Question

The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm minute. The rate of change of lateral surface when the radius = 7 cm and altitude = 24 cm, is

• A. $54\pi \text{ cm}^2/\text{min}$
• B. $7\pi \text{ cm}^2/\text{min}$
• C. $27\pi \text{ cm}^2/\text{min}$
• D. None of these

Hint:

For related rates problems involving roots, it is often less prone to error if you define the intermediate variable (like slant height $l$) and evaluate its value first before plugging it into the differentiated formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are dealing with related rates. We need to find the rate of change of the lateral surface area of a cone given the rates of change of its radius and altitude.
Step 2: Key Formula or Approach:
The lateral surface area $S$ of a right circular cone is given by: \[ S = \pi r l = \pi r \sqrt{r^2 + h^2} \] We need to differentiate this equation with respect to time $t$ to find $\frac{dS}{dt}$.
Step 3: Detailed Explanation:
Given values:
$\frac{dr}{dt} = 3 \text{ cm/min}$ (increasing)
$\frac{dh}{dt} = -4 \text{ cm/min}$ (decreasing)
At the given instant: $r = 7 \text{ cm}$ and $h = 24 \text{ cm}$.
First, let's find the slant height $l$ at this instant: \[ l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm} \] Now, differentiate $S = \pi r \sqrt{r^2 + h^2}$ with respect to $t$ using the product and chain rules: \[ \frac{dS}{dt} = \pi \left[ \frac{dr}{dt} \cdot \sqrt{r^2 + h^2} + r \cdot \frac{d}{dt}\left(\sqrt{r^2 + h^2}\right) \right] \] \[ \frac{dS}{dt} = \pi \left[ \frac{dr}{dt} \cdot l + r \cdot \left( \frac{1}{2\sqrt{r^2 + h^2}} \cdot \left(2r\frac{dr}{dt} + 2h\frac{dh}{dt}\right) \right) \right] \] \[ \frac{dS}{dt} = \pi \left[ l \cdot \frac{dr}{dt} + \frac{r}{l} \left( r\frac{dr}{dt} + h\frac{dh}{dt} \right) \right] \] Substitute the known values into the derivative: \[ \frac{dS}{dt} = \pi \left[ 25(3) + \frac{7}{25} \left( 7(3) + 24(-4) \right) \right] \] \[ \frac{dS}{dt} = \pi \left[ 75 + \frac{7}{25} \left( 21 - 96 \right) \right] \] \[ \frac{dS}{dt} = \pi \left[ 75 + \frac{7}{25} (-75) \right] \] \[ \frac{dS}{dt} = \pi [ 75 + 7(-3) ] \] \[ \frac{dS}{dt} = \pi [ 75 - 21 ] = 54\pi \text{ cm}^2/\text{min} \] Step 4: Final Answer:
The rate of change of the lateral surface area is $54\pi \text{ cm}^2/\text{min}$.