Question

The volume of a cube is increasing at the rate of 6 cm³/s. How fast is the surface area of cube increasing, when the length of an edge is 8 cm?

Answer 1

Given

`(dv)/dt` = 6 cm³/s 

l = 8 cm

find, `(ds)/dt`

V = volume of cube = l³

⇒ `(dv)/dt = 3l^2 (dl)/dt`

⇒ `6 = 3l^2 (dl)/dt`

⇒ `2/l^2 = (dl)/dt`   ...(i)

Surface Area of cube

S = 6l²

⇒ `(ds)/dt = 12l  (dl)/dt`   ...(ii)

Now put eqn (i) in enq (ii)

⇒ `(ds)/dt = 12lxx2/l^2`

⇒ `(ds)/dt = 24/8`

⇒ `(ds)/dt` = 3 cm²/s