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Application of Differentiation Rate of Change

Example
A snow ball is melting and its volume is 3 decreasing at the rate of 12 cm / min . When its radius is 6 cm, find a) the

rate of change of its radius, b) the rate of change of its surface area.

Let Vcm3 , Acm 2 and r cm be the volume, surface area and radius of the snow ball.

4 3 V = πr 3 dV 4 2 dr 2 dr = π (3r ) = 4πr dt 3 dt dt dV Q = ?12 and r = 6 dt 2 dr ? 12 = 4π (6) dt dr 1 =? dt 12π

1 cm/ min The radius is decreasing at a rate of 12π
b) A = 4πr dA dr dr = 4π (2r ) = 8πr dt dt dt dA 1 = 8π (6)(? ) When r = 6, dt 12π = ?4
2

The surface area is decreasing at a rate of

4cm / min

2

Example
The surface area of a soap bubble increases at 2 a constant rate of 4 cm / s . Find the rate at which the volume is increasing when its radius is 2 cm.

solution : A = 4πr 2 dA dr dr = 4π (2r ) = 8πr dt dt dt dr 1 dA ∴ = dt 8πr dt

4 3 V = πr 3 dV 4 2 dr 2 dr = π (3r ) = 4πr dt 3 dt dt
=

1 dA = 4πr 8πr dt r dA
2

dr 1 dA Q = dt 8πr dt

2 dt dA = 4, r = 2 dt dV 2 ∴ = ( 4) = 4 dt 2

The volume is increasing at a rate of 4cm / s

3

Example
An inverted hollow circular cone of semi-vertical angle 30 and height 30 cm is completely filled with water. The water then begins to leak through a small hole at the 30° vertex. Initially, just as the water begins to leak, the depth decreases at a rate of 1 cm/s. a) (i) Show that the volume of water in the cone is given π h 3 , where h is the depth of water. by V = 9 (ii) Find the rate at which the volume of water is decreasing initially.

a) (i) Let r cm be the radius of water surface. 1 2 r cm V = πr h 3

1 h 2 πh 30° So, V = π ( ) h = 3 9 3 2 dV π πh dh 2 dh (ii) = (3h ) = dt 9 dt 3 dt dh dV π 2 as = ?1, = (30 )(?1) = ?300π dt dt 3 3 The volume of water is decreasing at 300π cm / s

r Q = tan 30° h

h ∴r = 33

h cm

30°

b) Assume the volume of water continues to decrease at the same rate. When the depth is 10 cm, find the rate of decrease of (i) the depth; and (ii) the area of water surface.

dV π h 2 dh b) (i) = dt 3 dt dV Now, = ? 300 π , h = 10 dt 2 π (10 ) dh ? 300 π = 3 dt

The depth is decreasing at a rate of 9 cm / s 2 (ii) Let A cm be the area of water surface.
h 2 1 2 A = πr = π ( ) = πh 3 3 dA 1 dh = π ( 2h) dt 3 dt
2

dh ∴ = ?9 dt

The area of water surface is decreasing at a rate of 60π cm 2 / s.

dh Now, = ? 9 , h = 10 dt dA 2 = π (10)(?9) = ?60π
dt 3

Example
An ice pillar in the shape of cylinder is melting so that its diameter and height are decreasing at the rates of 1 cm/h and 3 cm/h respectively. Find the rate of change of its volume when its diameter and height are 2 m and 13 m respectively.

dr = ?0.005 m/h dt

dh = ?0.03 m/h dt

V = πr h
2

dV dr ? 2 dh ? = π ?2rh + r ? dt dt dt ? ? dV 2 = π 2(1)(13)(?0.005) + (1) (?0.03) dt = ?0.16π The volume is decreasing at a rate of

[

]

0.16π m / h
3

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