The volume of a sphere is increasing at the rate of $1200 \text{ cm}^3/\text{s}$. The rate of increase in its surface area when the radius is $10 \text{ cm}$ is: (in $\text{ cm}^2/\text{s}$)

The point on the curve $y^2 = 18x$,at which the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate is . . . . . . . $\left(\frac{dx}{dt} \neq 0\right)$

Consider an expanding sphere of instantaneous radius $R$ whose total mass remains constant. The expansion is such that the instantaneous density $\rho$ remains uniform throughout the volume. The rate of fractional change in density $\left(\frac{1}{\rho} \frac{d \rho}{dt}\right)$ is constant. The velocity $v$ of any point on the surface of the expanding sphere is proportional to

$A$ point is moving on the curve $y=x^3-3x^2+2x-1$ and the $y$-coordinate of the point is increasing at the rate of $6 \text{ units/sec}$. When the point is at $(2, -1)$,the rate of change of the $x$-coordinate of the point is:

$A$ particle moves in a straight line in such a way that its velocity at any point is given by $v^2 = 2 - 3x$,where $x$ is measured from a fixed point. The acceleration is

In $\Delta ABC$,sides $a$ and $b$ are kept constant such that there is an error $\alpha$ in angle $C$. What is the relative error in the area?