Problem 61 - Technical Mathematics with Calculus

Question

The approximate density of seawater at a depth of \(h\) miles is \(d=64.0 e^{0.00676 h} \mathrm{lb} / \mathrm{ft}^{3} .\) Find the rate of change of density, with respect to depth, at a depth of 1.00 mile.

Step-by-Step Solution

Answer

The rate of change of density at a depth of 1.00 mile is approximately \(0.4358 \mathrm{lb} / \mathrm{ft}^3 / \mathrm{mile}\).

Step 1: Write down the density function

The density of seawater is given by the function \(d(h) = 64.0e^{0.00676h} \, \mathrm{lb} / \mathrm{ft}^3\), where \(h\) is the depth in miles.

Step 2: Differentiate the density function

To find the rate of change of density with respect to depth, take the derivative of \(d(h)\) with respect to \(h\). This gives \(\frac{dd}{dh} = 64.0 \times 0.00676 \times e^{0.00676h}\).

Step 3: Evaluate the derivative at \(h = 1.00\) mile

Substitute the value of \(h = 1.00\) into the derivative to find the rate of change at that depth. \(\frac{dd}{dh} = 64.0 \times 0.00676 \times e^{0.00676 \times 1.00}\).

Step 4: Calculate the value of the derivative

Perform the calculation to find the rate of change of density at \(h = 1.00\) mile. \(\frac{dd}{dh} = 64.0 \times 0.00676 \times e^{0.00676} \approx 64.0 \times 0.00676 \times 1.0067973 \approx 0.4358 \, \mathrm{lb} / \mathrm{ft}^3 / \mathrm{mile}\).

Key Concepts

Exponential Functions

Exponential functions are fundamental in both pure and applied mathematics, especially when it comes to growth processes. These functions can be identified by their variable exponent, which in this case, is dependent on depth, denoted as 'h'. The general form of an exponential function is given by \(f(x) = a \cdot b^x\), where \(a\) is a constant term, \(b\) is the base of the exponential, and \(x\) is the exponent. In the context of seawater density, the function presented in the problem is \(d(h) = 64.0e^{0.00676h}\).
Here, \(e\) represents the base of the natural logarithm, approximately equal to 2.71828, which is a constant that often appears in applications involving growth or decay. Exponential functions show how quantities can increase or decrease very rapidly — in this case, density changes relative to the depth in the sea. \(64.0\) is the initial density value at the surface, and \(0.00676h\) tells us how density grows exponentially as depth increases.
A key property of exponential functions is that their rate of change is proportional to their current value, which is particularly relevant when discussing derivatives and rates of change — topics we'll delve into next.

Derivatives

Derivatives are a critical tool in calculus, used to determine the rate of change of a function with regards to one of its variables. Intuitively, the derivative measures how a function value changes as its input changes. Mathematically, it is represented as \( \frac{df}{dx} = \lim_{h \to 0}(\frac{f(x+h) - f(x)}{h} )\) for a function \(f(x)\). For our seawater density problem, the derivative of \(d(h)\) with respect to \(h\) indicates how the seawater's density changes as we go deeper into the ocean.
To find this derivative, we differentiate the function \(d(h) = 64.0e^{0.00676h}\) with respect to \(h\). We apply the rule of differentiation for exponential functions, which states that the derivative of \(e^{kx}\) is \( ke^{kx}\). Therefore, the rate of change of density, expressed as \( \frac{dd}{dh}\), becomes \(64.0 \times 0.00676 \times e^{0.00676h}\). This expression gives us a formula for the rate of change at any given depth, allowing us to identify how quickly the seawater density is increasing or decreasing at specific points.

Calculus Applications

Calculus applications extend across various science and engineering fields, and understanding them can provide insights into complex real-world problems. In the context of environmental science, for instance, calculus allows us to model and predict how physical quantities, like density, change under certain conditions.
Applying calculus to our seawater density equation, we've found the derivative function that predicts the rate of change of density with depth. To apply this knowledge practically, we can then evaluate the derivative at a specific depth, as was done in Step 3 of the provided solution.
By calculating \( \frac{dd}{dh} \) at \(h = 1.00\) mile, we can predict how density is expected to change near that point. The result, approximately \(0.4358 \mathrm{lb}/\mathrm{ft}^3/\mathrm{mile}\), tells us that for every additional mile we go deeper into the ocean, the density increases by this rate. This application of calculus is not only useful for theoretical studies but also for practical situations like designing submarines that must withstand pressure changes at varying depths.