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subject: Application Of Derivatives In Real Life [print this page]

In the branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; Let us see an example for application of derivatives in real life.

applications of derivatives in real life

Introduction to application of derivatives in real life:

In the branch of mathematics, the derivative is a measure of how a function changes as its input changes. A derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; Let us see an example for application of derivatives in real life.

Real Life Derivative Applications:

The applications of derivatives in day to day life are in the feild of engineering , scientific research, find the rate of change of things,find the tangents.

Examples of Application Derivatives Used in Real Life:

Ex 1:If radius of circle increasing at uniform rate of 3 cm/s, find rate of increasing of area of circle, at instant when radius is 30 cm.

Solution:

'(dr)/dt' = 3 cm/s

Area of circle = 'pi' *r * r

Differentiating w.r.t. to t,

'(dA)/dt' = 2'pi' *r '(dr)/dt'

'(dA)/dt' = 2'pi' *30*3 = 180'pi' cm2/s.

Ex 2: How ('(dy)/(dx)' )(x1,y1)= 0 when tangent is 11 to y-axis.

Solution: The slop of the line = tan 'theta' where 0 with x-axis in anticlockwise direction.

If tangent is 11 to y-axis

'(dy)/(dx)' = 0 or tan'theta' = 0

=> '(dy)/(dx)' = 0.

Ex 3:Find point on curve y = x2- 2x at which tangent is 11 to x-axis.

solution: Let the point be P(x1, y1) on curve y = x2- 2x .................................... (i)

'(dy)/(dx)' = '(d (x^2 - 2x))/(dx)'

'(d)/(dx)' (x2- 2x) = 2x1- 2

But tangent is 11 to y-axis

('(dy)/(dx)' ) = 0

2x11- 2 = 0

So we get, x1= 1 ........................................ (ii)

Since Q(x1, y1) lies on curve.

y1= x12- 2x1

For x1= 1, y1= 1 - 2 = - 1

For x1= - 1 y1= 1 + 2 = 3

Points are (1, - 1) & (- 1, 3)

Ex 4:Find eq. of tangent & normal to curve 3y = 4 - x2, at (1,1)

Solution:The given curve is 3y = 4 - x2................................... (i)

differentiating (i) w.r.t. x

3('(dy)/(dx)' ) = -3x => ('(dy)/(dx)' ) = -x

('(dy)/(dx)' )(x1,y1)(1, 1) = - 1

eq. of tangent at (1, 1) is y - 1 = - 1(x - 1) = x + y = 3

& eq. of normal at (1,1) is

y - 1 = 1(x - 1)

y - x = 0.

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Practical Uses of Derivatives:

In practical, derivatives can be used to find various characteristics of graph.

Derivatives can be used as rate measure

It is also used for graphing complicated equations.

It is used for errors and approximation

Theorems like Rolle's and Lagrange's uses derivatives

Maximum and minimum function is also one of the application of derivative.

Practically it is used to graph peaks, valley and slopes without graphing calculator.

These are the few practical uses of derivatives.

Example Problems for Practical Uses of Derivatives :

Example 1:

Compute the rate of change of a circle area with respect to its radius r when r = 16 cm.

Solution:

Let A be the area of a circle of radius r. Then,

Area, A = ?r2

By differentiating the above equation, we get

'(dA)/(dr)' = 'd/(dr)' ?r2

= 2?r

When radius of the circle, r = 16cm

'(dA)/(dr)' = 2? * 16

= 32?

Hence, the area is changing at the rate of 32? cm2/cm

Example 2:

Find absolute maximum and minimum of the function f(x) = sin x on the interval [0, ?].

Solution:

Clearly, the greatest value of the function, i.e., the absolute maximum, is f(?/2) = 1

Also, the smallest value of the function, i.e., the absolute minimum, is f(0) = f(?) = 0

Example 3:

Find maximum and minimum value of the function f(x) = - (x - 1)2 + 5 for all x e R.

Solution:

For this function, we obtain the absolute maximum when its negative part is 0, i.e., when (x - 1)2 is 0. So, the greatest value of the function is 5.

Clearly, it does not have a least value.

by: mathqa

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