To find how fast the area of the rectangle is changing at the moment the length is $10 \, \text{cm}$, we use the following approach:

Step 1: Define the variables

• Let $l$ represent the length of the rectangle (in cm), and it increases with time at the rate $\frac{dl}{dt} = 3 \, \text{cm/s}$.
• Let $w$ represent the width of the rectangle, which is constant at $w = 5 \, \text{cm}$.
• Let $A$ represent the area of the rectangle. The formula for the area is:

$A = l \cdot w$

Step 2: Differentiate the area equation with respect to time ($t$)

Since $w$ is constant, the derivative of $A$ with respect to $t$ is:

$\frac{dA}{dt} = w \cdot \frac{dl}{dt}$

Step 3: Substitute the known values

At the instant when $l = 10 \, \text{cm}$, $w = 5 \, \text{cm}$, and $\frac{dl}{dt} = 3 \, \text{cm/s}$, we calculate:

$\frac{dA}{dt} = 5 \cdot 3 = 15 \, \text{cm}^2/\text{s}$

Final Answer:

The area is increasing at a rate of $\mathbf{15 \, \text{cm}^2/\text{s}}$ when the length is $10 \, \text{cm}$.

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Would you like a detailed breakdown of this process or have any questions?
Here are five related questions for deeper understanding:

1. How does the rate of change of width affect the area if it were not constant?
2. What happens to the rate of area change if the length rate increases over time?
3. How is the derivative applied when both length and width change?
4. Can this concept be extended to shapes other than rectangles, such as triangles or circles?
5. What are some real-life applications of related rates in geometry?

Tip: Always identify constants and changing variables in a problem to simplify