The rate of change of the volume of a sphere with respect to its surface area,when the radius is $5 \ m$ is

$A$ stone is falling freely and describes a distance $s$ in $t$ seconds given by the equation $s = \frac{1}{2}g{t^2}$. The acceleration of the stone is

$A$ spherical iron ball of radius $10 \, cm$ is coated with a layer of ice of uniform thickness that melts at a rate of $50 \, cm^3/min$. When the thickness of the ice is $5 \, cm$,find the rate at which the thickness of the ice decreases.

The ordinate of a point describing the circle $x^2 + y^2 = 25$ decreases at the rate of $1 \, cm/sec$. Find the rate of change of the abscissa of the point when the ordinate is equal to $3 \, cm$ (Given $x > 0, y > 0$).

If $y=5x^2+6x+6$,$x=2$,and $\Delta x=0.001$,then the value of $dy$ is:

The radius of a circular plate is increasing at the rate of $0.01 \text{ cm/s}$ when the radius is $12 \text{ cm}$. Then,the rate at which the area increases,is