Problem 47 - Applied Calculus

Question

The time, \(\mathrm{T}\), in seconds for one complete oscillation of a pendulum is given by \(T=f(L)=1.111 \sqrt{L}\), where \(L\) is the length of the pendulum in feet. Find the following quantities, with units, and interpret in terms of the pendulum. (a) \(f(100)\) (b) \(f^{\prime}(100)\).

Step-by-Step Solution

Answer

(a) 11.11 seconds; (b) 0.05555 per foot. Increase the pendulum length increases swing time.

Step 1: Understand the Function

The function given is \( T = f(L) = 1.111 \sqrt{L} \), where \( L \) is the length of the pendulum in feet and \( T \) is the time in seconds for a complete oscillation. Our task is to find specific values of this function and its derivative.

Step 2: Calculate \( f(100) \)

Plug \( L = 100 \) into the function \( f(L) = 1.111 \sqrt{L} \) to find the time for one oscillation when \( L = 100 \) feet. \[ f(100) = 1.111 \times \sqrt{100} = 1.111 \times 10 = 11.11 \text{ seconds}\] Thus, \( f(100) = 11.11 \text{ seconds} \).

Step 3: Find the Derivative \( f'(L) \)

Differentiate \( f(L) = 1.111 \sqrt{L} = 1.111 L^{1/2} \) with respect to \( L \). Using the power rule, the derivative will be: \[ f'(L) = 1.111 \times \frac{1}{2} L^{-1/2} = \frac{1.111}{2\sqrt{L}} \]

Step 4: Calculate \( f'(100) \)

Plug \( L = 100 \) into the expression for the derivative \( f'(L) \). \[ f'(100) = \frac{1.111}{2\sqrt{100}} = \frac{1.111}{2 \times 10} = \frac{1.111}{20} = 0.05555 \text{ per foot} \] Hence, \( f'(100) = 0.05555 \text{ per foot} \).

Step 5: Interpret the Results

For part (a), \( f(100) = 11.11 \text{ seconds} \) means it takes approximately 11.11 seconds for the pendulum to complete one swing when its length is 100 feet. For part (b), \( f'(100) = 0.05555 \text{ per foot} \) indicates that if the length of the pendulum is increased by a small amount, the period of oscillation increases by approximately 0.05555 seconds per each additional foot.

Key Concepts

Function Differentiation

Differentiation is a handy tool in calculus that helps us understand how a function changes as its input changes. The original function we have is related to a pendulum's period, given by: \[ T = f(L) = 1.111 \sqrt{L} = 1.111 L^{1/2} \] Here, the input is the pendulum length \( L \), and the output is the period \( T \). Differentiating this function means finding \( f'(L) \), which tells us the rate of change of the pendulum period with respect to its length. To differentiate \( f(L) \), we use the power rule. The power rule states that if \( f(x) = ax^n \), then \( f'(x) = nax^{n-1} \). Applying this, the derivative is: \[ f'(L) = 1.111 \times \frac{1}{2} L^{-1/2} = \frac{1.111}{2\sqrt{L}} \] This derivative, \( f'(L) \), tells us how sensitive the oscillation period \( T \) is to small changes in \( L \). A higher derivative value means the period changes a lot with a small change in lengths.

Oscillation Period

The oscillation period of a pendulum refers to the time it takes to complete one full swing, from one side to the other and back. This is a key measurement for pendulums and is impacted by several factors including the length of the pendulum. In the provided function, \( f(L) = 1.111 \sqrt{L} \), \( L \) is the pendulum length and \( f(L) \) gives the oscillation period. When we plug in \( L = 100 \) feet into the function, we find: \[ f(100) = 1.111 \times \sqrt{100} = 11.11 \] Thus, the period for a 100-foot pendulum is 11.11 seconds. This means if you let a pendulum swing freely, it will take a little over 11 seconds to make one complete oscillation. The formula we use comes from the understanding that a longer pendulum swings slower, which is why period lengthens with longer \( L \).

Pendulum Length

Pendulum length \( L \) is a significant factor governing the oscillation period. The longer the pendulum, the longer it takes to complete a swing. This relationship is governed by the function \( f(L) = 1.111 \sqrt{L} \), where the length \( L \) is the only variable affecting the period \( T \). Notably, the square root relationship indicates that the period increases with the square root of the pendulum length. This means larger increases in length at larger sizes don't accelerate the period increase as much as smaller lengths do. Understanding this relationship is critical in contexts like timekeeping, where pendulum clocks were commonly used. If you want a clock to tick slower, you'd increase the length of its pendulum. Even small changes in length can affect the accuracy, which is highlighted by the derivative \( f'(L) \), showcasing how a minor adjustment in \( L \) leads to a proportional but significant change in \( T \). For example, \( f'(100) = 0.05555 \) per foot indicates that each extra foot added to a 100-foot pendulum results in a period increase of about 0.05555 seconds.