The volume of a spherical balloon is increasing at the rate of $40 \ cm^3/\min$. The rate of change of the surface area of the balloon at the instant when its radius is $8 \ cm$ is ........ $cm^2/\min$.

The distance $s$ moved by a particle in time $t$ is given by $s = f(t) = t^3 - 6t^2 + 9t$,where $s$ is in metres and $t$ is in seconds. The velocity of the particle at $t = 2 \ s$ is: (in $m/s$)

The height of a right circular cylinder is decreasing while its diameter is increasing at a rate of $4 \text{ cm/s}$ so as to keep its volume unchanged. The rate of change in its lateral surface area (in $\text{cm}^2/\text{s}$) at the instant when its diameter is $8 \text{ cm}$ and height is $12 \text{ cm}$, is (in $\pi$)

An object is moving in the clockwise direction around the unit circle $x^2+y^2=1$. As it passes through the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,its $y$-coordinate is decreasing at the rate of $3 \text{ units/sec}$. The rate at which the $x$-coordinate changes at this point is

$A$ particle moves along the curve $\frac{x^2}{16} + \frac{y^2}{4} = 1$. When the rate of change of abscissa is $4$ times that of its ordinate,then the quadrant in which the particle lies is

$A$ point $P$ is moving on the curve $x^3 y^4 = 2^7$. The $x$-coordinate of $P$ is decreasing at the rate of $8 \text{ units per second}$. When the point $P$ is at $(2, 2)$, the $y$-coordinate of $P$: