y = ƒ x be a function of x.

δx small change in x

δy corresponding small change in y

δx 0

δy dylim = =

δx dxRate of change of y with respect to x



Rate of Change of

Quantities

Similarly, rate of change of velocity with respect to time t,

represents acceleration.dv

dt

Hence, velocity of a point body is defined as the rate ofchange of displacement with respect to time t.

Velocity at a time t = t0 can be written as at t = t0 dsdt




If both x and y are functions of t, then

dy dy dx dx

= × = ƒ xdt dx dt dt

Rate of change of y with respect to t

dy=

dx

x rate of change of x with respect to t



Example - 1An edge of a variable cube is increasing at the

rate of 5 cm/s. How fast is the volume of thecube increasing when the edge is 6 cm long.



Solution

2dV=15x

dt2dV

=3x 5dt

2 3

x=6

dV=15 6 =540 cm /s

dt

Let x be the edge of the variable cube and

V be the volume at any time t.

2dV dx=3xdt dt

3 dxV=x and =5cm/s [Given]

dt



Example - 2The radius of a spherical soap bubble is

increasing at the rate of 0.2 cm/sec. Findthe rate of increase of its surface area,when the radius is 7 cm.



SolutionLet r be the radius of a spherical soap bubble

and S be the surface area at any time t.

2 drThen, S=4 r and =0.2 cm/s

dt

dA dr=8 rdt dt

dA

=8 r 0.2dt

x=7

dS 22=1.6× ×7dt 7

= 1.6 x 22 = 35.2 cm2 /sec



Tangent y = ƒ xLet be a continuous curve and let (x

0

, y0

)be a point on the curve.

The slope of the tangent to curve f (x) at (x0, y0) is

0 0

0x , y

dy

ƒ x or dx

The equation of the tangent to the curve at (x0, y0) is

0 0

0 0x , y

dyy - y = x - x

dx



Normal

Equation of normal to the curve at (x0 , y0) is

0 0

0 0

x , y

1y-y =- x- x

dy

dx

As normal is perpendicular to tangent at the point of contact

0 0x , y

1 1Slope of normal=- i.e.-

dySlope of tangent

dx



Example-3

Find the equation of the tangent and normal to the curve

y = x4 – 6x3 + 13x2 – 10x + 5 at (0,5).

Solution :

4 3 2y=x -6x +13x -10x+5

3 2dy=4x -18x +26x-10

dx

3 2

0,5

dy

= 4 0 -18 0 +26 0 -10=-10dx



Solution Cont.Equation of tangent at (0, 5) is y – 5 = -10 (x – 0)

y-5=-10x 10x+y-5=0

Slope of the normal at (0, 5)

1 1=- =

-10 10

Equation of normal at (0, 5) is

1

y-5= x -0 10y =x+5010



Example-4If the tangent to the curve at (1, -6) is parallel to

the line x – y + 5 = 0, find the values of a and b.

3y=x +ax+b

3Solution: Givencurve is y = x +ax+b

2dy

=3x +adx

2

1, -6

dyThe slope of the tangent at 1,- 6 = =3 1 +a

dx

=a+3



Con.

The tangent is parallel tox y 5 0 y x 5

a+3=1 a=-2

Therefore, the curve becomes 3y = x -2x+b ... i

(1, –6) lies on (i)

6 1 2 b b 5

a 2, b 5



Angle Between Two Curves

1 2θ=θ - θ

1 2tanθ=tan θ -θ

P (x, y)0 0y = g(x) y = f(x)

2

1

y

xO



Angle Between Two

Curves1 2

1 2

tanθ -tanθ

=1+tanθ tanθ

1 0 1 2 0 2Where tanθ =f' x =m and tanθ =g' x =m

The other angle is 1800 -

(1) Orthogonal curves: m1m2 = - 1

(2) Curves touch each other: m1 = m2



Example-5Show that the curves x2 = 4y and 4y + x2 = 8

intersect orthogonally at (2, 1).



SolutionWe have x2 = 4y and 4y + x2 = 8

dy

dx

dy2x = 4 and 4 +2x = 0

dx

x dy xdyand = -

dx 2 dx 2=

dy2 2dym = = =1 and m = = - = -11 2dx 2 2dx2,1 2,1

Hence, the curves intersect orthogonally at (2, 1).

m1m2 = 1 x (-1) = -1



Increasing Function

Increasinfunction

a x1 bx2

X

Y

f(x1)

f(x2)

O



Increasing FunctionA function is said to be a strictly

increasing function of x on (a, b).

1 2 1 2 1 2If x < x in a, b ƒ x < ƒ x for all x , x a, b

‘Strictly increasing’ is also referred to as ‘Monotonically increasing’.



Decreasing Function

Decreasingfunction

a x1 bx2

Y

f(x1)

f(x2)

O



Decreasing FunctionA function ƒ(x) is said to be a strictlydecreasing function of x on (a, b).

1 2 1 2 1 2If x < x in a, b ƒ x > ƒ x for all x , x a, b

‘Strictly decreasing’ is also referred toas ‘Monotonically decreasing’.



Use of Derivative

Let f(x) be a differentiable real functiondefined on an open interval (a, b).

(i) If ƒ x > 0 for all x (a, b) f(x) is increasing on (a,b).

(ii) If ƒ x < 0 for all x (a, b) f(x) is decresing on (a,b).



Use of Derivative (Con.)

Y = f(x) T

X

Y

O T' a

b Figure 1

P

Slope of tangent at any point in (a, b) > 0

As tanθ>0 for 0<θ<90°

dy

ƒ x 0dx

for all x in (a, b).



Use of Derivative (Con.)

Figure 2 T'

X

Y

T a

b P

O

Slope of tangent at any point in (a, b) < 0

As tanθ<0 for 90°<θ<180°

dy

ƒ x 0dx

for all x in (a, b).



Example-1

For the function f(x) = 2x3

– 8x2

+ 10x + 5,find the intervals where

(a)f(x) is increasing(b) f(x) is decreasing



SolutionWe have

3 2 ƒ(x) = 2x - 8x + 10x + 5

2 ƒ (x) = 6x - 16x + 10

2=2(3x - 8x +5)

=2(3x -5)(x -1)

ƒ (x)= 0 2(3x - 5) (x - 1)= 0

5x = , 1

3



Solution Cont.

5For 1< x < , ƒ (x) is negative

3

5For x > ,ƒ (x) is positive

3

For x < 1, is positive. ƒ (x)=3(3x - 5)(x - 1)

ƒ(x) is increasing for x < 1 and5

x>3

and it decreases for 5

1<x< 3



Example-2Find the intervals in which the function

in increases or decreases.

ƒ(x) = x + cosx

[0, 2 ]

Solution: We have ƒ(x) = x + cosx

ƒ (x) = 1 - sinx

As sinx is 1 for all x 0, 2

And sinx =1 for x =2

ƒ x > 0 for all x except x = 2

ƒ x is increasing for all x except x =2

.



Session

Applications of Derivatives - 2



Session Objectives

Rolle’s Theorem

Geometrical Meaning

Lagrange’s Mean Value Theorem

Geometrical Meaning

Approximation of Differentials

Class Exercise

Maximum and Minimum

Extreme and Critical points

Greatest and Least Values



Maximum and Minimum



Maximum and Minimum

e point a is called the point of maximum of the function f(x).

the figure, y = f(x) has maximum values at Q and S.

ƒ a > ƒ a+δIf and ƒ a > ƒ a- δ for all small values ofδ.

e point b is called the point of minimum of the function f(x).

the figure, y = f(x) has minimum values at R and T.

ƒ b < ƒ b +δIf and ƒ b < ƒ b - δ for all small values ofδ.

Let y = ƒ xbe a function



Extreme PointsThe points of maximum or minimum of

a function are called extreme points.

At these points, ƒ x = 0, if ƒ x exists.

X

Y

O (i)

P

f i n

c r e

a s i n

g f d

e c r e

a s i n

g

X

Y

O (ii)

Q f i n

c r e

a s i n

g f d e c r e

a s i n

g

At P and Q ƒ x does not exit.



Critical Pointshe points at which or at which

does not exist are called critical points.

ƒ x = 0 ƒ x

A point of extremum must be one of thecritical points, however, there may exista critical point, which is not a point of

extremum.



Theorem - 1et the function be continuous in some

interval containing x0 .

y = ƒ x

ƒ x > 0 ƒ x < 0(i) If when x < x0 andWhen x > x0 then f(x) has maximum

value at x = x0

ƒ x < 0 ƒ x > 0(ii) If when x < x0 andWhen x > x0 ,then f(x) has minimum

value at x = x0



Theorem - 2If x

0

be a point in the interval in which y= f(x) is

defined and if

0 0 ƒ x = 0 and ƒ x 0

if 0 0i ƒ x is a maximum ƒ x < 0

if 0 0ii ƒ x is a minimum ƒ x > 0



Greatest and Least Values

The greatest or least value of a continuousfunction f(x) in an interval [a, b] is

attained either at the critical points or atthe end points of the interval.

So, obtain the values of f(x) at these

points and compare them to determinethe greatest and the least value in the

interval [a, b].



Example-3

Find all the points of maxima and

minima and the correspondingmaximum and minimum values of

the function:

4 3 23 45f x = - x - 8x - x +105

4 2

(CBSE 1993)



Solution

4 3 23 45f x = - x - 8x - x +1054 2

3 2f' x = -3x - 24x - 45x

We have

2f' x = -3x x +8x+15

For maximum or minimumf’(x) = 0

2-3x x +8x+15 =0

-3x x +3 x +5 = 0

x=0,-3,-5



Solution Cont. 2f'' x = -9x - 48x - 45

At x = 0, f'' 0 = -45 < 0

f(x) is maximum at x = 0

The maximum value at x = 0 is f(0) = 105

f(x) is minimum at x = -3

The minimum value at x = -3 is

2

f'' -3 =-9 -3 - 48 -3 - 45=18 >0At x = -3,

4 3 23 45 231

f -3 = - -3 - 8 -3 - -3 +105 =4 2 4



Solution Cont.

The maximum value at x = -5 is

2

f'' -5 = -9 -5 - 48 -5 - 45 =-30 < 0

f(x) is maximum at x = -5At x = -5,

4 3 23 45 295

f -5 = - -5 - 8 -5 - -5 +105 =4 2 4



Example-4Show that the total surface area of a cuboid with a square

base and given volume is minimum, when it is a cube.

Solution: Let the cuboid has a square base of edge x andheight y.

2The volume of cuboid, V = x y

The surface area of cuboid, S =2 x×x+x×y+x×y

2= 2x + 4xy

2

2

V=2x + 4x.

x



Con.

2 2V

S=2 x + x

dSFor minimum surface area, =0

dx

2

2V2 2x - =0

x

3 3x - V =0 x = V

2

2 3

d S 4V=2 2 +

dx x



Con.

3

2V= 4 1+

x

3x = V

2

2

d S 2V= 4 1+ = 4×3 =12

Vdx

23

2

d VAs > 0 at x = V

dx

3x = V , surface area is minimum. At



Con.

33

x = V V = x2 3x y = x y = x

Cuboid is a cube.



Rolle’s Theorem Let f(x) be a real function defined in the closed

interval [a, b] such that

(i) f(x) is continuous in the closed interval [a, b]

(ii) f(x) is differentiable in the open interval (a, b).

(iii) f(a) = f(b)

Then, there is a point c in the open interval

(a, b), such that ƒ c = 0



Geometrical Meaning

There will be at least one point with in [a, b] wheretangent of the curve will be parallel to x-axis.

ƒ c = 0

f(a) = f(b)

Y

Xx = a x = b

f(a)

A

f(b)

B

Figure (1)O

f(a) = f(b)

f(a)

Y

O Xx = a x = b

I

II

IIIIV

f(b)B

Figure (2)



Example - 1

Verify Rolle’s theorem for the function

f(x) = x2 – 8x + 12 on the interval [2, 6].

Solution :

(1) Given function f(x) is polynomial function.

f(x) is continuous on [2, 6]

(2) f'(x) = 2x – 8 exists in (2, 6)

f(x) is differentiable in (2, 6)

We have f(x) = x2 – 8x + 12



Solution(3) f(2) = 22 – 8 x 2 + 12 = 0 and f(6) = 62 – 8 x 6 + 12 = 0

f(2) = f(6)

Three exists some such that f'(c) = 0 c 2,6

2c-8 =0 c = 4 2, 6

Hence, Rolle’s theorem is verified.

All the conditions of Rolle’s theorem is satisfied.



Example - 2Using Rolle’s theorem, find the points on the curve

where the tangent is parallel to x-axis.y = x(x - 4), x [0, 4],

Solution: (1) Being a polynomial function, the given function iscontinuous on [0, 4].

2 y =2(x - 2) exits in 0, 4 .

Function is differentiable in (0, 4).

x=0 x=43 y =0 and y =0



Con.

All conditions of Rolle’s theorem are satisfied.

So, x 0, 4 such that

ƒ (x) = 0 2(x - 2) = 0 x = 2

y =2(2 - 4) y =- 4

Required point is (2, –4)

Lagrange’s Mean Value



Lagrange s Mean Value

TheoremLet f(x) be a function defined on [a, b] such that

(i) it is continuous on [a, b].

(ii) it is differentiable on (a, b).

Then,there exists a real number c (a,b)

such that ƒ b - ƒ a

ƒ c =b-a



Geometrical Meaning

O D EX

Y

(a, f(a)) C

B(b, f(b))

(c, f(c))

A

F



Geometrical Meaning

By Lagrange’s Mean Value theorem,

ƒ b - ƒ a

ƒ c = tan = ƒ cb-a

Slope of the chord AB = Slope of the tangent at c, ƒ c

From the triangle AFB,

ƒ b - ƒ aBFtan = tan =

AF b-a



Example - 3Verify Lagranges Mean Value theorem for

the function f(x) = x2

– 3x + 2 on [-1, 2].Solution :

(1) The function f(x) being a polynomialfunction is continuous in [-1, 2].

(2) f'(x) = 2x – 3 exists in (-1, 2)

f(x) is differentiable in (-1, 2)



Solution

c -1, 2So, there exists at least one such that

f 2 - f -1f' c =

2- -1

0-62c -3=2+1

1

2c - 3 = -2 c = -1, 22

Hence, Lagrange's mean value theorem is verified.



Example - 4Using Lagrange’s mean value theorem, find the point on the curve

, where tangent is parallel to the chord joining (1, –2)and (2, 1).

3y = x - 3x

Solution: (1) The function being a polynomial function is continuous

on [1, 2].

22 y = 3x - 3 exists in 1, 2

Function is differentiable in (1, 2).

So, x 1, 2 such that tangent is parallel to chord joining(1, –2) and (2, 1)



Contd.

2 ƒ(2) - ƒ(1) 2 - (-2) ƒ (x) = 3 x -1 =

2 -1 1

2 24 7 7x - 1= x = x =±

3 3 3

3

27 7y = - 3× ±3 3

2 7y =+

3 3

7 2 7 7 2 7The points are , - , - , .

3 3 3 3 3 3

Approximation of



Approximation of

Differentials

As by the definition of

dy

dx

δx 0

y dylim =

x dx

Hence, for small increment in x, change in y will be

dyy = × x

dx

dyHence, y+ y = y+ x

dx.

l



Example - 5

Solution :

Using differentials, find the approximate

value of37.

Let y = x

Taking x = 36, Δx =1 x+ Δx =37

Δy = x + Δx - x

Δy= 37-6

37=6+Δy



Contd.dy

Δy = Δx

dx

dy1 1Δy= ×1 y = x =

dx2 x 2 x

1 1 0.082 6 12

37 6 y

= 6 + 0.08 = 6.08



Example - 6Using differentials, find the approximate

value of 1329

3Let y = x

Taking x = 27, Δx =2 x+Δx =29

3 3Δy= x+Δx- x

3Δy= 29-3

329=3+Δy

Solution :



Contd.2

-dy 1 3Δy= Δx= x ×23dx

1

32

3

dy 1y= x =dx

3x

2

-3

2= 27

3

2 2= = =0.074

3×9 27

329 =3+0.074=3.074



SolutionSo, there exists at least one where tangent is

parallel to chord joining (3, 0) and (5, 4).

c 3, 5

ƒ(5) - ƒ(3) ƒ (c)=

5- 3

4 - 02(c - 3)=2

c =4 3, 5

2y (4 3) 1

At x = 4

Required point is (4, 1)

l i



Class Exercise - 5Using differentials, find the approximate

value of up to 3 places of decimals. 1482

4Let y= x

Taking x=81, Δx=1 x+Δx=82

4 4Δy= x+Δx - x

4Δy= 82 -3

482=3+Δy

Solution :



Solution3

-4dy 1

Δy= Δx= x ×1dx 4

1dy 14y=x =

3dx44x

3

-4

1= 81

4

1=

4×27

25 0.926= = =0.00926

100×27 100

482 = 3 + 0.00926 = 3.00926 3.009



Thank you

application of derivatives class 12 isc

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