Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise: 60

Answer

(a) $I(x)lna$ (b) $-3.05\times10^{-8}$ (c) $0.41$

Work Step by Step

Given: $I=I_{0}a^{x}$ Where, $I_{0}$ is the light intensity at the surface and $a$ is a constant such that $0 < a < 1$. The rate of change of with respect to x is given as follows: $I'(x)=I_{0}a^{x}lna=I(x)lna$ ...(2) (b) As we are given $I_{0}=8$ and $a=0.38$ , and our objective is to find the rate of change of intensity with respect to depth at a depth of 20 m. $I(20)=8(0.38)^{20}\approx 3.15\times10^{-8}$ From equation 2, we have The rate of intensity is given as: $I'(20)=I(20)ln(0.38)= -3.05\times10^{-8}$ (c) The average light intensity between the surface and a depth of 20 m can be calculated as follows: $I_{avg}(x)=\frac{I_{0}(a^{20}-1)}{20lna}$ As we are given $I_{0}=8$ and $a=0.38$ $I_{avg}(x)=\frac{8((0.38)^{20}-1)}{20ln(0.38)}=0.41$
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