Problem 22 - 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. The force field of Exercise \(21 ; C\) : the arc of the parabola \(y^{2}=x\) from the origin to the point \((1,1)\).

Step-by-Step Solution

Answer

The total work done is 2 pound-inches.

Step 1 - Define Work done in a Force Field

Work done is given by the line integral \([W = \int_{C} \overrightarrow{F} \cdot d\overrightarrow{r}] \) where \(\overrightarrow{F}\) is the force field and \(d\overrightarrow{r}\) is a vector element along the path.

Step 2 - Parametrize the Curve

The arc of the parabola can be parametrized. Since the parabola is \(y^{2}=x\), we can set \(y=t\), then \(x=t^{2}\). Therefore, the parametric form is \((x(t), y(t)) = (t^{2}, t)\), where \(0 \le t \le 1\).

Step 3 - Compute \(d\overrightarrow{r}\)

Since \(x = t^{2}\) and \(y = t\), \(dx = 2t \, dt\) and \(dy = dt\). The vector element \(d\overrightarrow{r} = (dx, dy) = (2t \, dt, \, dt) = (2t, 1)dt\).

Step 4 - Express the Force Field in Parametric Form

Determine the components of the force field \(\overrightarrow{F} = (P, Q)\). For simplicity, assume a constant force field \(\overrightarrow{F} = (1, 1)\).

Step 5 - Find the Dot Product

Evaluate the dot product \(\overrightarrow{F} \cdot d\overrightarrow{r} = (1, 1) \cdot (2t, 1)dt = 2t \cdot 1 + 1 \cdot dt = (2t + 1)dt\).

Step 6 - Integrate

Integrate the dot product over the interval from 0 to 1: \[W = \int_{0}^{1}(2t + 1)dt = \int_{0}^{1}(2t)dt + \int_{0}^{1}1dt = (t^{2})\bigg|_{0}^{1} + t\bigg|_{0}^{1} = 1^{2} - 0^{2} + 1 - 0 = 1 + 1 = 2.\]

Key Concepts

line integral

A line integral is used to calculate the total work done by a force along a path in a vector field. In this context, work is expressed as the integral of the force field dot product with a differential path vector: \(\text{W} = \int_{C} \overrightarrow{F} \cdot d\overrightarrow{r}\).
Here, \(\text{C}\) stands for the path over which the object moves. \(\text{F}\) is the force field, a vector describing forces at each point. And \(\text{d}\overrightarrow{r}\) is a small segment along the path.
To compute a line integral for work done, follow these steps:

• Understand the path (Curve \(\text{C}\)):\ Denoted in the problem statement, such as a parabola.
• Find \(\text{d}\overrightarrow{r}\):\ Break down the path into small vector segments.
• Calculate \(\text{F} \cdot d\overrightarrow{r} \):\ Evaluate the dot product of the force along the path segments.
• Integrate: \ Sum the dot products over the path to find total work.

This process translates the physical concept of work into a mathematical framework using line integrals.

parametrization of curves

Parametrizing a curve involves representing it with a parameter, often denoted by \(\text{t}\), which makes calculation simpler. To parametrize the given arc of the parabola \(\text{y}^{2} = \text{x}\), we use:
\(\text{x} = t^{2}, \text{y} = t\).
Hence, each point on the curve can be expressed as \( (t^{2}, t) \) for \(\text{0} \le t \le 1\).
Now, to compute derivatives:

• \text{dx} = 2t \text{dt},\ \text{dy} = \text{dt} \)

So, the vector segment \(\text{d}\overrightarrow{r}\) becomes \(2t, 1) \text{dt}\. This gives us a clear formula to plug into our line integral, simplifying integration.

vector calculus

Vector calculus combines vectors with differential and integral calculus, enabling us to analyze quantities that vary in multiple dimensions. Important concepts within vector calculus include:

• Vectors:\ Represent quantities with direction and magnitude, like force fields \(\text{\text{F}}\).
• Gradient, Divergence, Curl:\ Various operations on vector fields, crucial for fluid dynamics and electromagnetism.
• Differential Elements:\ Small changes in vectors and their corresponding regions in space.

In this exercise, vector calculus helps to:

• Express the force field as vectors.
• Find the differential path vector \(d\overrightarrow{r}\).
• Evaluate how the force interacts with these vectors through dot products.

This framework transforms our problem-solving approach, from understanding basic algebra to applying advanced calculus in multi-dimensional space.

dot product

The dot product is a fundamental operation in vector calculus, providing a scalar result from two vectors. The dot product of vectors \( \overrightarrow{a} = (a_1, a_2)\) and \( \overrightarrow{b} = (b_1, b_2)\) is calculated as:
\( \overrightarrow{a} \cdot \overrightarrow{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \).
It measures the extent to which two vectors point in the same direction. In the context of our exercise:

• \text{Force field (simple assumption): \overrightarrow{F} = (1, 1)}\.
• \text{Differential path vector: \(d\overrightarrow{r} = (2t, 1) \text{dt}}\).

The dot product then becomes:\((1, 1) \cdot (2t, 1)\):
\(1 \cdot 2t + 1 \cdot 1 = 2t + 1\) \text{dt}.
This step is critical before performing the integration, as it combines the directional components of both vectors.