Let's solve the problem step by step.

Problem Recap:

• A triangular metal lamina expands such that the base length is always three times its height.
• The rate of increase of its area is $\frac{dA}{dt} = 0.27 \, \text{cm}^2/\text{sec}$.
• We are to find the rate of change of its height, $\frac{dh}{dt}$, when the height $h = 9 \, \text{cm}$.

Given:

• Base length $b = 3h$ (since base is three times the height).
• Area of the triangle $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3h \times h = \frac{3}{2} h^2$.
• Rate of increase of area $\frac{dA}{dt} = 0.27 \, \text{cm}^2/\text{sec}$.

Step-by-Step Solution:

1. Express the area in terms of height: $A = \frac{3}{2} h^2$

2. Differentiate the area $A$ with respect to time $t$: $\frac{dA}{dt} = \frac{d}{dt} \left(\frac{3}{2} h^2 \right)$ Using the chain rule: $\frac{dA}{dt} = \frac{3}{2} \times 2h \times \frac{dh}{dt} = 3h \frac{dh}{dt}$

3. Substitute the given values: We know $\frac{dA}{dt} = 0.27 \, \text{cm}^2/\text{sec}$ and $h = 9 \, \text{cm}$. Substitute these values into the differentiated equation: $0.27 = 3 \times 9 \times \frac{dh}{dt}$

4. Solve for $\frac{dh}{dt}$: $0.27 = 27 \times \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{0.27}{27} = 0.01 \, \text{cm/sec}$

Final Answer:

The rate of change of the height when the height is 9 cm is $\frac{dh}{dt} = 0.01 \, \text{cm/sec}$.

Would you like more details or have any questions? Here are 5 related questions you might find interesting:

1. How would the solution change if the base was twice the height instead of three times?
2. What if the rate of area change were different, say $0.5 \, \text{cm}^2/\text{sec}$?
3. How does this problem relate to real-world applications like material expansion?
4. What is the significance of the chain rule in this problem?
5. How could you extend this problem to find the rate of change of the base length?

Tip: Understanding how to apply the chain rule is crucial in solving problems involving rates of change, especially when variables are interdependent.