Question

Given the following vector fields and oriented curves \(C,\)evaluate \(\int_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=\langle x, y\rangle\) on the parabola \(\mathbf{r}(t)=\left\langle 4 t, t^{2}\right\rangle,\) for \(0 \leq t \leq 1\)

Short answer

Answer: The value of the line integral \(\int_{C} \mathbf{F} \cdot \mathbf{T} \, ds\) is \(\frac{17}{2}\).

Step-by-step solution

Compute the derivative of \(\mathbf{r}(t)\)

To find the derivative of \(\mathbf{r}(t)\), we differentiate each component with respect to \(t\): $$ \mathbf{r'}(t) = \frac{d}{dt} \left\langle 4t, t^{2}\right\rangle = \langle 4, 2t \rangle. $$

Compute the magnitude of \(\mathbf{r'}(t)\)

The magnitude of a vector is given by the square root of the sum of the squares of its components, so for \(\mathbf{r'}(t)\) we have: $$ \|\mathbf{r'}(t)\| = \sqrt{(4)^{2} + (2t)^{2}} = \sqrt{16 + 4t^{2}}. $$

Determine the unit tangent vector \(\mathbf{T}\)

The unit tangent vector \(\mathbf{T}\) is found by dividing the derivative of \(\mathbf{r}(t)\) by its magnitude: $$ \mathbf{T} = \frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|} = \frac{\langle 4, 2t \rangle}{\sqrt{16 + 4t^{2}}}. $$

Convert \(\mathbf{F}\) into its parametric form

Since \(\mathbf{F} = \langle x, y \rangle\), we will replace \(x\) and \(y\) with their parametric expressions in terms of \(t\) as given by \(\mathbf{r}(t)\). So, \(\mathbf{F}\) becomes $$ \mathbf{F}(t) = \langle 4t, t^{2} \rangle. $$

Compute the dot product \(\mathbf{F} \cdot \mathbf{T}\)

To evaluate the dot product, we multiply the corresponding components of \(\mathbf{F}(t)\) and \(\mathbf{T}\), then sum the products: $$ \mathbf{F}(t) \cdot \mathbf{T} = \frac{(4t)(4) + (t^{2})(2t)}{\sqrt{16 + 4t^{2}}} = \frac{16t + 2t^{3}}{\sqrt{16 + 4t^{2}}}. $$

Evaluate the line integral

Finally, to evaluate the line integral, we will integrate the dot product of \(\mathbf{F}\) and \(\mathbf{T}\) with respect to arc length \(ds\). We should multiply the scalar function found in step 5 by the differential of arc length \(ds = \|\mathbf{r'}(t)\| dt\), and integrate over the interval \([0, 1]\): \begin{align*} \int_{C}\mathbf{F} \cdot \mathbf{T} \, ds &= \int_{0}^{1} \frac{(16t + 2t^{3})}{\sqrt{16 + 4t^{2}}} \cdot \sqrt{16 + 4t^{2}} \, dt \\ &= \int_{0}^{1} (16t + 2t^{3}) \, dt \\ &= \left[8t^{2} + \frac{1}{2}t^{4}\right]_0^1 \\ &= 8 + \frac{1}{2} = \boxed{\frac{17}{2}}. \end{align*} So, the value of the line integral \(\int_{C} \mathbf{F} \cdot \mathbf{T} \, ds\) is \(\frac{17}{2}\).

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics concerned with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space. A vector field is represented by a vector-valued function that assigns vectors to points in space, often defining forces or velocities at those points.

Within vector calculus, line integrals are used to measure the effect along a curve or path in space. As seen in the exercise, the line integral of the vector field \textbf{F} over the curve C, denoted as \(\int_{C}\mathbf{F} \cdot \mathbf{T} d s\), captures how much the field \textbf{F} 'aligns' with the direction of the curve. This is especially useful in fields like physics and engineering to evaluate work done by a force along a path, or flow across a curve.

Tangent Vector

In the context of line integrals, the tangent vector (\textbf{T}) plays a pivotal role as it represents the direction in which the curve is 'heading' at any given point. Mathematically, the tangent vector to a curve at a particular point is the derivative of the position vector \textbf{r}(t) with respect to the parameter \(t\), evaluated at that point.

The unit tangent vector, which has a magnitude of 1, is found by normalizing this derivative. It's used as a key component in calculating line integrals because it provides directionality to the movement along the curve. When performing a line integral, the dot product of this unit tangent vector with the vector field in question yields the component of the field that's parallel to the curve.

Parametric Equations

Parametric equations define a set of functions to express the coordinates of the points that make up a geometric object, such as a curve. In the given exercise, the curve C is represented by the parametric equations \textbf{r}(t) which define the parabola in terms of the parameter \(t\). Parametric representations are powerful because they easily express curves and surfaces where traditional Cartesian coordinate functions fall short.

The equations \(\mathbf{r}(t)=\langle 4t, t^2\rangle\) provide the x and y components of every point on the parabola as functions of \(t\). When we differentiate \textbf{r}(t), we obtain the velocity vector, which after normalizing, gives us the tangent vector \textbf{T} crucial for computing the line integral.

Dot Product

The dot product, or scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is central to the computation of line integrals in vector calculus.

Given two vectors \textbf{A} and \textbf{B}, their dot product \textbf{A} \cdot \textbf{B} is calculated as \(A_xB_x + A_yB_y\) in two dimensions or \(A_xB_x + A_yB_y + A_zB_z\) in three dimensions, where \(A_x\), \(A_y\), and \(A_z\) are the components of \textbf{A} and similarly for \textbf{B}. In the scenario of line integrals, the dot product is used to project one vector onto the direction of another, thus measuring how much one vector 'contributes' to the other. In the provided exercise, the dot product of \textbf{F} and \textbf{T} reflects the magnitude of the vector field \textbf{F} in the direction of the curve's tangent.