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 \( l \). Given \( l = b^2 \). The area is \( A = l \times b = b^2 \times b = b^3 \). We are given \( \frac{dA}{dt} = 4 \, \text{cm}^2/\text{s} \) and need to find \( \frac{dl}{dt} \) when \( b = 4.5 \, \text{cm} \). Step 1: Differentiate the area equation with respect to time. \[ \frac{dA}{dt} = 3b^2 \frac{db}{dt} \] Substituting \( \frac{dA}{dt} = 4 \): \[ 4 = 3b^2 \frac{db}{dt} \] 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 \( \frac{db}{dt} \approx 0.79 \): \[ \frac{dl}{dt} = 2(4.5)(0.79) \approx 7.11 \, \text{cm/s} \] The rate at which the length of the rectangle is increasing when the breadth is 4.5 cm is approximately \( \boxed{7.11 \, \text{cm/s}} \).