Chapter 4.5, Problem 74PS

Solution

a.

To Calculate: which amount does the wind speed increase when the altitude doubles.

  $s=1.386$ (knots).

Given: the model of this function is given as follows:

  $s\left(h\right)=2\ln \left(100h\right)$   ...... (1)

A concept used: By the quotient property of the logarithm ${\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)$ ; Where, $a$ is the base of the logarithm

Calculation:

Now, simplify the above equation (1) as;

  $s\left(h\right)=2\ln \left(100h\right)$   ...... (2)

Now replacing on both sides $h$ by $2h$ in the above equation

  $s\left(2h\right)=2\ln \left(100\cdot 2h\right)$

  $s\left(2h\right)=2\ln \left(200h\right)$   ...... (3)

From equation (2) and equation (3), then;

Change in wind speed $s=s\left(2h\right)-s\left(h\right)$

  $\begin{array}{c}s=2\ln \left(200h\right)-2\ln \left(100h\right)\\ s=2\left[\ln \left(200h\right)-\ln \left(100h\right)\right]\\ s=2\left[\ln \left(\frac{200h}{100h}\right)\right]\left[\ln a-\ln b=\ln \left(\frac{a}{b}\right)\right]\\ s=2\ln 2\end{array}$

Thus,

  $\begin{array}{c}s=2\cdot 0.693\\ s=1.386\end{array}$

Their increase in wind speed is $s=1.386$ (knots) doubling the height.

Conclusion:

Hence, the increase in wind speed $s=1.386$ (knots) doubles the height.

b.

To prove: the given function is $s(h)=\frac{2}{\log e}\left(\log h+2\right)$ .

  $s(h)=\frac{2}{\log e}\left(\log h+2\right)$.

Given: the model of this function is given as :

  $s\left(h\right)=2\ln \left(100h\right)$ .

Concept used: Use the change of base property of the logarithm, if $a,b$ and $c$ are positive numbers with $b\ne 1$ and $c\ne 1$ then, ${\log }_{c}a=\frac{{\log }_{b}a}{{\log }_{b}c}$ .

Calculation: From the given function of the model $s\left(h\right)=2\ln \left(100h\right)$ .

  $s\left(h\right)=2\left[\frac{\log \left(100\right)h}{\log e}\right]\left[\ln x=\frac{\log x}{\log e}\right]$

Thus,

  $\begin{array}{c}s\left(h\right)=2\left[\frac{\log 100+\log h}{\log e}\right]\\ s\left(h\right)=\frac{2}{\log e}\left(\log h+\log 100\right)\left[\log mn=\log m+\log n\right]\\ s\left(h\right)=\frac{2}{\log e}\left(\log h+\log {10}^2\right)\\ s\left(h\right)=\frac{2}{\log e}\left(\log h+2\right)\left[{\log }_a{a}^b=1\right]\end{array}$

Conclusion: Hence, it is proved $s\left(h\right)=\frac{2}{\log e}\left(\log h+2\right)$ .