Question

The height of a cone with semi vertical angle $\pi/3$ is increasing at the rate of 2 units/min. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{3}$
• D. $\sqrt{2}$

Hint:

In related rates problems, first establish a static equation relating the geometric quantities. Then, differentiate this equation with respect to time, treating all variables as functions of time. Finally, substitute the given instantaneous values and rates to find the unknown rate.

Answer 1

The Correct Option is C

Step 1: Write down the formula for the volume of a cone.
The volume $V$ of a cone is given by $V = \frac{1}{3}\pi r^2 h$, where $r$ is the radius and $h$ is the height. The problem states that the volume is kept fixed, so $V$ is a constant.

Step 2: Differentiate the volume equation with respect to time.
Since $r$ and $h$ are functions of time $t$, we use the product rule for differentiation. \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{3}\pi r^2 h\right). \] Since $V$ is constant, $\frac{dV}{dt} = 0$. \[ 0 = \frac{1}{3}\pi \left( (2r \frac{dr}{dt})h + r^2 (\frac{dh}{dt}) \right). \] This simplifies to $2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0$.

Step 3: Use the given semi-vertical angle to relate r and h.
The problem states that at the instant under consideration, the semi-vertical angle $\alpha = \pi/3$. The relationship is $\tan\alpha = \frac{r}{h}$. \[ \tan(\pi/3) = \frac{r}{h} \implies \sqrt{3} = \frac{r}{h} \implies r = h\sqrt{3}. \]

Step 4: Solve for the required rate $\frac{dr{dt}$.}
From the equation in Step 2, we can write: \[ \frac{dr}{dt} = -\frac{r^2}{2rh} \frac{dh}{dt} = -\frac{r}{2h} \frac{dh}{dt}. \] Now substitute the relationship $r=h\sqrt{3}$ from Step 3: \[ \frac{dr}{dt} = -\frac{h\sqrt{3}}{2h} \frac{dh}{dt} = -\frac{\sqrt{3}}{2} \frac{dh}{dt}. \] We are given that the height is increasing at 2 units/min, so $\frac{dh}{dt} = 2$. \[ \frac{dr}{dt} = -\frac{\sqrt{3}}{2} (2) = -\sqrt{3}. \] The negative sign indicates that the radius is decreasing. The rate of decrease is the magnitude of this value. \[ \text{Rate of decrease} = \boxed{\sqrt{3}}. \]