Chapter 6.5, Problem 45E

a.

To determine

To determine the amount with which wind speed increase when the altitude doubles.

Answer to Problem 45E

The amount by which wind speed increased is $2\ln 2\approx 1.39\text{knots}$ .

Explanation of Solution

Given:

  $s\left(h\right)=2\ln 100h$

Calculation:

  $s\left(h\right)=2\ln 100h$

When altitude doubles, the wind speed is $s\left(2h\right)$ .

  $\begin{array}{c}s\left(2h\right)=2\ln 100\left(2h\right)\\ =2\ln 200h\end{array}$

The amount by which wind speed increased is given by:

  $\begin{array}{c}{s}_a\left(h\right)=s\left(2h\right)-s\left(h\right)\\ =2\ln 200h-2\ln 100h\\ =2\left(\ln 200h-\ln 100h\right)\\ =2\left(\ln 200+\ln h-\ln 100-\ln h\right)\\ =2\left(\ln \frac{200}{100}\right)\\ =2\ln 2\end{array}$

Hence, the amount by which wind speed increased is $2\ln 2\approx 1.39\text{knots}$ .

b.

To determine

To show the given function can be written as the expression shown.

Answer to Problem 45E

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

Explanation of Solution

Given:

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

Calculation:

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

  $\begin{array}{c}s\left(h\right)=2\left(\ln 100+\ln h\right)\\ =2\left(\ln {10}^2+\ln h\right)\\ =2\left(2\ln 10+\ln h\right)\\ =2\left(2{\log }_{e}10+{\log }_{e}h\right)\end{array}$

Use change of base formula.

  $\begin{array}{c}s\left(h\right)=2\left(2\left(\frac{{\log }_{10}10}{{\log }_{10}e}\right)+\frac{{\log }_{10}h}{{\log }_{10}e}\right)\\ =\frac{2}{\log e}\left(2+\log h\right)\end{array}$

Recall that ${\log }_{10}a=\log a$ .

Hence, the result.