A reservoir's water depth, in meters, is modeled by \(\mathrm{h(t) = 12.4 - 0.30(t - 5)}\) for 5 leq t...

1. TRANSLATE the problem information

• Given information:
  • Function: \(\mathrm{h(t) = 12.4 - 0.30(t - 5)}\) where \(\mathrm{t}\) is hours after midnight
  • Domain: \(\mathrm{5 \leq t \leq 20}\)
  • Find: Water depth at 5:00 p.m.

• Key insight: Need to convert 5:00 p.m. to hours after midnight format

2. TRANSLATE the time conversion

• 5:00 p.m. in 24-hour format = 17:00
• Since \(\mathrm{t}\) represents hours after midnight: \(\mathrm{t = 17}\)

3. SIMPLIFY by substituting and evaluating

• Substitute \(\mathrm{t = 17}\) into \(\mathrm{h(t) = 12.4 - 0.30(t - 5)}\):

\(\mathrm{h(17) = 12.4 - 0.30(17 - 5)}\)

\(\mathrm{h(17) = 12.4 - 0.30(12)}\)

\(\mathrm{h(17) = 12.4 - 3.6}\)

\(\mathrm{h(17) = 8.8}\)

Answer: (B) 8.8

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Incorrectly converting 5:00 p.m. to hours after midnight

Students might think "5:00 p.m." means \(\mathrm{t = 5}\), since they see the number 5 in the time. This leads them to calculate:

\(\mathrm{h(5) = 12.4 - 0.30(5 - 5)}\)

\(\mathrm{= 12.4 - 0.30(0)}\)

\(\mathrm{= 12.4 - 0}\)

\(\mathrm{= 12.4}\)

This may lead them to select Choice (D) (12.4).

The Bottom Line:

This problem tests whether students can correctly convert between standard time notation (a.m./p.m.) and mathematical time variables. The function evaluation itself is straightforward once the correct time value is identified, but the time conversion is the critical first step that determines success or failure.