Question

An edge of a variable cube is increasing at the rate of 10cm/sec. How fast the volume and also its surface area is increasing when the edge is 5cm long.

Answer 1

Given \(\frac{dx}{dt}\) = 10 cm/sec, x = 5 cm, \(\frac{dv}{dt}\) = ? \(\frac{ds}{dt}\)= ? 

(i) V = x³ 

\(\frac{dv}{dt}\) = 3x² \(\frac{dx}{dt}\) = 3 × (5²) × 10 = 750cm³/sec. 

(ii) S = 6x² . 

\(\frac{ds}{dt}\) = 12 × \(\frac{dx}{dt}\) = 12.5 .10 = 600cm² /sec.