The relationship between the number of decibels $\beta$ and the intensity of a sound I in watts per square centimeter is given by $\beta = 10 \log _ { 10 } \left( \frac { I } { 10 ^ { - 16 } } \right)$ . Find the rate of change in the number of decibels when the intensity is $10 ^ { - 5 }$ watt per square centimeter. Round your answer to the nearest decibel.
A)43,429 decibels per watt per square cm
B)4,342,945 decibels per watt per square cm
C)434,294 decibels per watt per square cm
D)4,342,944,819 decibels per watt per square cm
E)434,296 decibels per watt per square cm

Question

Quizplus · Accepted Answer

To find the rate of change of the number of decibels with respect to the intensity, we differentiate the given equation with respect to $I$. The derivative of $\beta = 10 \log_{10}\left(\frac{I}{10^{-16}}\right)$ with respect to $I$ is $\frac{d\beta}{dI} = \frac{10}{I \ln(10)}$. Substituting $I = 10^{-5}$ gives $\frac{d\beta}{dI} = \frac{10}{10^{-5} \ln(10)} = \frac{10^6}{\ln(10)}$. Since $\ln(10) \approx 2.302585$, the rate of change is approximately $\frac{10^6}{2.302585} \approx 434,294$ decibels per watt per square cm.

The Relationship Between the Number of Decibels β\betaβ And the Intensity of a Sound I in Watts Per

The Relationship Between the Number of Decibels $\beta$ And the Intensity of a Sound I in Watts Per