The area of an expanding rectangle at the rate of `48cm^(2)//s.` the length of rectangle is always equal to square of the breadth. At which rate the length is increasing at the instant when th ebreadth is 4 cm?

Answer

To solve the problem step by step, we need to find the rate at which the length of the rectangle is increasing when the breadth is 4 cm. Let's break this down:

### Step 1: Define the variables
Let:
- \( B \) = breadth of the rectangle (in cm)
- \( L \) = length of the rectangle (in cm)
- \( A \) = area of the rectangle (in cm²)

### Step 2: Establish the relationship between length and breadth
According to the problem, the length \( L \) is always equal to the square of the breadth \( B \):
\[ L = B^2 \]

### Step 3: Write the formula for the area of the rectangle
The area \( A \) of the rectangle can be expressed as:
\[ A = L \times B = B^2 \times B = B^3 \]

### Step 4: Differentiate the area with respect to time
We know that the area is changing with time. Therefore, we differentiate the area \( A \) with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(B^3) = 3B^2 \frac{dB}{dt} \]

### Step 5: Use the given rate of change of area
We are given that the area is expanding at a rate of \( \frac{dA}{dt} = 48 \, \text{cm}^2/\text{s} \). Thus, we can set up the equation:
\[ 3B^2 \frac{dB}{dt} = 48 \]

### Step 6: Solve for \( \frac{dB}{dt} \) when \( B = 4 \, \text{cm} \)
Substituting \( B = 4 \, \text{cm} \) into the equation:
\[ 3(4^2) \frac{dB}{dt} = 48 \]
\[ 3(16) \frac{dB}{dt} = 48 \]
\[ 48 \frac{dB}{dt} = 48 \]
\[ \frac{dB}{dt} = 1 \, \text{cm/s} \]

### Step 7: Find the rate of change of length
Now, we need to find the rate at which the length \( L \) is increasing, which is given by \( \frac{dL}{dt} \). We differentiate \( L = B^2 \):
\[ \frac{dL}{dt} = 2B \frac{dB}{dt} \]

### Step 8: Substitute \( B = 4 \, \text{cm} \) and \( \frac{dB}{dt} = 1 \, \text{cm/s} \)
Substituting the values:
\[ \frac{dL}{dt} = 2(4)(1) = 8 \, \text{cm/s} \]

### Final Answer
The rate at which the length of the rectangle is increasing when the breadth is 4 cm is:
\[ \frac{dL}{dt} = 8 \, \text{cm/s} \]