if the surface area of a cube increases at the rate of 24 cm^2 s^-1, then find the rate at which its volume increases w.r.t. time side of cube is given to be 2 cm.
Answers
Given :-
▪ The surface area of a cube increases at the rate of 24 cm/s²
▪ Side of cube is given to be 2 cm.
To Find :-
▪ Rate at which the volume of the cube increases with respect to time or dV / dt.
Solution :-
Given us the rate of change of the surface area of the cube as,
⇒ dS / dt = 24 cm²/s
But, we have to find the rate of change of volume of the cube, so first, we should find the rate at which the side of the cube increases. We know,
⇒ Surface Area, S = 6a²
Differentiate both sides w.r.t t
⇒ dS / dt = d(6a²) / dt
⇒ 24 = 6 × d(a²) / dt
⇒ 4 = d(a²) / da × da/dt
⇒ 4 = 2a × da/dt
⇒ da / dt = 2 / a
So, We found the rate of change of side of the side of the cube to be 2 / a, where
- a = side of cube
Now, We know,
⇒ Volume of cube, V = a³
Differentiate both sides w.r.t t
⇒ dV / dt = d(a³) / dt
⇒ dV / dt = d(a³)/da × da/dt
⇒ dV / dt = 3a² × 2 / a [ da/dt = 2 / a ]
⇒ dV / dt = 6a
Now, The volume will increase at the rate of 6a cm³/s
But, we have to find the rate of change of volume only when the side of the cube is given to be 2 cm. so put a = 2, we get
⇒ dV/dt = 6×2
⇒ dV/dt = 12 cm³ / s
Hence, The volume will increase at the rate of 12 cm³/s