   Chapter 4.5, Problem 63E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Sound Intensity The relationship between the number of decibels β and the intensity of a sound I (in watts per square centimeter) is given by β = 10 log 10 ( I 10 − 16 ) .Find the rate of change in the number of decibels when the intensity is 10 − 4 watt per square centimeter.

To determine

To calculate: The rate of change in the number of decibels when the intensity is 104 watt per square centimeter and the relation between the number of decibels β and the intensity of sound I is provided by the equation β=10log10(I1016).

Explanation

Given information:

The intensity is 104 watt per square centimeter and the relation between the number of decibels β and the intensity of sound I is provided by the equation β=10log10(I1016).

Formula used:

As with the natural logarithm function if x and y is a positive number such that x1 and y1 then logxy=lnxlny.

As with the natural logarithm function if a and x is a positive number such that a1 and x1 then, as per change-of -base formula logax=lnxlna.

Let u be a differentiable function of x then, ddx[lnx]=1x,x>0 and ddx(lnu)=1ududx,u>0.

Calculation:

Consider the intensity is 104 watt per square centimeter and the relation between the number of decibels β and the intensity of sound I is provided by the equation β=10log10(I1016)

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