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If the surface area of a cube is increasing at a rate of 3.6 cm2/sec, retaining its shape; then the rate of change of its volume (in cm3/sec), when the length of a side of the cube is 10 cm, is : (1) 9 (2) 18 (3) 10 (4) 20 |
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Answer» (a) 9 \(\frac{d}{dt}\)(6a2) = 3.6 ⇒ 12a \(\frac{da}{dt}\) = 3.6 a\(\frac{da}{dt}\) = 0.3 \(\frac{dv}{dt}\) = \(\frac{d}{dt}\)(a3) = 3a \((a\frac{da}{dt})\) = 3 x 10 x 0.3 = 9 |
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