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CBSE Class 11 » CBSE Class 11 Study Materials » Mathematics » RATE OF CHANGE
CBSE

RATE OF CHANGE

Differentiation is the rate of change of a function with respect to another variable of a function. The rate of change in differentiation is one of the applications of differentiation.

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If derivative dsdt is written, then it is meant that the rate of change of distances with respect to the time t. When one quantity y varies with other quantity x, then dydx or (f’(x)) denotes the rate of change of y with respect to x. 

If two variables x and y are varying with respect to another variable t then, if x=f(t) and y=g(t), then by chain rule, 

                         dydx = dydt/ dxdt, if dxdt ≠ 0

The derivative as the rate of change of function has many applications which include acceleration in physics, marginal function in economics, and population growth rates also. 

DEFINE RATE OF CHANGE-

Let us take two variables x and y. If x and y are varying with respect to another variable t, i.e., if x=f(t)and y=g(t). Using the chain rule, 

                     dydx = dydt/ dxdt, if dxdt ≠ 0

Hence, the rate of change of y with respect to x can be given by the rate of change of y and that of x both with respect to t. 

dydx is positive if y increases as x increases and will be negative if y decreases as x increases. 

To explain this, let us understand it with an example-

Find the rate of change of the area of the circle per second with respect to its radius r when r=10cm. 

Area A of the circle with radius r is given by A= r2. 

The rate of change of area A with respect to radius r = dAdr= ddr(r2)=2Πr.

r= 10 cm , dAdr=20

Hence, the  Area of the circle is changing at the rate of 20 cm2/s. 

 

RATE OF CHANGE IN DIFFERENTIATION –

We will use differentiation to know whether a function is increasing or decreasing. Let I be an interval contained in the domain of real value function f. Then f is said to be 

(i) increasing on I if x1<x2 in I which implies f(x1) < f(x2) for all x1, x2 I.  

(ii) decreasing on I, if x1,x2 in I which implies f(x1) < f(x2) for all x1, x2 I.

(iii) constant on I, if f(x)=c for all x I, where c is constant. 

(iv) decreasing on I if  x1<x2 in I which implies f(x1) ≥ f(x2) for all x1, x2 I.

(v) strictly decreasing on I if   x1<x2 in I which implies f(x1) > f(x2) for all x1, x2 I.

The increasing or decreasing of a function can be defined in the other way also. Let x0 be any point in the domain of real value function f. Then, f is said to be increasing, decreasing at x0 if there exists an open interval I which contains x0 such that f is increasing, decreasing in I. 

Let f be the continuous on [a, b] and differentiable on the open interval (a, b). Then, 

(i) f is increasing in [a, b] if f'(x) > 0 for each x (a, b)

(ii) f is decreasing in [a, b] if f'(x)<  0   for each x (a, b)

(iii) f is constant function in [a, b] if  f'(x)= 0   for each x (a, b)

 

RATE OF CHANGE EXAMPLE-

The examples related to the rate of change of differentiation is given below- 

1.The volume of a cube is increasing at the rate of  9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 20 centimeters? 

 Solution- Let x be the length of a side. 

                           V be the volume 

                           S be the surface area of the cube

                     Then,  V=x3 and S=6×2, where x is a function of t. 

                                dNdt= 9cm2/ s  (Given)

                      Therefore,  9= dNdt = ddt(x2 ). dxdt (By chain rule)

                                                         = 3×2 dxdt 

                                 Or          dxdt = 3×2

                        Now,      dSdt= ddt( 6×2) =ddt(6×2). dxdt = 12x(3×2) =36x

                        When x=10 cm, dSdt= 3.6 cm2/sec 

 

  1. Show the function f given by f(x)=x3-3×2+4x, x Ris increasing on R.

           Solution-       f'(x)= 3×2-6x +4

                                             = 3( x2- 2x+1)+1  

                                             = 3(x-1)2+1 >0, in every interval on R.

                                   Therefore, the function f is increasing on R.  

CONCLUSION-

The rate of change of a function is the rate of change of one variable with respect to another variable. It is one of the important applications of derivatives. Above we have discussed how to identify the function if it is increasing, decreasing, or constant. We know that rate of change of function dydx is positive if y increases as x increases and will be negative if y decreases as x increases. 

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Get answers to the most common queries related to the CBSE 11th Examination Preparation.

What is the rate of change of function?

Ans.  Let us take two variables x and y. If x and y are varying with respect...Read full

What are the applications of derivatives?

Ans:–     The applications of the derivative are- ...Read full

What is the equation of tangent of the curve y=f(x) at any given point (x0, y0)?

Ans.  The equation of a tangent to a curve y=f(x) at (x0, y0) is given by ...Read full

What is the equation of normal when dy/dx is zero at point (x0, y0)?

Ans :   If dy...Read full

What is the critical point of a function?

 Ans :A point c in the domain of function f at which either f'(c)=0...Read full

Ans.  Let us take two variables x and y. If x and y are varying with respect to another variable t, i.e., if x=f(t)and y=g(t). Using the chain rule, 

                     dydx = dydt/ dxdt, if dxdt ≠ 0

     Hence, the rate of change of y with respect to x can be given by the rate of change of y and that of x both with respect to t. 

Ans:–     The applications of the derivative are-

  1. To determine the rate of change of quantities
  2. To find the equation of tangent and normal to the curve at a point
  3. To find turning points on the graph a function that helps in locating points on the smallest value of a function occurs. 

 

Ans.  The equation of a tangent to a curve y=f(x) at (x0, y0) is given by

                               y-y0 = dydx|(x0,y0)  (x- x0)     

 

Ans :   If dydx at any point (x0, y0) is zero, then the equation of normal is x= x0. 

 

 Ans :A point c in the domain of function f at which either f'(c)=0 or f is not differentiable is called critical point of a function. 

 

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