$A$ 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

$A$ 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

The volume of a sphere is increasing at the rate of $4 \pi \text{ cm}^3/\text{sec}$. When its volume is $288 \pi \text{ cm}^3$, the rate of increase (in $\text{cm/sec}$) in its radius is

Let $x$ be the length of each of the equal sides of an isosceles triangle and $\theta$ be the angle between these sides. If $x$ is increasing at the rate $\frac{1}{12} \text{ m/hour}$ and $\theta$ is increasing at the rate $\frac{\pi}{180} \text{ rad/hour}$,then the rate at which the area of the triangle is increasing when $x=12 \text{ m}$ and $\theta=\frac{\pi}{4}$ is:

$A$ rod of length $41 \ ft$ with an end $A$ on the floor and another end $B$ on the wall perpendicular to the floor is sliding away horizontally from the wall at the rate of $3 \ ft/min$. When the end $B$ is at a height of $9 \ ft$ from the floor, the rate at which the area of the triangle formed by the rod with the wall and the floor changes at that instant is (in $ft^2/min$):

$A$ 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