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

Rate of Change of Quantities

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.dvdt

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

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

dsdt

Rate of Change of Quantities

If both x and y are functions of t, then

dy dy dx dx= × = ƒ x

dt 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 - 1

An edge of a variable cube is increasing at the rate of 5 cm/s. How fast is the volume of the cube 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=3x

dt dt

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

dt

Example - 2

The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area,when the radius is 7 cm.

Solution

Let 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 r

dt dt

dA=8 r 0.2

dt

x=7

dS 22=1.6× ×7

dt 7

= 1.6 x 22 = 35.2 cm2/sec

Tangent

y = ƒ xLet be a continuous curve and let (x0 , y0) be a point on the curve.

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

0 0

0x , y

dyƒ 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

dydx

As normal is perpendicular to tangent at the point of contact

0 0x , y

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

dySlope of tangentdx

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=-10

dx

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

1y - 5= x - 0 10y =x+50

10

Example-4

If 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: Given curve is y = x +ax +b

2dy= 3x +a

dx

21, -6

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

dx

= a+3

Con.

The tangent is parallel to x 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)

21

y

xO

Angle Between Two Curves

1 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-5

Show that the curves x2 = 4y and 4y + x2 = 8 intersect orthogonally at (2, 1).

Solution

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

dydx

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

Thank you