Question

Solve b, I understand a and c so I included it for reference

Answer 1

Given:

$T=10\times2^{-0.274a}$

To Determine: The average rate of change of T with respect to a over the interval of 24000 and 30000

Solution

Let us determine the value of a at the interval given

$\begin{gathered} a=0\text{ correspond to 22000ft} \\ at\text{ 24000ft,} \\ a=\frac{24000-22000}{1000} \\ a=\frac{2000}{1000} \\ a=2 \\ at\text{ 30000ft} \\ a=\frac{30000-22000}{1000} \\ a=\frac{8000}{1000} \\ a=8 \end{gathered}$

Let us determine the value of T at the given interval using the values of a

$\begin{gathered} a=2 \\ T=10\times2^{-0.274(2)} \\ T=10\times2^{-0.548} \\ T=10\times0.683968 \\ T=6.83968 \end{gathered}$$\begin{gathered} a=8 \\ T_8=10\times2^{-0.274\times8} \\ T_8=10\times2^{-2.192} \\ T_8=10\times0.218848 \\ T_8=2.18848 \end{gathered}$

The rate of change would be

$\begin{gathered} r=\frac{T_8-T_2}{a_8-a_2} \\ r=\frac{2.18488-6.83968}{8-2} \\ r=-\frac{4.6512}{6} \\ r=-0.7752min-per-1000ft \\ r=-0.7752\times60sec-per-1000ft \\ r=-46.512s-per-1000ft \end{gathered}$

Hence, the rate of change of T with respect to a is -46.512 seconds per 1000ft