Question

The area spherical balloon of radius 6 cm increases at the rate of 2 then find the rate of increase in the volume. 

Answer 1

We are asked to determine the rate of change of the volume of a sphere, given that the surface area is increasing at a specific rate.

Step 1: The formulas for the surface area and volume of a sphere:
The surface area \( A \) of a sphere with radius \( r \) is given by: \[ A = 4\pi r^2 \] The volume \( V \) of a sphere with radius \( r \) is given by: \[ V = \frac{4}{3} \pi r^3 \]

Step 2: Given information:
We are told that the surface area is increasing at a rate of 2 cm²/sec: \[ \frac{dA}{dt} = 2 \, \text{cm}^2/\text{sec} \] We are also asked to find the rate of change of the volume when the radius is 6 cm.

Step 3: Relating the rate of change of surface area to the rate of change of radius:
Differentiate the surface area formula \( A = 4\pi r^2 \) with respect to time \( t \): \[ \frac{dA}{dt} = 8\pi r \frac{dr}{dt} \] This relates the rate of change of the surface area to the rate of change of the radius (\( \frac{dr}{dt} \)).

Step 4: Relating the rate of change of volume to the rate of change of radius:
Similarly, differentiate the volume formula \( V = \frac{4}{3} \pi r^3 \) with respect to time \( t \): \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] This relates the rate of change of the volume to the rate of change of the radius (\( \frac{dr}{dt} \)).

Step 5: Substituting the known values:
We are given that \( \frac{dA}{dt} = 2 \) cm²/sec, and we need to find \( \frac{dV}{dt} \) when the radius is 6 cm.

Using the surface area equation: \[ 2 = 8\pi (6) \frac{dr}{dt} \] Simplifying: \[ 2 = 48\pi \frac{dr}{dt} \] Solving for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{2}{48\pi} = \frac{1}{24\pi} \, \text{cm/sec} \]

Step 6: Calculating the rate of change of volume:
Now, substitute \( \frac{dr}{dt} = \frac{1}{24\pi} \) into the volume rate of change formula: \[ \frac{dV}{dt} = 4\pi (6)^2 \left( \frac{1}{24\pi} \right) \] Simplifying: \[ \frac{dV}{dt} = 4\pi (36) \left( \frac{1}{24\pi} \right) \] \[ \frac{dV}{dt} = 144\pi \times \frac{1}{24\pi} \] \[ \frac{dV}{dt} = 6 \, \text{cm}^3/\text{sec} \]

Final Answer:
Therefore, the rate of change of the volume when the radius is 6 cm and is increasing at a rate of \( 2 \) cm/sec is \( \frac{dV}{dt} = 6 \, \text{cm}^3/\text{sec} \).