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. Here’s how we can approach the solution:

### Step 1: Define Variables
Let:
- \( y \) = breadth of the rectangle
- \( x \) = length of the rectangle
- The area \( A \) of the rectangle is given by \( A = x \cdot y \).

### Step 2: Relationship Between Length and Breadth
According to the problem, the length of the rectangle is always equal to the square of the breadth:
\[ x = y^2 \]

### Step 3: Differentiate the Area with Respect to Time
The area \( A \) can be expressed in terms of \( y \):
\[ A = x \cdot y = y^2 \cdot y = y^3 \]
Now, differentiate both sides with respect to time \( t \):
\[ \frac{dA}{dt} = 3y^2 \frac{dy}{dt} \]

### Step 4: Given Rate of Change of Area
We are given that the area is expanding at a rate of:
\[ \frac{dA}{dt} = 48 \, \text{cm}^2/\text{s} \]

### Step 5: Substitute Known Values
We need to find \( \frac{dx}{dt} \) when \( y = 4 \, \text{cm} \). First, calculate \( x \) when \( y = 4 \):
\[ x = y^2 = 4^2 = 16 \, \text{cm} \]

### Step 6: Find \( \frac{dy}{dt} \)
Substituting \( y = 4 \) into the differentiated area equation:
\[ 48 = 3(4^2) \frac{dy}{dt} \]
\[ 48 = 3(16) \frac{dy}{dt} \]
\[ 48 = 48 \frac{dy}{dt} \]
\[ \frac{dy}{dt} = 1 \, \text{cm/s} \]

### Step 7: Differentiate the Length with Respect to Time
Now differentiate the equation \( x = y^2 \) with respect to time:
\[ \frac{dx}{dt} = 2y \frac{dy}{dt} \]

### Step 8: Substitute Known Values to Find \( \frac{dx}{dt} \)
Substituting \( y = 4 \) and \( \frac{dy}{dt} = 1 \):
\[ \frac{dx}{dt} = 2(4)(1) = 8 \, \text{cm/s} \]

### Conclusion
The rate at which the length is increasing when the breadth is 4 cm is:
\[ \frac{dx}{dt} = 8 \, \text{cm/s} \]

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