Problem 68 - Calculus: Early Transcendental Functions

Question

If \(C\) has parametric equations \(x=x(t), y=y(t), z=z(t)\) \(a \leq t \leq b,\) for differentiable functions \(x, y\) and \(z\) show that \(\int_{C} \mathbf{F} \cdot \mathbf{T} d s=\int_{a}^{b}\left[F_{1}(x, y, z) x^{\prime}(t)+F_{2}(x, y, z) y^{\prime}(t)+\right.\) \(\left.F_{3}(x, y, z) z^{\prime}(t)\right] d t,\) which is the work line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}.\)

Step-by-Step Solution

Answer

This problem demonstrates how the work done by a vector force \( \mathbf{F} \) along a curve \(C\) can be expressed as the integral over either the path \(C\) itself or an integral over its parameter \(t\). This is a fundamental concept in the field of line integrals in vector calculus.

Step 1: Express the Vector T

The unit tangent vector \( \mathbf{T} \) of a curve \(C\) parametrized by \(t\) can be expressed by differentiating the position vector component-wise and normalizing it. So, \( \mathbf{T} = \frac{d\mathbf{r}}{ds} = \frac{dx}{ds}\mathbf{i} + \frac{dy}{ds}\mathbf{j} + \frac{dz}{ds}\mathbf{k} \) where \( \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \)

Step 2: Express the Arc Length ds

The arc length \(ds\) can be expressed in terms of \(t\) using the Pythagorean theorem, as \(ds = \sqrt{(dx)^2 + (dy)^2 + (dz)^2} = \sqrt{(x'(t)dt)^2 + (y'(t)dt)^2 + (z'(t)dt)^2} = dt \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\).

Step 3: Subsitute the Unit Tangent Vector and the Arc Length into the Line Integral

Now, we substitute \( \mathbf{T} \) and \( ds \) into the line integral and we express it as follows \( \int_{C}\mathbf{F} \cdot \mathbf{T}ds = \int_{C}\mathbf{F} \cdot \frac{d\mathbf{r}}{ds} ds =\int_{C}\mathbf{F} \cdot d\mathbf{r} \)

Step 4: Equation Transform

Then the form integrand over \(t\) will look like this \( \int_{a}^{b} [F_{1}(x, y, z) x^{\prime}(t) + F_{2}(x, y, z) y^{\prime}(t) + F_{3}(x, y, z) z^{\prime}(t)] dt \)

Step 5: Proving the Equivalence

By looking at the transformation done in step 3 on the line integral over \(C\) and the integral over \(t\) in step 4, it can be clearly seen that the two expressions are equivalent. Thus, we have proved the given equation that a line integral can be expressed in the form of an integral over one parameter \(t\).

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in calculus, used to describe a curve by expressing the coordinates of the points on the curve as functions of a parameter, often denoted as \( t \). Instead of defining \( y \) directly in terms of \( x \) or vice versa, you define both \( x(t) \) and \( y(t) \) independently.
This method is particularly useful for representing complex curves that are difficult to express using a single function.
In three-dimensional space, a curve \( C \) can be represented parametrically with equations \( x = x(t) \), \( y = y(t) \), and \( z = z(t) \). These provide a flexible way to explore the path of the curve as \( t \) varies over an interval \([a, b]\).
By utilizing parametric equations, it's easier to perform various calculations on the curve, such as finding the tangent vector, arc length, or analyzing integrals over the curve, leading to simplified and elegant solutions.

Unit Tangent Vector

The unit tangent vector is a fundamental concept in calculus that allows us to understand the direction in which a curve is heading at a given point. Vectors provide not only magnitude but also direction, which makes them quite useful in multi-dimensional calculus.
To compute the unit tangent vector \( \mathbf{T}(t) \) for a parametric curve, we start by finding the derivative of the position vector \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \), which gives the tangent vector \( \mathbf{r}'(t) = x'(t)\mathbf{i} + y'(t)\mathbf{j} + z'(t)\mathbf{k} \).
However, to convert this into a unit tangent vector, which is dimensionless and has a magnitude of 1, you divide by its magnitude \( ||\mathbf{r}'(t)|| \).
Thus, the unit tangent vector \( \mathbf{T}(t) \) is defined as:

• \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \)

This provides a standardized way to express the direction of the curve in space at any point \( t \).

Arc Length

Arc length is a crucial measure in calculus that represents the total distance along a curve, from one endpoint to another. It involves integrating the magnitude of the derivative of the curve's position vector.
For a parametric curve \( C \) with derivatives \( x'(t), y'(t), \) and \( z'(t) \), the arc length \( ds \) in differential form is given by:

• \( ds = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt \)

This formula essentially uses the Pythagorean theorem in three dimensions to account for tiny distances along the curve.
The actual total arc length \( S \) between two points \( t = a \) and \( t = b \) can be calculated by integrating:

• \( S = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt \)

This integral shows how much the curve stretches and helps in various applications such as physics where distance traveled is important.

Vector Calculus

Vector calculus extends the concepts of calculus to vector fields, which are functions that assign a vector to each point in a subset of space. It is an essential tool for dealing with physical phenomena that involve directions and magnitudes, like fluid flow and electromagnetism.
In the context of line integrals, vector calculus is used to compute the work done by a force field \( \mathbf{F} \) as it moves along a path or curve \( C \).
The work line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) essentially sums up the component of the force field in the direction of movement along the curve, weighted by the infinitesimal distances \( d\mathbf{r} \).
The line integral of a vector field is calculated using:

• \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} [F_{1}(x, y, z) x'(t) + F_{2}(x, y, z) y'(t) + F_{3}(x, y, z) z'(t)] \, dt \)

This tells us precisely how much of the force is applied in the direction of the curve, providing insights crucial for engineering and physics.