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

For the curve $y=5x-2x^{3}$,if $x$ increases at the rate of $2 \text{ units/sec}$,then how fast is the slope of the curve changing when $x=3$?

If a particle moves such that the displacement $s$ is proportional to the square of the velocity $v$, then its acceleration $a$ is

$A$ ladder of length $17 \,m$ rests with one end against a vertical wall and the other on the level ground. If the lower end slips away at the rate of $1 \,m/sec$, then when it is $8 \,m$ away from the wall, its upper end is coming down at the rate of

$A$ man $2 \text{ m}$ tall walks at the rate of $1 \frac{2}{3} \text{ m/s}$ towards a street light which is $5 \frac{1}{3} \text{ m}$ above the ground. The rate at which the length of his shadow is changing when he is $3 \frac{1}{3} \text{ m}$ away from the base of the light is . . . . . . (in $\text{ m/s}$)

The volume $V$ and depth $x$ of water in a vessel are connected by the relation $V = 5x - \frac{x^2}{6}$. If the volume of water is increasing at the rate of $5 \, cm^3/sec$,find the rate at which the depth of water is increasing when $x = 2 \, cm$.