Question

The area of an expanding rectangle is increasing at the rate of \( 4 \, \text{cm}^2/\text{s} \). The length of the rectangle is always square of its breadth. At what rate is the length of the rectangle increasing at an instant, when breadth \( = 4.5 \, \text{cm} \)?

Hint:

To find rates of change involving related quantities, first express the relationship between the quantities, then use differentiation with respect to time.

Answer 1

Let the breadth of the rectangle be \( b \) and the length be \( l \). We are given that the length is always the square of the breadth: \[ l = b^2. \] The area \( A \) of the rectangle is: \[ A = l \times b = b^2 \times b = b^3. \] We are given that the rate of change of the area is \( \frac{dA}{dt} = 4 \, \text{cm}^2/\text{s} \). We need to find the rate at which the length is changing, i.e., \( \frac{dl}{dt} \), at the instant when \( b = 4.5 \, \text{cm} \). Step 1: Differentiate the area equation with respect to time \[ \frac{dA}{dt} = 3b^2 \frac{db}{dt}. \] We know \( \frac{dA}{dt} = 4 \), so: \[ 4 = 3b^2 \frac{db}{dt}. \] Step 2: Solve for \( \frac{db}{dt} \) Substitute \( b = 4.5 \): \[ 4 = 3(4.5)^2 \frac{db}{dt}. \] \[ 4 = 3(20.25) \frac{db}{dt}. \] \[ 4 = 60.75 \frac{db}{dt}. \] \[ \frac{db}{dt} = \frac{4}{60.75} \approx 0.79 \, \text{cm/s}. \] Step 3: Find \( \frac{dl}{dt} \) Since \( l = b^2 \), differentiate \( l \) with respect to time: \[ \frac{dl}{dt} = 2b \frac{db}{dt}. \] Substitute \( b = 4.5 \) and \( \frac{db}{dt} \approx 0.79 \): \[ \frac{dl}{dt} = 2(4.5)(0.79) \approx 7.11 \, \text{cm/s}. \] Thus, the rate at which the length of the rectangle is increasing when the breadth is 4.5 cm is approximately \( \boxed{7.11 \, \text{cm/s}} \).