Application Of Derivatives In Real Life
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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