The edge of a cube is equal to the radius o the sphere. If the rate at which the volume of the cube is increasing is equal to `lambda`, then the rate of increase of volume of the sphere, is

Options

A. `(4pi lambda)/(3)`
B. `4pi lambda`
C. `(lambda)/(3)`
D. none of these

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the relationship between the cube and the sphere
Given that the edge of the cube (denote it as \( a \)) is equal to the radius of the sphere (denote it as \( r \)), we have:
\[ r = a \]

### Step 2: Write the formula for the volume of the cube
The volume of the cube \( V_c \) is given by:
\[ V_c = a^3 \]

### Step 3: Differentiate the volume of the cube with respect to time
To find the rate of change of the volume of the cube, we differentiate \( V_c \) with respect to time \( t \):
\[ \frac{dV_c}{dt} = \frac{d}{dt}(a^3) = 3a^2 \frac{da}{dt} \]
We know from the problem that the rate at which the volume of the cube is increasing is equal to \( \lambda \):
\[ \frac{dV_c}{dt} = \lambda \]
Thus, we can equate:
\[ 3a^2 \frac{da}{dt} = \lambda \quad \text{(Equation 1)} \]

### Step 4: Write the formula for the volume of the sphere
The volume of the sphere \( V_s \) is given by:
\[ V_s = \frac{4}{3} \pi r^3 \]
Since \( r = a \), we can rewrite the volume of the sphere in terms of \( a \):
\[ V_s = \frac{4}{3} \pi a^3 \]

### Step 5: Differentiate the volume of the sphere with respect to time
Now, we differentiate \( V_s \) with respect to time \( t \):
\[ \frac{dV_s}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi a^3\right) = \frac{4}{3} \pi \cdot 3a^2 \frac{da}{dt} = 4\pi a^2 \frac{da}{dt} \]
This gives us the rate of increase of the volume of the sphere:
\[ \frac{dV_s}{dt} = 4\pi a^2 \frac{da}{dt} \quad \text{(Equation 2)} \]

### Step 6: Substitute from Equation 1 into Equation 2
From Equation 1, we have \( 3a^2 \frac{da}{dt} = \lambda \). We can solve for \( \frac{da}{dt} \):
\[ \frac{da}{dt} = \frac{\lambda}{3a^2} \]
Now, substitute \( \frac{da}{dt} \) back into Equation 2:
\[ \frac{dV_s}{dt} = 4\pi a^2 \left(\frac{\lambda}{3a^2}\right) \]
This simplifies to:
\[ \frac{dV_s}{dt} = \frac{4\pi \lambda}{3} \]

### Final Result
Thus, the rate of increase of the volume of the sphere is:
\[ \frac{dV_s}{dt} = \frac{4\pi \lambda}{3} \]