If the displacement $S$ of a particle travelling along a straight line in $t$ seconds is given by $S = 2t^3 + 2t^2 - 2t - 3$,then the time taken (in seconds) by the particle to change its direction 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 the rate of $0.7 \, cm/s$. What is the rate of increase of its circumference?

$A$ is a point on the circle with radius $8$ and centre at $O$. $A$ particle $P$ is moving on the circumference of the circle starting from $A$. $M$ is the foot of the perpendicular from $P$ on $OA$ and $\angle POM = \theta$. When $OM = 4$ and $\frac{d\theta}{dt} = 6 \text{ radians/sec}$,then the rate of change of $PM$ is (in units/sec)

If the distance $s$ traveled by a particle in time $t$ is $s = a \sin t + b \cos 2t$,then the acceleration at $t = 0$ is

The rate of change of the surface area of a sphere of radius $r$ when the radius is increasing at the rate of $2 \text{ cm/sec}$ is proportional to