The rate of change of the volume of a sphere of radius $r$ with respect to its surface area is:

If $pv = 81$,then $\frac{dp}{dv}$ at $v = 9$ is equal to

$A$ ladder $5 \ m$ long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of $2 \ m/sec$. How fast is its height on the wall decreasing when the foot of the ladder is $4 \ m$ away from the wall?

$A$ point is moving on $y=4-2x^2$. The $x$-coordinate of the point is decreasing at the rate of $5 \text{ units/s}$. Then, the rate at which the $y$-coordinate of the point is changing when the point is at $(1, 2)$ is

$A$ stone is dropped in a quiet lake and it is observed that waves move in circles. If the radius of a circular wave increases at the rate of $2 \text{ cm/sec}$, then the rate of increase in its area at the instant when its radius is $10 \text{ cm}$, is in $\text{cm}^2\text{/sec}$: (in $\pi$)

The displacement $x$ of a particle at time $t$ is given by $x = At^2 + Bt + C$, where $A, B, C$ are constants and $v$ is the velocity of the particle. Then the value of $4Ax - v^2$ is: