A sphere of 10cm radius has a uniform thickness of ice around it. Ice is melting at rate `50 cm^3//min` when thickness is 5cm then rate of change of thickness

Answer

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

### Step 1: Understand the problem
We have a sphere with a radius of 10 cm, and it has a uniform thickness of ice around it. The ice is melting at a rate of 50 cm³/min when the thickness of the ice is 5 cm. We need to find the rate of change of the thickness of the ice.

### Step 2: Define the variables
Let:
- \( r = 10 \) cm (radius of the sphere)
- \( x \) = thickness of the ice (in cm)
- \( V \) = volume of the sphere with the ice
- \( \frac{dV}{dt} = -50 \) cm³/min (the volume is decreasing, hence the negative sign)

### Step 3: Write the formula for the volume of the sphere with ice
The radius of the sphere with the ice is \( R = r + x = 10 + x \). The volume \( V \) of the sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (10 + x)^3 \]

### Step 4: Differentiate the volume with respect to time
To find the rate of change of volume with respect to time, we differentiate \( V \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi (10 + x)^3 \right) \]
Using the chain rule:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3(10 + x)^2 \cdot \frac{dx}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi (10 + x)^2 \frac{dx}{dt} \]

### Step 5: Substitute known values
We know that \( \frac{dV}{dt} = -50 \) cm³/min and \( x = 5 \) cm. Substituting these values into the equation:
\[ -50 = 4 \pi (10 + 5)^2 \frac{dx}{dt} \]
This simplifies to:
\[ -50 = 4 \pi (15)^2 \frac{dx}{dt} \]
Calculating \( (15)^2 = 225 \):
\[ -50 = 4 \pi \cdot 225 \frac{dx}{dt} \]
\[ -50 = 900 \pi \frac{dx}{dt} \]

### Step 6: Solve for \( \frac{dx}{dt} \)
Now, we can isolate \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{-50}{900 \pi} = \frac{-1}{18 \pi} \]
Thus, the rate of change of thickness of the ice is:
\[ \frac{dx}{dt} = -\frac{1}{18 \pi} \text{ cm/min} \]

### Conclusion
The rate of change of thickness of the ice when the thickness is 5 cm is \(-\frac{1}{18 \pi}\) cm/min.