Question

Pressure under Water The pressure \(p\) experienced by a diver under water is related to the diver's depth \(d\) by an equation of the form \(p=k d+1(k\) a constant). When \(d=0\) meters, the pressure is 1 atmosphere. The pressure at 100 meters is 10.94 atmospheres. Find the pressure at 50 meters.

Short answer

The pressure at a depth of 50 meters under water is 5.97 atmospheres.

Step-by-step solution

Determine the constant \(k\)

Using the given condition that the pressure at 100 meters is 10.94 atmospheres, we can substitute \(d = 100\) and \(p = 10.94\) into the equation \(p=k \cdot d + 1\) to get \(10.94 = k \cdot 100 + 1\). Solving this equation for \(k\) gives \(k = (10.94 - 1) / 100 = 0.0994\) atmospheres per meter.

Find the pressure at 50 meters

With \(k = 0.0994\), we can now substitute \(d = 50\) into the original equation to get \(p = 0.0994 \cdot 50 + 1\), which gives \(p = 5.97\) atmospheres.

Key concepts

Differential Equations

Differential equations are mathematic tools that relate functions with their derivatives. They serve to model the behaviors of dynamic systems, including the variation of pressure with depth, like in our example. Although the original problem does not involve differential calculus directly, understanding how differential equations are constructed is fundamental when it comes to more complex scenarios of pressure variation where rate of change plays a role.

For example, if we had the rate of pressure change concerning depth as a function, say the derivative \( dp/dd \) (where \( p \) is pressure and \( d \) is depth), we might have a differential equation to solve for \( p \). The basics of mastering differential equations can help in comprehending how such relationships are derived from real-world physical laws, in this case, hydrostatic pressure.

Unit Conversion

Unit conversion is a pivotal skill in problem-solving that involves translating different measurements into a common unit to make calculations consistent and understandable. Our example required understanding the concept of atmospheres and how it relates to meters of depth.

Typically, pressure under water increases by 1 atmosphere for every 10 meters of descent, with 1 atmosphere being the pressure at sea level. Efficiently using unit conversion, students can bridge the gap between abstract mathematical formulas and practical measurements they may encounter, for example, when engaged in activities like diving.

Problem-Solving in Calculus

Problem-solving in calculus involves identifying the mathematical principles relevant to a scenario and applying them systematically to find a solution. In our example, the key steps included isolating the variable \( k \) after recognizing it as a proportionality constant, and then applying this constant to a new situation (50 meters depth). Calculus problems often require paying attention to the relationships between variables, and this problem suggests an underlying direct proportionality between depth and pressure, a starting concept that can lead to more complex calculus applications.

Real-world Application of Calculus

Calculus has diverse applications in the real world, ranging from engineering to environmental science. In our scenario, calculus helps to understand how pressure changes with depth in water - a concept that can be crucial in fields such as marine biology and oceanography. For an ocean engineer, calculating the pressure on a submarine's hull at various depths using calculus principles is crucial for design and safety. This exercise demonstrates that even simple linear models often serve as the introductory step to engaging with the natural world through mathematical lenses.