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


Application of derivatives in real life,Real Life Derivative Applications

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)

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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.

by: Smith




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