Physics, asked by reetukumari68, 6 months ago

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

Answered by DrNykterstein
4

Given :-

The surface area of a cube increases at the rate of 24 cm/

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 =

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

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