The volume of a spherical balloon is increasing at the rate of $2 \ cm^3/sec$. When its radius is $4 \ cm$,the rate of change of its surface area (in $cm^2/sec$) is

The rate of change of the diagonal $R$ of a square with respect to its area $A$ is:

If a particle moves in a straight line according to the law $x = a \sin (\sqrt{\lambda} t + b)$,then the particle will come to rest at two points whose distance is [symbols have their usual meaning]

The equation of motion of a particle moving along a straight line is $s = 2t^3 - 9t^2 + 12t$,where the units of $s$ and $t$ are $cm$ and $sec$. The acceleration of the particle will be zero after:

The volume of a sphere increases at the rate of $\pi \text{ cm}^3/\text{s}$. When the radius is $2 \text{ cm}$,the rate at which the radius increases is . . . . . . $\text{cm/s}$.

$A$ right circular cone has height $9 \text{ cm}$ and radius of base $5 \text{ cm}$. It is inverted and water is poured into it. If at any instant, the water level rises at the rate $\frac{\pi}{A} \text{ cm/sec}$, where $A$ is the area of the water surface at that instant, then the cone is completely filled in: (in $\text{ sec}$)