$A$ triangle has two fixed vertices $A(a, 0)$ and $B(0, b)$. Let its third vertex $C$ move along the line $x = y$. If $s$ is the area of triangle $ABC$, then $\frac{ds}{dx} =$

If water is poured into a cylindrical tank of radius $3.5 \ ft$ at the rate of $1 \ ft^3/min$, then the rate at which the level of the water in the tank increases (in $ft/min$) is

$A$ vessel in the shape of an inverted cone of height $10 \ ft$ and semi-vertical angle $30^{\circ}$ is full of water. Due to a hole at the vertex, the slant height of the water in the vessel is decreasing at a constant rate of $\frac{1}{\sqrt{3}} \ ft/min$. The rate (in $cu. \ ft/min$) at which the volume of water in the vessel is decreasing, when the volume of water is $\frac{8 \pi}{\sqrt{3}} \ cu. \ ft$, is

What will be the rate of change of the volume of a sphere with radius $r$,with respect to its diameter?

Let $B \equiv (0,3)$ and $C \equiv (4,0)$. The point $A$ is moving on the line $y=2x$ at the rate of $2 \text{ units/second}$. The area of $\triangle ABC$ is increasing at the rate of

$A$ stone is dropped into a still pond,creating ripples that move at a speed of $3.5 \text{ cm/sec}$. At the instant when the radius of the ripples is $7.5 \text{ cm}$,how fast is the enclosed area increasing?