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Difficulty: Very hardInterpreting Linear Relationships in Context

An oceanographic research vessel measures hydrostatic pressure during deep-sea dives. During a dive in temperate waters, the pressure PP, in atmospheres (atm), at a depth of dd meters below the surface is modeled by the linear equation P=ρd+1.0P = \rho d + 1.0, where ρ\rho is a constant representing the rate of change of pressure with respect to depth. At a depth of 150150 meters, the pressure is 16.016.0 atm. During a second dive in a high-salinity region, the rate of change of pressure with respect to depth is 25%25\% greater than the rate in the temperate waters. According to the model for the second dive, at what depth, in meters, is the pressure 26.026.0 atm?

Answer: 200 meters

Answer

200
To find the depth on the second dive where the pressure is 26.026.0 atm, we first find the slope of the linear model for the first dive. Given P=ρd+1.0P = \rho d + 1.0, substituting the point (150,16.0)(150, 16.0) gives 16.0=150ρ+1.016.0 = 150\rho + 1.0, which yields ρ=0.1\rho = 0.1 atm/m. The slope for the second dive is 25%25\% greater, so the new slope is 0.1×1.25=0.1250.1 \times 1.25 = 0.125 atm/m. The equation for the second dive is P=0.125d+1.0P = 0.125d + 1.0. Substituting P=26.0P = 26.0 yields 26.0=0.125d+1.026.0 = 0.125d + 1.0. Solving for dd gives 25.0=0.125d25.0 = 0.125d, which results in a depth of 200200 meters.

Step-by-Step Solution

1
Find the rate of change of pressure with respect to depth in temperate waters.
ρ=0.1\rho = 0.1 atm/m
Substitute the depth d=150d = 150 and pressure P=16.0P = 16.0 into the linear equation P=ρd+1.0P = \rho d + 1.0 to get 16.0=150ρ+1.016.0 = 150\rho + 1.0. Solving for ρ\rho yields 15.0=150ρ15.0 = 150\rho, which means ρ=0.1\rho = 0.1.
2
Calculate the rate of change of pressure with respect to depth for the second dive.
ρnew=0.125\rho_{\text{new}} = 0.125 atm/m
The rate of change for the second dive is 25%25\% greater than the rate of 0.10.1 atm/m from the first dive. Thus, ρnew=0.1×1.25=0.125\rho_{\text{new}} = 0.1 \times 1.25 = 0.125.
3
Calculate the depth where the pressure is 26.026.0 atm using the new rate of change.
d=200d = 200 meters
Set up the equation for the second dive: P=0.125d+1.0P = 0.125d + 1.0. Substitute P=26.0P = 26.0 to get 26.0=0.125d+1.026.0 = 0.125d + 1.0. Subtracting 1.01.0 from both sides gives 25.0=0.125d25.0 = 0.125d. Solving for dd gives d=25.00.125=200d = \frac{25.0}{0.125} = 200.

Key Concept

Interpreting and manipulating slope in a linear relationship context
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Topics

Interpreting Linear Relationships in Context

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