Tutorial Exercise The length of a rectangle is increasing at a rate of $6 \mathrm{~cm} / \mathrm{s}$ and its width is increasing at a rate of $3 \mathrm{~cm} / \mathrm{s}$. When the length is $14 \mathrm{~cm}$ and the width is $10 \mathrm{~cm}$, how fast is the area of the rectangle increasing? Step 1 Using $A=I w$, we have \[ \frac{d A}{d t}=(\square) \cdot \frac{d w}{d t}+(\square) \frac{d t}{d t} \] Submit Skip (you cannot come back)

Step 1 ：The area of a rectangle is given by the product of its length and width. The rate of change of the area with respect to time can be found by differentiating the area with respect to time. This will give us the sum of two terms: the first term is the product of the rate of change of length and the current width, and the second term is the product of the current length and the rate of change of width.

Step 2 ：Given that the length of the rectangle is increasing at a rate of \(6 \, \text{cm/s}\), the width is increasing at a rate of \(3 \, \text{cm/s}\), the current length is \(14 \, \text{cm}\), and the current width is \(10 \, \text{cm}\).

Step 3 ：Substitute the given values into the equation to find the rate of change of the area: \(da/dt = l \cdot dw/dt + w \cdot dl/dt = 14 \cdot 3 + 10 \cdot 6 = 102 \, \text{cm}^2/\text{s}\).

Step 4 ：Final Answer: The rate at which the area of the rectangle is increasing is \(\boxed{102 \, \text{cm}^2/\text{s}}\).