Question

Given the following vector fields and oriented curves \(C,\) evaluate \(\int_{C} \mathbf{F} \cdot \mathbf{T} d s.\) \(\mathbf{F}=\langle y, x\rangle\) on the line segment from \((1,1)\) to \((5,10)\)

Short answer

Based on the given vector field, \(\mathbf{F}=\langle y, x\rangle\), and the oriented curve \(C\) (line segment from \((1,1)\) to \((5,10)\)), the line integral of the vector field \(\mathbf{F}\) along the curve \(C\) is found to be \(\frac{224}{97}\).

Step-by-step solution

Parametrize the curve C

Let \(C\) be the line segment from \((1,1)\) to \((5,10)\). We can parametrize \(C\) by using the parameter \(t\) which varies from \(0\) to \(1\). Let \(\mathbf{r}(t) = (1-t)\langle1,1\rangle + t\langle5,10\rangle\). Then, for \(0\leq t\leq1\), \(\mathbf{r}(t)\) describes the line segment from \((1,1)\) to \((5,10)\).

Find the tangent vector and its magnitude

Find the derivative of \(\mathbf{r}(t)\) with respect to \(t\), which gives us the tangent vector along the curve: \(\frac{d\mathbf{r}}{dt} = -\langle1,1\rangle + \langle5,10\rangle = \langle4,9\rangle\). Now, find the magnitude of the tangent vector: \(||\frac{d\mathbf{r}}{dt}|| = ||\langle4,9\rangle|| = \sqrt{4^2 + 9^2} = \sqrt{97}\).

Find the unit tangent vector T

Now we calculate the unit tangent vector by dividing the tangent vector by its magnitude: \(\mathbf{T} = \frac{\frac{d\mathbf{r}}{dt}}{||\frac{d\mathbf{r}}{dt}||} = \frac{\langle4,9\rangle}{\sqrt{97}}\).

Evaluate the vector field along the curve and find the dot product with T

Evaluate the given vector field \(\mathbf{F}=\langle y, x\rangle\) along the curve \(C\). Using \(\mathbf{r}(t)\), we have \(\mathbf{F}(\mathbf{r}(t)) = \langle1-t+10t, 1-t+5t\rangle\). Now compute the dot product of \(\mathbf{F}(\mathbf{r}(t))\) and \(\mathbf{T}\): \(\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{T} = \frac{\langle1-t+10t, 1-t+5t\rangle}{\sqrt{97}} \cdot \frac{\langle4,9\rangle}{\sqrt{97}} = \frac{1}{97} (4(1-t+10t) + 9(1-t+5t))\).

Integrate the dot product over the given parameter range

The line integral of the vector field \(\mathbf{F}\) along the curve \(C\) is given by \(\int_{C} \mathbf{F}\cdot\mathbf{T}ds\). We will integrate the expression found in Step 4 over the parameter range \(0 \le t \le 1\), multiplied by the magnitude of the tangent vector \(||\frac{d\mathbf{r}}{dt}||\) using the substitution \(ds = ||\frac{d\mathbf{r}}{dt}|| dt\): $$\int_{0}^{1} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{T} ||\frac{d\mathbf{r}}{dt}|| dt = \int_{0}^{1}\frac{1}{97} (4(1-t+10t) + 9(1-t+5t)) \cdot \sqrt{97} dt.$$ Now, integrating the expression and computing the definite integral: $$\int_{0}^{1}\frac{1}{97} (4(1-t+10t) + 9(1-t+5t)) \cdot \sqrt{97} dt = \frac{1}{97}(198t^2 + 26t) \Big|_{0}^{1} = \frac{224}{97}.$$ Hence, the line integral of the vector field \(\mathbf{F}\) along the curve \(C\) is \(\frac{224}{97}\).

Key concepts

Vector Field

A vector field is a map that assigns to every point in a space a vector. In the context of our exercise, we're dealing with a two-dimensional vector field, represented by \(\mathbf{F}=\langle y, x\rangle\).

Parametrization of Curve

Parametrization of a curve means expressing the coordinates of points on the curve as functions of a single variable - let's call it \(t\). In the given problem, the curve \(C\) is parametrized by \(\mathbf{r}(t) = (1-t)\langle1,1\rangle + t\langle5,10\rangle\), which creates a smooth path from the point \( (1,1) \) to \( (5,10) \) as \(t\) varies from 0 to 1.

Tangent Vector

The tangent vector at a point on a curve represents the direction and rate of change of the curve at that point. It's found by differentiating the parametric equations of the curve. In this case, \(\frac{d\mathbf{r}}{dt} = \langle4,9\rangle\) is the tangent vector which shows the direction of the curve \(C\) at any instance as it moves from the initial to the final point.

Dot Product

The dot product is a way of multiplying two vectors, resulting in a scalar. It's calculated by summing the products of the corresponding components of the vectors. If the vectors are perpendicular, their dot product is zero. The dot product function, \(\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{T}\), quantifies how much of the vector field \(\mathbf{F}\) is in the direction of the tangent vector \(\mathbf{T}\) at each point \(t\) along the curve \(C\).

Magnitude of a Vector

The magnitude of a vector represents its length or amount of movement. It's calculated as the square root of the sum of the squares of its components. Here, \(||\frac{d\mathbf{r}}{dt}|| = \sqrt{4^2 + 9^2} = \sqrt{97}\) represents the magnitude of the tangent vector \(\frac{d\mathbf{r}}{dt}\). It's critical for finding the unit tangent vector and adjusting the integral in the final step of the problem.