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=x3 Now by differentiating V with respect to t, we get
⇒=. ⇒==3x2 By substituting the value of dV/dx in dV/dt we get
⇒=3x2. Now by substituting the dV/dt = 7 cm
3/sec in the above equation we get,
⇒7=3x2.⇒= As we know that, surface area of a cube is given by:
6a2 , where a is the length of each side of the cube.
i.e S =
6a2 Now by differentiating S with respect to t we get,
⇒=. ⇒==12x Now by substituting the value of dS/dx in dS/dt, we get
⇒=12x. By substituting the value of
dx∕dt=7∕3x2 in the abve equation we get
⇒=12x.= Now by substituting x = 12 cm in the above equation we get,
⇒[]x=12==2cm2∕sec