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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?
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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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