Quiz 1.3:

Question:

The following table gives te temperature-depth profile for, \(T=f(d)\), in a borehole in Belleterre, Quebec, where \(T\) is the average temperature at a depth, \(d\).

\(d\), depth (m) 25 50 75 100 125 150 175 200 225 250 275 300
\( T \), temp (\(^\circ \)C ) 5.50 5.20 5.10 5.10 5.30 5.50 5.75 6.00 6.25 6.50 6.75 7.00

Could \(f\) be linear? Explain.

Give your observations about \(f\) for \( d \geq 150 \)

Solution:

The function \(f\) could not be linear because the rate of change is not constant.
\( \frac{5.20-5.50}{50-25}=\frac{-.30}{25}=-.012 ^\circ \mbox{C}/ \mbox{m} \)
\( \frac{5.10-5.20}{75-50}=\frac{-.10}{25}=-.004 ^\circ \mbox{C}/ \mbox{m} \)
\( \frac{5.10-5.10}{100-75}=\frac{0}{25}= 0 ^\circ \mbox{C}/ \mbox{m} \)

etc.


Once we reach \(d=150\), the rate of change does become constant though,
\( \frac{5.75-5.50}{175-150}=\frac{6.00-5.75}{200-175}=\frac{6.25-6.00}{225-200}= \cdots = \frac{.25}{25}=.01^\circ \mbox{C}/ \mbox{m} \)
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