The volume of sphere is increasing at the rate of 1200c.cm/sec. the rate increase in its surface area when the radius is 10cm is
dv/dt = 4πr^2 dr/dt
da/dt = 8πr dr/dt
(dv/dt)/(da/dt) = r/2
so, when r=10, da/dt = 240 cm^2/s
To find the rate of increase in the surface area of a sphere, we need to differentiate the surface area formula with respect to time.
The formula for the volume of a sphere is given by V = (4/3)πr³.
Differentiating both sides of the equation with respect to time, we get dV/dt = 4πr²(dr/dt), where dV/dt represents the rate of change of volume with respect to time and dr/dt represents the rate of change of the radius with respect to time.
Given that dV/dt = 1200 cm/sec and the radius is 10 cm, we can substitute these values into the equation to find dr/dt. Using r = 10 cm and dV/dt = 1200 cm/sec, we get:
1200 = 4π(10)²(dr/dt)
dr/dt = 1200 / (4π(10)²)
dr/dt = 1200 / (4π(100))
dr/dt = 1200 / (400π)
dr/dt = 3 / π
Therefore, the rate of increase in the surface area of the sphere when the radius is 10 cm is 3/π square cm/sec.