Problem 45 - Thomas Calculus

Question

Salt water with a density of \(\delta=0.25 \mathrm{g} / \mathrm{cm}^{2}\) flows over the curve \(\mathbf{r}(t)=\sqrt{t} \mathbf{i}+t \mathbf{j}, 0 \leq t \leq 4,\) according to the vector field \(\mathbf{F}=\delta \mathbf{v},\) where \(\mathbf{v}=x y \mathbf{i}+(y-x) \mathbf{j}\) is a velocity field measured in centimeters per second. Find the flow of F over the curve \(\mathbf{r}(t).\)

Step-by-Step Solution

Answer

The flow is \( \frac{5}{3} \) cm³/s.

Step 1: Determine Vector Field F

Given that the vector field \(\mathbf{F} = \delta \mathbf{v}\), where \(\delta = 0.25\) is the density of the saltwater, and \(\mathbf{v} = x y \mathbf{i} + (y - x) \mathbf{j}\) is the velocity field, we can express \(\mathbf{F}\) as follows: \[\mathbf{F} = 0.25(x y \mathbf{i} + (y - x) \mathbf{j}) = (0.25 x y) \mathbf{i} + (0.25(y - x)) \mathbf{j}.\]

Step 2: Parametrize the Curve r(t)

The curve is given by \(\mathbf{r}(t) = \sqrt{t} \mathbf{i} + t \mathbf{j}\). This means the coordinates \((x, y)\) on the curve at time \(t\) are \((\sqrt{t}, t)\). Substitute \(x = \sqrt{t}\) and \(y = t\) into \(\mathbf{v}\) to get these on the curve.

Step 3: Calculate \(\mathbf{F}(\mathbf{r}(t))\)

Substitute \(x = \sqrt{t}\) and \(y = t\) into the expression for \(\mathbf{F}\):\[\mathbf{F}(\mathbf{r}(t)) = 0.25(\sqrt{t} \cdot t) \mathbf{i} + 0.25(t - \sqrt{t}) \mathbf{j} = 0.25 t^{3/2} \mathbf{i} + 0.25(t - t^{1/2}) \mathbf{j}.\]

Step 4: Find the tangent vector \(\mathbf{r}'(t)\)

Differentiate \(\mathbf{r}(t) = \sqrt{t} \mathbf{i} + t \mathbf{j}\) with respect to \(t\):\[\mathbf{r}'(t) = \left(\frac{1}{2\sqrt{t}}\right) \mathbf{i} + \mathbf{j}.\]

Step 5: Set up the line integral for flux calculation

The flow of \(\mathbf{F}\) over \(\mathbf{r}(t)\) from \(t=0\) to \(t=4\) is given by the line integral of \( \mathbf{F} \cdot \mathbf{r}'(t) \). Substitute \(\mathbf{F}(\mathbf{r}(t))\) and \(\mathbf{r}'(t)\) into the dot product:\[\int_{0}^{4} \left((0.25 t^{3/2}) \cdot \frac{1}{2\sqrt{t}} + 0.25(t - t^{1/2}) \right) \, dt.\]

Step 6: Simplify the integrand

Simplify the integrand expression: \[0.25 t^{3/2} \cdot \frac{1}{2\sqrt{t}} = 0.125 t \] and \[0.25(t - t^{1/2}) = 0.25t - 0.25 t^{1/2}.\] The expression inside the integral is \[0.125t + 0.25t - 0.25t^{1/2} = 0.375t - 0.25t^{1/2}.\]

Step 7: Evaluate the integral

Calculate \[\int_{0}^{4} (0.375t - 0.25t^{1/2}) \, dt\]. This equals \[0.375 \int_{0}^{4} t \, dt - 0.25 \int_{0}^{4} t^{1/2} \, dt.\] Solve each integral separately:\[0.375 \left[\frac{t^2}{2}\right]_{0}^{4} = 0.375 \times \frac{16}{2} = 3,\] and \[0.25 \left[\frac{2}{3}t^{3/2}\right]_{0}^{4} = 0.25 \times \frac{2}{3} \times 8 = \frac{4}{3}.\] Subtract the results: \[3 - \frac{4}{3} = \frac{9}{3} - \frac{4}{3} = \frac{5}{3}.\]

Step 8: Conclusion

The flow of water with density 0.25 g/cm² through the curve is \( \frac{5}{3} \) cm³/s.

Key Concepts

Vector Calculus

Understanding vector calculus is essential when dealing with line integrals, such as the exercise you encountered. This field of mathematics provides tools and techniques to analyze vectors and their interactions in space. These vectors can represent various quantities, from forces to fluid flow.
In the given problem, the vector field \( \mathbf{F} \) depicts the flow of saltwater along a curve, combining density and velocity. Calculus allows us to describe the change and behavior of these vectors over time, particularly as we move along a given path or curve.
When working with vector fields, it's important to comprehend notions like divergence and curl. These concepts might not be directly calculated in this exercise, but they provide deeper insights into how vectors behave and interact over a particular region. Vector calculus forms the backbone for performing the line integrals needed here, enabling us to calculate quantities like flux efficiently.

Flux Calculation

The main objective of this exercise is to calculate the flux of the vector field \( \mathbf{F} \) across the given curve \( \mathbf{r}(t) \). Flux calculation is essentially measuring how much of a vector field passes through a certain surface or path.
In this specific instance, the line integral of \( \mathbf{F} \cdot \mathbf{r}'(t) \) from \( t = 0 \) to \( t = 4 \) represents the flux of the field through the curve. The dot product between \( \mathbf{F} \) and the tangent vector \( \mathbf{r}'(t) \) is crucial, as it ensures we're only considering the component of the vector field that is parallel to the curve.
After simplifying the integral, the resulting value tells us the volume rate, in cubic centimeters per second, at which the saltwater flows through the curve. Breaking down the steps of simplifying the integral and substitution is critical, as it directly affects how accurately you measure the flux.

Velocity Field

In vector calculus, the concept of a velocity field represents how fluid moves through space. In this task, the velocity field \( \mathbf{v} = xy \mathbf{i} + (y-x) \mathbf{j} \) plays a central role in determining the field \( \mathbf{F} \) by incorporating the fluid's density.
This velocity field indicates that fluid velocity depends on both \( x \) and \( y \) coordinates, showcasing differing movements in the \( \mathbf{i} \) and \( \mathbf{j} \) directions. Understanding this behavior is crucial, as it shapes how the vector field interacts and flows over the given curve.
Velocity fields help predict fluid dynamics, providing insights into how quickly fluid elements move and the direction of this movement. In this problem, accurately substituting \( x \) and \( y \) from the parametrization \( \mathbf{r}(t) \) is key for appropriately determining the field \( \mathbf{F} \) over the entire curve.

Parametrization

Parametrization is about representing a curve or surface using parameters, making complex paths manageable and calculable. In this problem, the curve \( \mathbf{r}(t) = \sqrt{t} \mathbf{i} + t \mathbf{j} \) is expressed in terms of the parameter \( t \).
By setting \( x = \sqrt{t} \) and \( y = t \), you're able to relate these coordinates back to the velocity field and, in turn, calculate \( \mathbf{F}(\mathbf{r}(t)) \). This process is indispensable for simplifying and solving line integrals along curves.
Parametrization not only aids in calculations but also offers a systematic approach to break down complex curves and surfaces into more simple, manageable segments. This can hugely benefit anyone dealing with vector fields or performing intricate flux calculations across non-linear paths.