Notes On Rate of Change of Quantities - CBSE Class 12 Maths
Velocity is the rate of change of distance with respect to time. Velocity = v = ds/dt Acceleration is defined as the rate of change in velocity with respect to time. Acceleration at time t = dv/dt = d/dt(v) = d/dt(ds/dt) = d2s/dt2 Rate of change: The change in a quantity with respect to time is known as rate of change. If a quantity y varies with respect to another quantity x, satisfying, y = f(x) then dy/dx represents the rate of change of y with respect to x. Let x = f(t) and y = g(t) be two distinct functions of time. Then dy/dx = (dy/dt)/(dx/dt), where dx/dt ≠ 0 ⇒ dy/dx = dy/dt x dt/dx The value of dy/dx at x = x0, i.e. (dy/dx)x=x0 represents the rate of change of y with respect to x at x = x0. Example : The side of a square is increasing at the rate of 0.02 cm/sec. What is the rate at which the area is increasing, when the side of the square is 6 centimetres? Sol : Let us A be the area of the square. Given that, the side of the square is increasing at the rate of 0.02 cm /sec. ⇒ ds/dt = 0.02 Area of square, A = s2 and The rate of change of area of square A = dA/dt. ⇒ dA/dt = d/dt(S2)    ⇒ dA/dt = 2S(ds/dt) Given that the side of square, s = 6 cm and ds/dt = 0.02 ⇒ dA/dt = 2(6)(0.02) = 0.24 The required rate of change of area of square = 0.24 cm2/sec. Note : The positive sign of dy/dx indicates that y increases as x increases. The negative sign of dy /dx indicates that y decreases as x decreases. Example : For what values of x is the rate of increase of x3 - 5x2 + 5x + 8 is twice the rate of Increase of x? Sol : Let y = x3 - 5x2 + 5x + 8 Differentiating both sides with respect to time "t", we get dy.dt = d(x3 - 5x2 + 5x + 8)/dt ⇒ dy/dt = (3x2 - 10x + 5)(dx/dt) .......(1) But given dy/dt = 2 (dx/dt) ⇒ (3x2 - 10x + 5) (dx/dt) = 2 (dx/dt) ⇒ (3x2 - 10x + 5) = 2 ⇒ 3x2 - 10x + 3 = 0 ⇒ (3x - 1)(x - 3) = 0 ⇒ x = 1/3 or 3 The values of x are 1/3 or 3. Example : The volume of a sphere is increasing at the rate of 20 cubic centimetres per second. Find the rate of change of its surface area at the instant when its radius is 8 cm. Sol : Let radius of the sphere = r cm, volume = V cm3 and surface area = S cm2 Given that dV/dt = 20 cm3/sec Volume of sphere, V = 4/3 πr3 ⇒ dV/dt = (4/3)(3πr2)(dr/dt) = 4πr2(dr/dt) ⇒ 20 = (4πr2) x dr/dt ⇒ dr/dt = 20/(4πr2) = 5/(πr2) ........(1) Surface area of sphere, S = 4pr2 ⇒ dS/dt = 8πr(dr/dt) ⇒ dS/dt = (8πr)x(5/πr2) ⇒ dS/dt = 40/r ⇒ dS/dt = 40/8 = 5 The rate of change of the surface area of the sphere when its radius is 8 cm is 5 cm2 / sec.

#### Summary

Velocity is the rate of change of distance with respect to time. Velocity = v = ds/dt Acceleration is defined as the rate of change in velocity with respect to time. Acceleration at time t = dv/dt = d/dt(v) = d/dt(ds/dt) = d2s/dt2 Rate of change: The change in a quantity with respect to time is known as rate of change. If a quantity y varies with respect to another quantity x, satisfying, y = f(x) then dy/dx represents the rate of change of y with respect to x. Let x = f(t) and y = g(t) be two distinct functions of time. Then dy/dx = (dy/dt)/(dx/dt), where dx/dt ≠ 0 ⇒ dy/dx = dy/dt x dt/dx The value of dy/dx at x = x0, i.e. (dy/dx)x=x0 represents the rate of change of y with respect to x at x = x0. Example : The side of a square is increasing at the rate of 0.02 cm/sec. What is the rate at which the area is increasing, when the side of the square is 6 centimetres? Sol : Let us A be the area of the square. Given that, the side of the square is increasing at the rate of 0.02 cm /sec. ⇒ ds/dt = 0.02 Area of square, A = s2 and The rate of change of area of square A = dA/dt. ⇒ dA/dt = d/dt(S2)    ⇒ dA/dt = 2S(ds/dt) Given that the side of square, s = 6 cm and ds/dt = 0.02 ⇒ dA/dt = 2(6)(0.02) = 0.24 The required rate of change of area of square = 0.24 cm2/sec. Note : The positive sign of dy/dx indicates that y increases as x increases. The negative sign of dy /dx indicates that y decreases as x decreases. Example : For what values of x is the rate of increase of x3 - 5x2 + 5x + 8 is twice the rate of Increase of x? Sol : Let y = x3 - 5x2 + 5x + 8 Differentiating both sides with respect to time "t", we get dy.dt = d(x3 - 5x2 + 5x + 8)/dt ⇒ dy/dt = (3x2 - 10x + 5)(dx/dt) .......(1) But given dy/dt = 2 (dx/dt) ⇒ (3x2 - 10x + 5) (dx/dt) = 2 (dx/dt) ⇒ (3x2 - 10x + 5) = 2 ⇒ 3x2 - 10x + 3 = 0 ⇒ (3x - 1)(x - 3) = 0 ⇒ x = 1/3 or 3 The values of x are 1/3 or 3. Example : The volume of a sphere is increasing at the rate of 20 cubic centimetres per second. Find the rate of change of its surface area at the instant when its radius is 8 cm. Sol : Let radius of the sphere = r cm, volume = V cm3 and surface area = S cm2 Given that dV/dt = 20 cm3/sec Volume of sphere, V = 4/3 πr3 ⇒ dV/dt = (4/3)(3πr2)(dr/dt) = 4πr2(dr/dt) ⇒ 20 = (4πr2) x dr/dt ⇒ dr/dt = 20/(4πr2) = 5/(πr2) ........(1) Surface area of sphere, S = 4pr2 ⇒ dS/dt = 8πr(dr/dt) ⇒ dS/dt = (8πr)x(5/πr2) ⇒ dS/dt = 40/r ⇒ dS/dt = 40/8 = 5 The rate of change of the surface area of the sphere when its radius is 8 cm is 5 cm2 / sec.

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