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⁄dₓ.
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 ᵈ^{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 = πr^{2}
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⁄dᵣ = π(2r)
ᵈᴬ⁄dᵣ = 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 = πr^{2}h
Rate of change of the volume V with respect to radius r :
ᵈⱽ⁄dᵣ = π(2r)h
ᵈⱽ⁄dᵣ = 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 = πr^{2}
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 cm^{2}/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 + 0.0000004θ^{2}
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 + 0.0000004θ^{2}
ᵈˣ⁄dθ = 0 + 0.00005(1) + 0.0000004(2θ)
ᵈˣ⁄dθ = 0.00005 + 0.0000008θ
Part (i) :
When θ = 100°C,
ᵈˣ⁄dθ = 0.00005 + 0.0000008(100)
ᵈˣ⁄dθ = 0.00005 + 0.00008
ᵈˣ⁄dθ = 0.00013 m/°C
To change meters to millimeters, multiply meters by 1000.
ᵈˣ⁄dθ = (0.00013 ⋅ 1000) mm/°C
ᵈˣ⁄dθ = 0.13 mm/°C
Part (ii) :
When θ = 400°C,
ᵈˣ⁄dθ = 0.00005 + 0.0000008(400)
ᵈˣ⁄dθ = 0.00005 + 0.00032
ᵈˣ⁄dθ = 0.00037 m/°C
ᵈˣ⁄dθ = (0.00037 ⋅ 1000) mm/°C
ᵈˣ⁄dθ = 0.37 mm/°C
Example 6 :
The luminous intensity I candelas of a lamp at varying voltage V is given by : I = 4 ⋅ 10^{-4}V^{2}. 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 ᵈᶦ⁄dᵥ.
I = 4 ⋅ 10^{-4}V^{2}
ᵈᶦ⁄dᵥ = 4 ⋅ 10^{-4}(2V)
ᵈᶦ⁄dᵥ = 8 ⋅ 10^{-4}V
Given : The light is increasing at 0.6 candelas per volt.
Substitute ᵈᶦ⁄dᵥ = 0.6 into the above equation and solve for V.
0.6 = 8 ⋅ 10^{-4}V
V = 750 volts
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