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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) = d^{2}s/dt^{2}

**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 = x_{0}, i.e. (dy/dx)_{x=x0} represents the rate of change of y with respect to x at x = x_{0}.

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 = s^{2} and The rate of change of area of square A = dA/dt.

⇒ dA/dt = d/dt(S^{2}) ⇒ 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 cm^{2}/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 x^{3} - 5x^{2 }+ 5x + 8 is twice the rate of Increase of x?

Sol : Let y = x^{3} - 5x^{2} + 5x + 8

Differentiating both sides with respect to time "t", we get

dy.dt = d(x^{3} - 5x^{2 }+ 5x + 8)/dt

⇒ dy/dt = (3x^{2} - 10x + 5)(dx/dt) .......(1)

But given dy/dt = 2 (dx/dt)

⇒ (3x^{2} - 10x + 5) (dx/dt) = 2 (dx/dt)

⇒ (3x^{2} - 10x + 5) = 2

⇒ 3x^{2} - 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 cm^{3} and surface area = S cm^{2}

Given that dV/dt = 20 cm^{3}/sec

Volume of sphere, V = 4/3 πr^{3}

⇒ dV/dt = (4/3)(3πr^{2})(dr/dt) = 4πr^{2}(dr/dt)

⇒ 20 = (4πr^{2}) x dr/dt

⇒ dr/dt = 20/(4πr^{2}) = 5/(πr^{2}) ........(1)

Surface area of sphere, S = 4pr^{2}

⇒ dS/dt = 8πr(dr/dt)

⇒ dS/dt = (8πr)x(5/πr^{2}) ⇒ 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 cm^{2} / sec.

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) = d^{2}s/dt^{2}

**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 = x_{0}, i.e. (dy/dx)_{x=x0} represents the rate of change of y with respect to x at x = x_{0}.

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 = s^{2} and The rate of change of area of square A = dA/dt.

⇒ dA/dt = d/dt(S^{2}) ⇒ 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 cm^{2}/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 x^{3} - 5x^{2 }+ 5x + 8 is twice the rate of Increase of x?

Sol : Let y = x^{3} - 5x^{2} + 5x + 8

Differentiating both sides with respect to time "t", we get

dy.dt = d(x^{3} - 5x^{2 }+ 5x + 8)/dt

⇒ dy/dt = (3x^{2} - 10x + 5)(dx/dt) .......(1)

But given dy/dt = 2 (dx/dt)

⇒ (3x^{2} - 10x + 5) (dx/dt) = 2 (dx/dt)

⇒ (3x^{2} - 10x + 5) = 2

⇒ 3x^{2} - 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 cm^{3} and surface area = S cm^{2}

Given that dV/dt = 20 cm^{3}/sec

Volume of sphere, V = 4/3 πr^{3}

⇒ dV/dt = (4/3)(3πr^{2})(dr/dt) = 4πr^{2}(dr/dt)

⇒ 20 = (4πr^{2}) x dr/dt

⇒ dr/dt = 20/(4πr^{2}) = 5/(πr^{2}) ........(1)

Surface area of sphere, S = 4pr^{2}

⇒ dS/dt = 8πr(dr/dt)

⇒ dS/dt = (8πr)x(5/πr^{2}) ⇒ 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 cm^{2} / sec.