A meteorologist measures the atmospheric pressure at altitude
.
(a) Plot the points and use the regression capabilities of the graphing utility to find a linear model for the revised data points.
Construct a table with and
values to calculate
values.
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Graph:
\Plot the ponts .
Using regression capabilities graph thepoints.
\Observe the graph:
\The linear model equation is .
\
h \ | \
\
p \ | \
\
ln(p) \ | \
\
h*ln(p) \ | \
\
| \
\
0 \ | \
\
10332 \ | \
\
9.24 \ | \
\
0 \ | \
\
0 \ | \
\
5 \ | \
\
5583 \ | \
\
8.62 \ | \
\
43.1 \ | \
\
25 \ | \
\
10 \ | \
\
2376 \ | \
\
7.77 \ | \
\
77.7 \ | \
\
100 \ | \
\
15 \ | \
\
1240 \ | \
\
7.12 \ | \
\
106.8 \ | \
\
225 \ | \
\
20 \ | \
\
517 \ | \
\
6.25 \ | \
\
125 \ | \
\
400 \ | \
\
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The linear Regression is in the form of ln P = ah + b
\where P is the Pressure and h is the altitude.
\To calculate the values of a and b
\Therefore the linear regression is ln P = -0.1496h + 9.296
\Graph
\(b)
\The linear model equation is .
Apply Exponent on both side
\The exponential model equation is . \ \
(c)
\Graph the Original Data and the Exponential Regression
\(d)
\Pressure
Apply derivative on each side with respect to .
\ \
If h = 5 then Rate of change of Pressure
\If height h = 5 km then Rate of change of Pressure is -771.40.
\If h = 18 then Pressure
\If height h = 18 then Rate of change of Pressure is -110.39.
\
Therefore,
\(a) Linear Regression is ln(P) = -0.1496h + 9.296
\(b) Exponential Regression is
(d) Rate of change of Pressure is -771.40 if h = 5 km.
\Rate of change of Pressure is -110.39 if h = 18.