UPSC NDA Math Model Paper 9

At any instant t, let the length of each side of the cube be x, V be its volume and S be its surface area. Given: dV/dt = 7 cm$^{3}$/sec As we know that, volume of a cube is given by: a , where a is the length of each side of the cube. i.e $V = x^{3}$ Now by differentiating V with respect to t, we get $\Rightarrow \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}$ $\Rightarrow \frac{dV}{dx} = \frac{d(x^{3})}{dx} = 3x^{2}$ By substituting the value of dV/dx in dV/dt we get $\Rightarrow \frac{dV}{dt} = 3x^{2} \cdot \frac{dx}{dt}$ Now by substituting the dV/dt = 7 cm$^{3}$/sec in the above equation we get, $\Rightarrow 7 = 3x^{2} \cdot \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{7}{3x^{2}}$ As we know that, surface area of a cube is given by: $6a^{2}$ , where a is the length of each side of the cube. i.e S = $6a^{2}$ Now by differentiating S with respect to t we get, $\Rightarrow \frac{dS}{dt} = \frac{dS}{dx} \cdot \frac{dx}{dt}$ $\Rightarrow \frac{dS}{dx} = \frac{d(6x^{2})}{dx} = \;12x$ Now by substituting the value of dS/dx in dS/dt, we get $\Rightarrow \frac{dS}{dt} = 12x \cdot \frac{dx}{dt}$ By substituting the value of $\frac{dx}{dt} = \frac{7}{3x^{2}}$ in the abve equation we get $\Rightarrow \frac{dS}{dt} = 12x \cdot \; \frac{7}{3x^{2}} = \frac{28}{x}$ Now by substituting x = 12 cm in the above equation we get, $\Rightarrow \left[ \frac{dS}{dt} \right]_{x=12} = \frac{28}{12} = \; 2\frac{1}{3} \; \text{cm}^{2}/\text{sec}$