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Question
The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution
Let at any instant of time t, the edge of the cube be x, surface area be S and the volume be V, then,
V = x3 and S = 6x2 ...(i)
Differentiating (i) w.r.t. t, we get
`= (dV)/dt = 3x^2 dx/dt` ....(ii)
and, `(dS)/dt = 6 (2x) dx/dt` ....(iii)
`(dV)/dt = 8cm^3 //sec` ...(Given)
`= 3x^2 dx/dt = 8 cm^ 3//sec` .... (using ii)
`= 3 (12 cm)^2 dx/dt = 8 cm^3// sec` ....(∴ x= 12 cm)
`= dx/dt = 8/432` cm/sec
`= 1/54` cm/sec
Subsituting this value of `dx/dt` in (iii), we get,
`(dS)/dt = 12 (12 cm) (1/54 cm//sec)` ....(∵ x = 12 cm)
`= 8/3` cm3/sec
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