$A$ sphere increases its volume at the rate of $\pi \text{ cm}^3/\text{s}$. The rate at which its surface area increases when the radius is $1 \text{ cm}$ 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:

The radius of a circle is increasing at a rate of $0.7 \text{ cm/s}$. The rate at which the circumference of the circle is increasing is . . . . . . .

If the velocity of a moving particle is directly proportional to the square root of the distance it has covered,what is its acceleration?

The law of motion of a body moving along a straight line is $x = \frac{1}{2} vt$,where $x$ is its distance from a fixed point on the line at time $t$ and $v$ is its velocity. Then:

$A$ spherical balloon is being inflated at the rate of $35 \, cm^3/min$. The rate of increase in the surface area (in $cm^2/min$) of the balloon when its diameter is $14 \, cm$,is