Problem 42 - Calculus - AP Edition

Question

A new design for a wind turbine adjusts the length of the turbine blade to keep the generated power constant even if the wind speed changes by a small amount. Assume that the power \(P\) (in watts) generated by the turbine can be expressed as \(P=0.87 \ell^{2} \nu^{3},\) where \(\ell\) is the length of the blade in meters and \(\nu\) is the speed of the wind in meters per second. Suppose the wind speed is increasing at a constant rate of \(0.01 \mathrm{m} / \mathrm{s}^{2}\) and that the length of the blade adjusts to keep the generated power constant. Determine how quickly \(\ell\) is changing at the moment when \(\ell=16\) and \(\nu=4\)

Step-by-Step Solution

Answer

\(\frac{d\ell}{dt} \approx -0.06\) m/s when the wind speed is increasing at 0.01 m/s², \(\ell=16\), and \(\nu=4\).

Step 1: Understand the Relationship

The power generated by the turbine is given by the equation \(P = 0.87 \ell^{2} u^{3}\), where \(\ell\) is the blade length and \(u\) is the wind speed. Here, we need to find the rate of change of \(\ell\) with respect to time (\(\frac{d\ell}{dt}\)) when the other parameters are specified.

Step 2: Differentiate with Respect to Time

Since the power \(P\) is constant over time, we set the derivative of \(P\) with respect to time to zero. Differentiating \(P = 0.87 \ell^{2} u^{3}\) with respect to \(t\), we get:\[0 = 0.87 \left( 2\ell \frac{d\ell}{dt} \cdot u^{3} + \ell^{2} \cdot 3u^{2} \frac{du}{dt} \right)\]Simplifying gives us the relationship between these derivatives.

Step 3: Substitute Known Values

Substitute \(\ell = 16\), \(u = 4\), and \(\frac{du}{dt} = 0.01\) m/s² into the equation obtained from Step 2:\[0 = 0.87 \left( 2 \times 16 \cdot \frac{d\ell}{dt} \cdot 4^{3} + 16^{2} \cdot 3 \times 4^{2} \times 0.01 \right)\]

Step 4: Solve for \\(\frac{d\ell}{dt}\\)

Simplify the expression obtained in Step 3:\[0 = 0.87 \left( 2 \times 16 \times \frac{d\ell}{dt} \times 64 + 256 \times 3 \times 16 \times 0.01 \right)\]\[0 = 0.87 \left( 2048 \cdot \frac{d\ell}{dt} + 122.88 \right)\]Set the inner expression equal to zero and solve for \(\frac{d\ell}{dt}\):\[2048 \cdot \frac{d\ell}{dt} = -122.88\]\[\frac{d\ell}{dt} = \frac{-122.88}{2048} \approx -0.06 \text{ m/s}\]

Step 5: Interpret the Result

The result means that the length of the turbine blade is decreasing at a rate of approximately \(-0.06\) meters per second when \(\ell = 16\) meters and \(u = 4\) m/s.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It is a mathematical process used to calculate the rate at which one quantity changes with respect to another. In simpler terms, it's the tool we use to understand how things change over time, or in response to other variables.

In the problem given, differentiation is used to determine how the length of the wind turbine blade, denoted as \(\ell\), changes as the wind speed \(u\) varies. This change is represented by \(\frac{d\ell}{dt}\). The differentiation process involves taking the derivative of the equation for power with respect to time. This is because the problem asks us to find the rate at which the length changes over time with a constant power output.

• Power Formula: \(P = 0.87 \ell^{2} u^{3}\)
• Set derivative of power \((P)\) with respect to time \((t)\) to zero when power is constant
• Differentiate: \[0 = 0.87\left(2\ell \frac{d\ell}{dt} \cdot u^{3} + \ell^{2} \cdot 3u^{2} \frac{du}{dt}\right)\]

Understanding this concept helps solve the problem by illustrating how the length adjusts to maintain constant power as wind speed changes. Differentiation is crucial for solving related rates problems like this.

Calculus Problem-Solving

Calculus problem-solving involves applying calculus concepts to find solutions to real-world problems. Related rates problems, such as this turbine blade problem, are exercises in understanding dynamic relationships between multiple variables.

In our exercise, we know the power equation. We need to find how quickly the blade's length changes. For these problems, we utilize the following steps:

• Identify what is given and what needs to be found, notably the change rate \(\frac{d\ell}{dt}\).
• Use differentiation to express these rates in relation to each other.
• Substitute known values into the differentiated equation.
• Solve for the unknown rate.

Execution of these steps turns a complex real-world scenario into manageable calculus equations, allowing for precise answers like the blade’s change rate of \(-0.06\) m/s. Mastering such steps is critical for efficiently tackling similar problems in academics and professional practice.

Physics and Calculus Integration

Physics and calculus integration highlights how calculus tools can model and solve physics-based problems. In this specific exercise, physics variables such as wind speed \(u\) and mechanical power \(P\) play significant roles.

Calculus helps bridge the conceptual gaps between physical principles and mathematical descriptions. The formula \(P = 0.87 \ell^{2} u^{3}\) is derived from physics, showing how power relates to blade dimensions and wind influence.

• The derivative of power with respect to time examines the interaction of natural elements like wind force, mathematically accounting for change.
• This synthesis informs practical engineering decisions, such as adjusting turbine configurations without compromising operational output.
• By solving for \(\frac{d\ell}{dt}\), we understand adjustments for maintaining efficiency amidst environmental variations.

This effective use of calculus within physical contexts underscores powerful analytical capabilities, harmonizing numerical insights with tangible engineering challenges.