2.5. Vector Fields and Line integral

 

Vector Fields

vector field assigns a vector to every point in space. Examples include hemodynamics, where vector fields are used to understand blood flow patterns and detect abnormalities such as stenosis or aneurysms, and respiratory dynamics, where vector fields model airflow in the lungs to study ventilation efficiency. 

Let \(\mathbf{r} = (x, y, z)\). A vector field is defined as: $$F(\mathbf{r}) = F_x(\mathbf{r}) \hat{i} + F_y(\mathbf{r}) \hat{j} + F_z(\mathbf{r}) \hat{k},$$ where: 

  • \( F_x(\mathbf{r}) \): The \(x\)-component of the vector field. 
  • \( F_y(\mathbf{r}) \): The \(y\)-component of the vector field. 
  • \( F_z(\mathbf{r}) \): The \(z\)-component of the vector field. 
  • \( \hat{i}, \hat{j}, \hat{k} \): Unit vectors in the \(x\)-, \(y\)-, and \(z\)-directions, respectively. 
Example 1: Blood Velocity Field  In a cylindrical blood vessel, the velocity of blood flow can be described as a vector field. For laminar flow, the velocity at any point is given by: \[ \mathbf{v}(r, z) = v_z(r) \hat{k}, \quad v_z(r) = v_{\text{max}} \left(1 - \frac{r^2}{R^2}\right), \] where \(v_{\text{max}}\) is the maximum velocity at the center of the vessel, \(R\) is the radius of the vessel, \(r\) is the radial distance from the center, and \(\hat{k}\) is the unit vector along the direction of blood flow. This model demonstrates that the velocity of blood is highest at the center of the vessel and decreases towards the walls due to the no-slip condition, which ensures that the fluid has zero velocity at the vessel's boundary.

Line Integral Along a Vector Field

The line integral of a vector field \(\mathbf{F}(\mathbf{r})\) along a curve \(C\) measures the work done by the field along the path. It is given by: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (F_x \, dx + F_y \, dy + F_z \, dz), \] where:

    • \(C\) is the curve along which the integral is evaluated.
    • \(\mathbf{F}(\mathbf{r}) = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}\) is the vector field. 
    • \(d\mathbf{r} = dx \, \hat{i} + dy \, \hat{j} + dz \, \hat{k}\) is the differential element of the curve. \end{itemize} 
  •  To reduce this line integral to a 1D integral, parameterize the curve \(C\) by a parameter \(t\), such that: \[ \mathbf{r}(t) = (x(t), y(t), z(t)), \quad t \in [a, b]. \] The differentials \(dx\), \(dy\), and \(dz\) are expressed as: \[ dx = \frac{dx}{dt} \, dt, \quad dy = \frac{dy}{dt} \, dt, \quad dz = \frac{dz}{dt} \, dt. \] Substituting these into the line integral, we have: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \left[ F_x(x(t), y(t), z(t)) \frac{dx}{dt} + F_y(x(t), y(t), z(t)) \frac{dy}{dt} + F_z(x(t), y(t), z(t)) \frac{dz}{dt} \right] dt. \] 
    •  Example: Work Done Along a Helix. Let \(\mathbf{F}(\mathbf{r}) = z \, \hat{i} + x \, \hat{j} + y \, \hat{k}\), and consider the helix parameterized by: \[ \mathbf{r}(t) = (\cos t, \sin t, t), \quad t \in [0, 2\pi]. \] Then: \[ \frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t, \quad \frac{dz}{dt} = 1. \] Substituting these into the line integral: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left[ z(-\sin t) + x(\cos t) + y(1) \right] dt, \] where \(x = \cos t\), \(y = \sin t\), and \(z = t\). Simplifying: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left[ t(-\sin t) + \cos t (\cos t) + \sin t (1) \right] dt. \] Further simplifications yield the final 1D integral: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left( -t\sin t + \cos^2 t + \sin t \right) dt. \] 
    • Example: Blood Flow Work Along an ArteryThe line integral can quantify the work done by blood pressure as blood moves along a curved artery. Suppose the pressure field \(\mathbf{F}(x, y) = -y \hat{i} + x \hat{j}\) acts along a path \(C\), parameterized as \(\mathbf{r}(t) = t \hat{i} + t^2 \hat{j}\), \(0 \leq t \leq 1\). The line integral is: \[\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt.\] Substitute:\[\mathbf{F}(\mathbf{r}(t)) = -t^2 \hat{i} + t \hat{j}, \quad \frac{d\mathbf{r}}{dt} = \hat{i} + 2t \hat{j}.\] The dot product is:\[\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = (-t^2)(1) + (t)(2t) = -t^2 + 2t^2 = t^2.\]The integral becomes:\[\int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3}.\]The work done by the pressure field along the artery is \( \frac{1}{3} \, \text{units} \).



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2.1 Vectors

2.4. Taylor Expansion and Approximation