The surface area of a spherical ball is increasing at the rate of $4\pi \text{ cm}^2/\text{s}$. The rate at which the radius is increasing when the surface area is $16\pi \text{ cm}^2$ is (in $\text{ cm/s}$)

If a spherical balloon has a variable diameter $d = 3x + \frac{9}{2}$ units, then the rate of change of its volume $V$ with respect to $x$ is:

$A$ rod of length $41 \ ft$ with an end $A$ on the floor and another end $B$ on the wall perpendicular to the floor is sliding away horizontally from the wall at the rate of $3 \ ft/min$. When the end $B$ is at a height of $9 \ ft$ from the floor, the rate at which the area of the triangle formed by the rod with the wall and the floor changes at that instant is (in $ft^2/min$):

The radius of a sphere is measured to be $20 \, cm$ with a possible error of $0.02 \, cm$. The consequent error in the surface area of the sphere is ....... $sq \, cm$.

The displacement of a particle in time $t$ is given by $s = 2t^2 - 3t + 1$. The acceleration is

The point on the curve $y^2 = 18x$,at which the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate is . . . . . . . $\left(\frac{dx}{dt} \neq 0\right)$