Problem 66 - Thomas Calculus

Question

In Exercises \(63-68,\) use a CAS to perform the following steps for finding the work done by force F over the given path: $$\begin{array}{l}{\mathbf{F}=2 x y \mathbf{i}-y^{2} \mathbf{j}+z e^{x} \mathbf{k} ; \quad \mathbf{r}(t)=-t \mathbf{i}+\sqrt{t} \mathbf{j}+3 t \mathbf{k}} \\ {1 \leq t \leq 4}\end{array}$$

Step-by-Step Solution

Answer

Calculate the work by evaluating the integral:\[ W = \int_{1}^{4} (2t^{3/2} - \frac{t^{1/2}}{2} + 9te^{-t}) \, dt \]

Step 1: Evaluate the Force Vector Field at r(t)

First, substitute the path parameterized by \( t \) into the force \( \mathbf{F} = 2xy \mathbf{i} - y^2 \mathbf{j} + ze^x \mathbf{k} \).Given: \[ \mathbf{r}(t) = -t \mathbf{i} + \sqrt{t} \mathbf{j} + 3t \mathbf{k} \]Substituting, we find:\[ \mathbf{F}(t) = 2(-t)(\sqrt{t}) \mathbf{i} - (\sqrt{t})^2 \mathbf{j} + (3t)e^{-t} \mathbf{k} \]

Step 2: Simplify the Force Expression

Simplify the force components with respect to parameter \( t \). The force components become:\[ \mathbf{F}(t) = -2t^{3/2} \mathbf{i} - t \mathbf{j} + 3te^{-t} \mathbf{k} \]

Step 3: Find the Derivative of the Path Function

Compute the derivative of the path \( \mathbf{r}'(t) \):\[ \mathbf{r}'(t) = -1 \mathbf{i} + \frac{1}{2\sqrt{t}} \mathbf{j} + 3 \mathbf{k} \]

Step 4: Calculate the Dot Product

Find the dot product of \( \mathbf{F}(t) \) and \( \mathbf{r}'(t) \).\[ \mathbf{F} \cdot \mathbf{r}' = (-2t^{3/2})(-1) + (-t)\left(\frac{1}{2\sqrt{t}}\right) + (3te^{-t})\cdot(3) \]Simplifying the dot product gives:\[ F \cdot \mathbf{r}' = 2t^{3/2} - \frac{t}{2\sqrt{t}} + 9te^{-t} \]

Step 5: Further Simplify the Dot Product

Simplify the term \( \frac{t}{2\sqrt{t}} \):\[ \frac{t}{2\sqrt{t}} = \frac{t^{1/2}}{2} \]So the complete dot product becomes:\[ F \cdot \mathbf{r}' = 2t^{3/2} - \frac{t^{1/2}}{2} + 9te^{-t} \]

Step 6: Calculate the Definite Integral

Find the work done by computing the definite integral from \( t=1 \) to \( t=4 \) of the dot product:\[ W = \int_{1}^{4} (2t^{3/2} - \frac{t^{1/2}}{2} + 9te^{-t}) \, dt \]

Step 7: Evaluate the Integral

Use a computer algebra system (CAS) to evaluate:\[ W = \int_{1}^{4} (2t^{3/2} - \frac{t^{1/2}}{2} + 9te^{-t}) \, dt \]**Let the CAS perform the integration to find W as a numerical value or simplified symbolic result.

Key Concepts

Force Vector Field

A force vector field is like a map that indicates the direction and strength of a force at various points in space. Imagine it as a collection of tiny arrows spread over a region, each pointing in the direction of the force's action and having a length proportional to its magnitude. In this exercise, the force vector field is given by \( \mathbf{F} = 2xy \mathbf{i} - y^2 \mathbf{j} + ze^x \mathbf{k} \). The components \( 2xy \mathbf{i} \), \( -y^2 \mathbf{j} \), and \( ze^x \mathbf{k} \) represent how the force varies along the x, y, and z directions respectively.

To evaluate how this field acts along a certain path, the force expression is substituted with the parameterized path \( \mathbf{r}(t) = -t \mathbf{i} + \sqrt{t} \mathbf{j} + 3t \mathbf{k} \). This shows how the force changes as it acts on an object moving along this path.

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar (a single number). This is particularly useful in physics for calculating work, where we determine how much of the force vector contributes to the movement along a path. The dot product is calculated as \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \) in this problem, where \( \mathbf{F}(t) \) is the force vector evaluated at the path, and \( \mathbf{r}'(t) \) is the derivative of that path.

• The negative sign in components of a dot product indicates that force opposes the direction of movement in some parts of its trajectory.
• Here, the expressions get combined and simplified, giving \( 2t^{3/2} - \frac{t^{1/2}}{2} + 9te^{-t} \), which reflects the scalar projection of force along the path.

This dot product is then used in the integral to find out how much work is done by the force.

Definite Integral

A definite integral calculates the accumulation of quantities, such as area under a curve. In the context of physics, it's perfect for finding total work done, which is the integral of force over a distance/path. With the dot product \( \mathbf{F} \cdot \mathbf{r}' \) calculated, the definite integral \( \int_{1}^{4} (2t^{3/2} - \frac{t^{1/2}}{2} + 9te^{-t}) \, dt \) is used to sum up all of the little pieces of work done as the force moves the object from \( t=1 \) to \( t=4 \).

This gives us the total work done over this section of the path. You'll often use a computer algebra system (CAS) to compute definite integrals involving complex functions, which simplifies tedious calculations into easy results.

Path Parameterization

Path parameterization involves representing a path or curve using parameters, often denoted as \( t \), which can be thought of as a type of coordinate that traces the path as it varies. In our exercise, the path is parameterized by \( \mathbf{r}(t) = -t \mathbf{i} + \sqrt{t} \mathbf{j} + 3t \mathbf{k} \).

This means at every point along the path, its position is defined in terms of \( t \). When \( t \) increases, it changes the x, y, and z coordinates in the i, j, and k directions according to the path's equations.

• This simplification makes it easier to handle vector calculus on curves and paths by converting spatial problems into parameter problems.
• It translates the physical trajectory into mathematical equations which can then be used to analyze physical phenomena like work done by forces.

Through parameterization, operations such as integration and differentiation become more straightforward, allowing problem-solving to occur in a manageable framework.