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Given Part: Clearly list all known variables, constants, or conditions provided in the problem.
To Find Part: Clearly state what needs to be determined or calculated.
Formulas Used: Write out all relevant formulas that will be used to solve the problem. Ensure formulas are correct and applicable.
Concept Used: Explain the main concept or principle that applies to the problem, including any theories or laws involved. The description should be clear and concise, providing a good understanding of the concept.
Complete Calculation: Show every step of the calculation without skipping any part. Each step should include the following:
Substitution of values into the formulas
Arithmetic or algebraic operations performed
Simplification or rearrangement of equations where necessary
Explanation: After every step, include a brief explanation of what was done and why it was done. This ensures that each step is clear and logically connected to the next.
Final Answer: Provide a clear and crisp final answer, emphasizing the key result or outcome of the calculation.
Additional Requirements:
Avoid the use of pronouns and ing forms in the steps.
Structure the answer in a way that is easy to understand and follow.
Ensure accuracy in every step, calculation, and explanation.

### Given Part:
- The weight of the plant \( W \) in pounds is estimated by the formula: 
  \[
  W = 26h^2
  \]
  where \( h \) is the height of the plant in feet.
- The current height of the plant is \( h = 4 \) feet.
- The height is increasing at a rate of \( \frac{dh}{dt} = 2 \) feet per year.

### To Find Part:
- The rate at which the weight \( W \) of the plant is increasing when the height \( h = 4 \) feet. Specifically, find \( \frac{dW}{dt} \) when \( h = 4 \) feet.

### Formulas Used:
1. The weight of the plant:
   \[
   W = 26h^2
   \]
2. The derivative of \( W \) with respect to time \( t \):
   \[
   \frac{dW}{dt} = \frac{dW}{dh} \cdot \frac{dh}{dt}
   \]

### Concept Used:
The chain rule in calculus is used to differentiate the weight \( W \) with respect to time \( t \). The chain rule allows the differentiation of a function of a function, which in this case applies since \( W \) is a function of \( h \), and \( h \) is a function of \( t \).

### Complete Calculation:
1. Differentiate \( W = 26h^2 \) with respect to \( h \):
   \[
   \frac{dW}{dh} = \frac{d}{dh}(26h^2) = 52h
   \]
   *Explanation*: The power rule was applied to differentiate \( h^2 \), and the constant multiplier \( 26 \) remains as a coefficient.

2. Substitute the known values into the formula for \( \frac{dW}{dt} \):
   \[
   \frac{dW}{dt} = 52h \cdot \frac{dh}{dt}
   \]
   *Explanation*: The chain rule was applied to express \( \frac{dW}{dt} \) in terms of \( \frac{dW}{dh} \) and \( \frac{dh}{dt} \).

3. Substitute \( h = 4 \) feet and \( \frac{dh}{dt} = 2 \) feet/year into the equation:
   \[
   \frac{dW}{dt} = 52(4) \cdot 2
   \]
   *Explanation*: The specific values for \( h \) and \( \frac{dh}{dt} \) were substituted into the differentiated formula.

4. Perform the arithmetic:
   \[
   \frac{dW}{dt} = 208 \cdot 2 = 416
   \]
   *Explanation*: The multiplication was carried out step by step to determine the final value.

### Final Answer:
The weight of the plant is increasing at a rate of **416 pounds per year** when the height of the plant is 4 feet.

---

### Additional Requirements:
1. Why was the chain rule necessary in this problem?
2. How would the result change if the rate of height increase were different?
3. What is the physical interpretation of \( \frac{dW}{dh} \)?
4. How does the constant 26 in the weight formula influence the final answer?
5. Could this method be applied to a different type of growth model? How?

### Tip:
Always check the units after every calculation step to ensure consistency and correctness in physical problems.

Learn how to calculate the rate of change in the weight of a plant using calculus. This problem involves applying the chain rule to differentiate the weight function with respect to time. Suitable for advanced high school students or calculus learners.

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