Problem 34 - The Calculus with Analytic Geometry

Question

In Exercises 21 through 34 , find the total work done in moving an object along the given arc \(C\) if the motion is caused by the given force field. Assume the arc is measured in inches and the force is measured in pounds. \(\mathbf{F}(\boldsymbol{x}, y, z)=x \mathbf{i}+y \mathfrak{j}+(y z-x) \mathbf{k} ; C: \mathbf{R}(t)=2 t \mathbf{i}+t^{2} \mathbf{j}+4 t^{3} \mathbf{k}, 0 \leq t \leq 1\)

Step-by-Step Solution

Answer

The total work done is 2.5 inch-pounds.

Step 1: Parameterize the Path

The path is already parameterized by \(\boldsymbol{R}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + 4t^3 \mathbf{k}\), where \( 0 \leq t \leq 1 \).

Step 2: Calculate \( \frac{d\boldsymbol{R}}{dt} \)

Differentiate \(\boldsymbol{R}(t)\) with respect to \( t \): \[ \frac{d\boldsymbol{R}(t)}{dt} \ = \ 2 \mathbf{i} + 2t \mathbf{j} + 12t^2 \mathbf{k} \]

Step 3: Substitute \( \boldsymbol{R}(t) \) into the Force Field

Substitute \( x = 2t, \ y = t^2, \ z = 4t^3 \) into \( \boldsymbol{F}(x, y, z) \) to get: \[ \boldsymbol{F}(2t, t^2, 4t^3) = 2t \mathbf{i} + t^2 \mathbf{j} + (t^2 \times 4t^3 - 2t) \mathbf{k} = 2t \mathbf{i} + t^2 \mathbf{j} + (4t^5 - 2t) \mathbf{k} \]

Step 4: Find Dot Product \( \boldsymbol{F}(\boldsymbol{R}(t)) \cdot \frac{d\boldsymbol{R}(t)}{dt} \)

Calculate the dot product of \( \boldsymbol{F}(2t, t^2, 4t^3) \) and \( \frac{d\boldsymbol{R}(t)}{dt} \): \[ (2t \mathbf{i} + t^2 \mathbf{j} + (4t^5 - 2t) \mathbf{k}) \cdot (2 \mathbf{i} + 2t \mathbf{j} + 12t^2 \mathbf{k}) = 2t \times 2 + t^2 \times 2t + (4t^5 - 2t) \times 12t^2 \] Simplify: \[ 4t + 2t^3 + 48t^7 - 24t^3 = 4t - 22t^3 + 48t^7 \]

Step 5: Integrate the Dot Product

Integrate the dot product \( \boldsymbol{F}(\boldsymbol{R}(t)) \cdot \frac{d\boldsymbol{R}(t)}{dt} \) over the interval from 0 to 1: \[ \text{Work} = \int_{0}^{1} (4t - 22t^3 + 48t^7) \ dt \] \[ = \left[ 2t^2 - \frac{22t^4}{4} + \frac{48t^8}{8} \right]_0^1 \] Simplifying: \[ = (2 \times 1 - \frac{22}{4} \times 1 + \frac{48}{8} \times 1) - (0) \] \[ = (2 - 5.5 + 6) \] \[ = 2.5 \]

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that extends calculus to vector fields. While typical calculus deals with scalar functions, vector calculus focuses on functions that return vectors. These functions are often found in physics and engineering and are useful for describing physical phenomena such as gravitational fields, electric fields, and fluid flow.

Key concepts in vector calculus include:

• Gradient: Represents the rate and direction of change in a scalar field.
• Divergence: Measures the magnitude of a source or sink at a given point in a vector field.
• Curl: Describes the rotation or the twisting force within a vector field.

These operations help in investigating various behaviors of physical systems and in calculating important quantities like work done by a vector field.

Line Integral

A line integral is a type of integral that is used to find the work done by a force field in moving an object along a curve. Unlike a definite integral, which integrates over intervals on the real line, a line integral integrates over a curve in a vector field.

In mathematical terms, to compute the work done by a force field \( \boldsymbol{F}(\boldsymbol{x}, y, z) \) over a path \( C \) represented by a parameterization \( \boldsymbol{R}(t) \), you need to:

• Parameterize the curve \( C \).
• Calculate the derivative of the parameterization \( \frac{d \boldsymbol{R}(t)}{dt} \).
• Substitute the parameterized variables into the force field.
• Find the dot product \( \boldsymbol{F}(\boldsymbol{R}(t)) \bullet \frac{d \boldsymbol{R}(t)}{dt} \).
• Integrate the result over the given bounds.

The final integral gives you the total work done.

Force Field

A force field is a vector field that represents a distribution of force over a region of space. It means that every point within this field has a vector that indicates the direction and magnitude of the force at that point. Force fields are commonly used to model physical phenomena like gravitational attraction, electromagnetic fields, and fluid dynamics.

In our specific problem, the force field is given by \( \boldsymbol{F}( \boldsymbol{x}, y, z) = x \boldsymbol{i} + y \boldsymbol{j} + (yz - x) \boldsymbol{k} \). This tells us:

• Component along x-axis: The force in the direction of \(\boldsymbol{i} \) depends on the x-coordinate.
• Component along y-axis: The force in the direction of \(\boldsymbol{j} \) depends on the y-coordinate.
• Component along z-axis: The force in the direction of \(\boldsymbol{k} \) depends on a combination of y, z, and x-coordinates.

By evaluating this force field along a specific curve, we can determine the work done in moving an object through the field along that curve.

Parameterization

Parameterization is the process of expressing a curve or surface as a function of one or more parameters. It simplifies the computation of line integrals and other vector calculus operations by representing complex geometries in simpler mathematical terms.

In our exercise, the curve \( C \) is given by the parameterization \( \boldsymbol{R}(t) = 2t \boldsymbol{i} + t^2 \boldsymbol{j} + 4t^3 \boldsymbol{k} \) where \( 0 \leq t \leq 1 \). This means:

• For \( \boldsymbol{i} \):\( x = 2t \)
• For \( \boldsymbol{j} \): \( y = t^2 \)
• For \( \boldsymbol{k} \): \( z = 4t^3 \)

The parameter \( t \) runs from 0 to 1, tracing out the path of the curve. This parameterization is vital as it transforms the problem into a format where calculus operations become straightforward.