DERIVATIVE AS A RATE MEASURE

If a quantity y depends on and varies with a quantity x then the rate of change of y with respect to x is ᵈʸ⁄d.

Thus for example, the rate of change of pressure p with respect to height h is ᵈᵖ⁄dh.

A rate of change with respect to time is usually called as ‘the rate of change’, the ‘with respect to time’ being assumed.

Thus for example, a rate of change of current i is ᵈⁱ⁄dt and a rate of change of temperature θ is θ⁄dt and so on.

Example 1 :

Find the rate of change of the area of a circle with respect to its radius.

Solution :

If r be the radius and A be the area of a circle, then the formula to find the area of the circle is

A = πr2

To find the rate of change of the area of a circle with respect to its radius, we have to find the derivative A in the above formula with respect to r.

ᵈᴬ⁄d = π(2r)

ᵈᴬ⁄d = 2πr

Example 2 :

Find the rate of change of the volume of a cylinder with respect to its radius.

Solution :

If r be the radius, h be the height and  V be the volume of a cylinder, then the formula to find the area of the circle is

V = πr2h

Rate of change of the volume V with respect to radius r :

ᵈⱽ⁄d = π(2r)h

ᵈⱽ⁄d = 2πrh

Example 3 :

Find the rate of change of the circumference C of a circle of radius r with respect to time t, if the radius is increasing at a rate of 2 cm per second. 

Solution :

Formula to find the circumference of a circle :

C = 2πr

Rate of change of the circumference with respect to time t.

ᵈᶜ⁄dt = 2πᵈʳ⁄dt

Subsitute ᵈʳ⁄dt = 2.

ᵈᶜ⁄dt = 2π(2)

ᵈᶜ⁄dt = 4π cm/sec.

When the radius of a circle is increasing at a rate of 2 cm per second, its circumference is increasing at a rate of 4π cm per second.

Example 4 :

Find the rate at which the area A of a circle of radius r is increasing, when the radius is 5 cm and the radius is increasing at a rate of 0.5 cm per second.

Solution :

Formula to find the area of a circle :

A = πr2

Rate of change of the area with respect to time t.

ᵈᴬ⁄dt = π(2r)ᵈʳ⁄dt

ᵈᴬ⁄dt = 2πrᵈʳ⁄dt

Substitute r = 5 and ᵈʳ⁄dt = 0.05.

ᵈᴬ⁄dt = 2π(5)(0.5)

ᵈᴬ⁄dt = π cm2/sec

Wen the radius of a circle is 5 cm and the radius is increasing at a rate of 0.5 cm per second, area of the circle is increasing at a rate of π cm square per second.

Example 5 :

The length x meters of a certain metal rod at temperature θ°C is given by

x = 1 + 0.00005θ + 0.0000004θ2

Determine the rate of change of length in mm/°C when the temperature is

(i) 100°C and

(ii) 400°C.

Solution :

To get rate of change of length with respect to temperature, we have to find the derivative x with resepct to θ.

x = 1 + 0.00005θ + 0.0000004θ2

ᵈˣ⁄ = 0 + 0.00005(1) + 0.0000004(2θ)

ᵈˣ⁄ = 0.00005 + 0.0000008θ

Part (i) :

When θ = 100°C,

ᵈˣ⁄ = 0.00005 + 0.0000008(100)

ᵈˣ⁄ = 0.00005 + 0.00008

ᵈˣ⁄ = 0.00013 m/°C

To change meters to millimeters, multiply meters by 1000.

ᵈˣ⁄ = (0.00013 ⋅ 1000) mm/°C

ᵈˣ⁄ = 0.13 mm/°C

Part (ii) :

When θ = 400°C,

ᵈˣ⁄ = 0.00005 + 0.0000008(400)

ᵈˣ⁄ = 0.00005 + 0.00032

ᵈˣ⁄ = 0.00037 m/°C

ᵈˣ⁄ = (0.00037 ⋅ 1000) mm/°C

ᵈˣ⁄ = 0.37 mm/°C

Example 6 :

The luminous intensity I candelas of a lamp at varying voltage V is given by : I = 4 ⋅ 10-4V2. Determine the voltage at which the light is increasing at a rate of 0.6 candelas per volt.

Solution :

The rate of change of light with respect to voltage is given by d.

I = 4 ⋅ 10-4V2

dᵥ 4 ⋅ 10-4(2V)

dᵥ = 8 ⋅ 10-4V

Given : The light is increasing at 0.6 candelas per volt.

Substitute dᵥ = 0.6 into the above equation and solve for V.

0.6 = 8 ⋅ 10-4V

V = 750 volts

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