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

Find the point on the parabola $y^2 = 18x$ where the rate of change of the $y$-coordinate is twice the rate of change of the $x$-coordinate.

$A$ ball thrown vertically upwards falls back on the ground after $6 \, s$. Assuming that the equation of motion is of the form $s = ut - 4.9t^2$,where $s$ is in metres and $t$ is in seconds,find the initial velocity $u$ at $t = 0$ in $m/s$.

If the volume of a sphere is increasing at the rate of $12 \text{ cm}^3/\text{sec}$,then the rate (in $\text{cm}^2/\text{sec}$) at which its surface area is increasing,when the diameter of the sphere is $12 \text{ cm}$,is:

If the volume of a sphere increases at the rate of $2 \pi \text{ cm}^3/\text{s}$,then the rate of increase of its radius (in $\text{cm}/\text{s}$),when the volume is $288 \pi \text{ cm}^3$,is

$A$ ladder $20 \ ft$ long leans against a vertical wall. The top end slides downwards at the rate of $2 \ ft/sec$. The rate at which the lower end moves on a horizontal floor when it is $12 \ ft$ from the wall is