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 is 7 cm and altitude is 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:

To find the rate of change of surface area, use the chain rule to differentiate the equation with respect to time, and substitute the given rates of change.

Answer 1

The Correct Option is A

The lateral surface area of a cone is given by: \[ A = \pi r \ell \] where \( r \) is the radius and \( \ell \) is the slant height of the cone. Using the Pythagorean theorem: \[ \ell = \sqrt{r^2 + h^2} \] where \( h \) is the height. Now, differentiate with respect to time \( t \): \[ \frac{dA}{dt} = \pi \left( r \frac{d\ell}{dt} + \ell \frac{dr}{dt} \right) \] Substitute the given values of \( r = 7 \), \( h = 24 \), \( \frac{dr}{dt} = 3 \), and \( \frac{dh}{dt} = -4 \), then calculate \( \frac{dA}{dt} \). Thus, the rate of change of the lateral surface is \( 54 \pi \, \text{cm}^2/\text{min} \).