Problem 50 - Calculus and its applications

Question

The intensity of a sound is given by \(I=I_{0} 10^{0.1 L},\) where \(L\) is the loudness of the sound measured in decibels and \(I_{0}\) is the minimum intensity detectable by the human ear. a) Find \(I\), in terms of \(I_{0}\), for the loudness of a power mower, which is 100 decibels. b) Find \(I\), in terms of \(I_{0}\), for the loudness of a just audible sound, which is 10 decibels. c) Compare your answers to parts (a) and (b). d) Find the rate of change \(d I / d L\). e) Interpret the meaning of \(\mathrm{dI} / \mathrm{dL}\).

Step-by-Step Solution

Answer

a) \( 10^{10} I_{0} \), b) \( 10 I_{0} \), c) The power mower is \( 10^9 \) times more intense, d) \( 0.1 I_{0} \ln(10) \cdot 10^{0.1 L} \), e) It shows exponential intensity growth with loudness.

Step 1: Substitute the loudness for a power mower into the formula

The formula is given by \( I = I_{0} 10^{0.1 L} \). For a power mower, \( L = 100 \) decibels. Substitute this into the formula:\[ I = I_{0} 10^{0.1 \times 100} = I_{0} 10^{10} \].

Step 2: Calculate the intensity for a power mower

After substitution, we have \( I = I_{0} \times 10^{10} \). Therefore, in terms of \( I_{0} \), the intensity of a power mower is \( 10^{10} I_{0} \).

Step 3: Substitute the loudness for a just audible sound into the formula

Use \( L = 10 \) for a just audible sound. Substitute into the formula:\[ I = I_{0} 10^{0.1 \times 10} = I_{0} 10^{1} \].

Step 4: Calculate the intensity for a just audible sound

After substitution, we have \( I = I_{0} \times 10^{1} \). Thus, the intensity of a just audible sound is \( 10 I_{0} \).

Step 5: Compare the results from parts (a) and (b)

For a power mower, \( I = 10^{10} I_{0} \) and for a just audible sound, \( I = 10 I_{0} \). The intensity of a power mower is \( 10^9 \) times greater than that of a just audible sound.

Step 6: Find the rate of change of intensity with respect to loudness, \( dI/dL \)

To find \( dI/dL \), differentiate the intensity function \( I = I_{0} 10^{0.1 L} \) with respect to \( L \):\[ \frac{dI}{dL} = I_{0} \cdot (\ln(10) \cdot 0.1) \cdot 10^{0.1 L} \]. Therefore, \( \frac{dI}{dL} = 0.1 I_{0} \ln(10) \cdot 10^{0.1 L} \).

Step 7: Interpret the meaning of \( \frac{dI}{dL} \)

The rate \( \frac{dI}{dL} \) describes how the intensity \( I \) changes as the loudness \( L \) changes. It tells us for each unit increase in decibels, the intensity increases by a factor of \( 0.1 I_{0} \ln(10) \times 10^{0.1L} \), indicating exponential growth in intensity as loudness increases.

Key Concepts

Decibels

Decibels are a unit used to express the intensity or loudness of sound. They are based on a logarithmic scale, specifically a base-10 logarithm. This means that an increase of 10 decibels represents a tenfold increase in intensity. For instance, a sound at 100 decibels is not just twice as loud as a sound at 50 decibels, but many times more intense. The formula given in the exercise expresses this relationship by establishing that the intensity of sound, \( I \), is proportional to \( 10^{0.1L} \), where \( L \) is the loudness in decibels.

• The base level for sound intensity is \( I_0 \), the smallest detectable sound by the human ear.
• Because decibels follow a logarithmic scale, even small numeral changes can have significant impacts on perceived loudness.
• The exponential function in the formula helps convert the linear decibel measurement into a practical intensity measure for varied sounds.

Understanding decibels is crucial in fields like acoustics and sound engineering because it helps in quantifying sound levels accurately in varied environments.

Differentiation

Differentiation is a mathematical tool used to compute how a function changes as its input changes. In this exercise, the differentiation of the function \( I = I_0 10^{0.1L} \) with respect to \( L \) tells us how the intensity of sound changes as the loudness level changes. In simpler terms, it answers the question: "If the loudness increases by one decibel, how much does the intensity increase?"

• This is achieved by taking the derivative of the function with respect to \( L \).
• For \( I = I_0 10^{0.1L} \), the differentiation process involves the natural logarithm \( \ln(10) \) due to the exponential function.
• The result, \( \frac{dI}{dL} = 0.1 I_{0} \ln(10) \cdot 10^{0.1L} \), captures the rate of change of intensity per unit loudness.

By interpreting this derivative, we understand that small changes in loudness can lead to exponential changes in intensity, emphasizing the sensitivity of human hearing to variations in sound levels.

Exponential Functions

Exponential functions are used when quantities grow or shrink rapidly at a constant relative rate. In the context of the sound intensity formula, , the function \( I=I_{0} 10^{0.1L} \) exemplifies how sound intensity grows as loudness increases. The base of the exponent is 10, and the exponent itself is a fraction of the decibel scale (0.1 times the loudness \( L \)).

• The exponential growth here means that even a small increase in \( L \) results in a multiplicative increase in \( I \).
• The exponent 0.1 acts as a scaling factor that links the decibel scale with exponential growth in intensity.
• Exponential functions are crucial in accurately modeling how natural phenomena, like sound, can escalate quickly with small changes.

By comprehending exponential functions, students can appreciate how powerful these mathematical structures are in describing real-world scenarios, such as the vast range of human sound perception.