Problem 67 - 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: $$\begin{array}{l}{\mathbf{F}=(2 y+\sin x) \mathbf{i}+\left(z^{2}+(1 / 3) \cos y\right) \mathbf{j}+x^{4} \mathbf{k}} \\ {\mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j}+(\sin 2 t) \mathbf{k}, \quad-\pi / 2 \leq t \leq \pi / 2}\end{array}$$

Step-by-Step Solution

Answer

The work done is given by the definite integral of the dot product over the interval. Use CAS to find this value.

Step 1: Find the Derivative of the Position Vector

The first step is to find the derivative of the position vector \( \mathbf{r}(t) \). Given that \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (\sin 2t) \mathbf{k} \), we find:\[ \mathbf{r}'(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} + (2\cos 2t) \mathbf{k} \]

Step 2: Substitute Path into Force Field

Substitute the components of \( \mathbf{r}(t) \) into \( \mathbf{F} = (2y + \sin x) \mathbf{i} + (z^2 + \frac{1}{3}\cos y) \mathbf{j} + x^4 \mathbf{k} \). The substitution gives:\[ \mathbf{F}(t) = 2(\cos t) + \sin(\sin t) \mathbf{i} + ((\sin 2t)^2 + \frac{1}{3} \cos(\cos t)) \mathbf{j} + (\sin^4 t) \mathbf{k} \]

Step 3: Compute the Dot Product

Compute the dot product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \):\[\left(2\cos t + \sin(\sin t)\right)\cos t + \left(\sin^2(2t) + \frac{1}{3}\cos(\cos t)\right)(-\sin t) + (\sin^4 t)(2\cos 2t)\]Simplify this expression as needed.

Step 4: Integrate the Dot Product

Find the definite integral of the simplified dot product over the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \):\[\int_{-\pi/2}^{\pi/2} \left[ (2\cos t + \sin(\sin t))\cos t + (-\sin t)(\sin^2(2t) + \frac{1}{3}\cos(\cos t)) + 2(\cos 2t)(\sin^4 t) \right] \, dt\]

Step 5: Compute Using a CAS

Use a Computer Algebra System (CAS) to compute the integral from Step 4. This will give you the total work done, which is a definite numerical value.

Key Concepts

Work Done by Force

The concept of work done by force in vector calculus is an important one, particularly when dealing with vectors and motion along a specific path. In simple terms, the work done by a force is the energy transferred when an object is moved by a force along a path. It is mathematically given by the integral of the force vector along the path over which it acts.

• To calculate the work done, one must know both the force vector and the path along which it acts.
• The work is positive if the force and movement occur in the same direction, indicating that the force aids the movement.

In our exercise, the force vector is given by a complex function of multiple variables, which need to be evaluated along a specific path described by a position vector. Understanding these vectors and how they interact is key to solving such problems.

Dot Product

The dot product, often called the scalar product, is a fundamental operation in vector calculus. It is used to calculate the work done by a force along a path. The dot product between two vectors results in a scalar and reflects how much one vector extends in the direction of another.

• Formula: If you have vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), their dot product is \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \).
• The result is a single number (a scalar) that can tell us about the cosine of the angle between the vectors.

In the exercise, we find the dot product between the force vector, \( \mathbf{F}(t) \), and the derivative of the position vector, \( \mathbf{r}'(t) \), to set up the integral for calculating work. This step is crucial because it sets the stage for integrating along the path described by the position vector.

Definite Integral

The definite integral plays a vital role in Calculus, particularly when computing the work done along a path. When we integrate the dot product of force and derivative of the position vector over a given interval, we calculate the total work done.

• Formula: For a continuous function \( f(t) \), the definite integral from \( a \) to \( b \) is \( \int_{a}^{b} f(t) \,dt \), which computes the net area under the curve of \( f(t) \) from \( t = a \) to \( t = b \).
• The interval \([-\pi/2, \pi/2] \) indicates the specific path over which we are calculating the work.

In this exercise, the integral evaluates how the force contributes to the movement from start to end of the interval. It supplies the net energy transferred, which is vital for understanding the force system's effectiveness over the path.

Position Vector

The position vector describes the path along which the movement occurs. It is crucial because it provides the trajectory that the force vector acts upon, and it is parameterized by the variable \( t \).

• Position vectors describe locations in space and can be considered as directions from an origin point.
• In our exercise, the position vector \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (\sin 2t) \mathbf{k} \) gives a dynamic path, moving in a specific trajectory as \( t \) varies from \( -\pi/2 \) to \( \pi/2 \).

Understanding the role of the position vector means seeing how different components (in this case, sine and cosine functions) predictably change as \( t \) changes. A clear grasp of this is needed to effectively map how the force vector interacts with space through the given path.