Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the work done by the force field \(\mathbf{F}\) in moving a particle along the curve \(C\). \(\mathbf{F}(x, y)=\left(x^{3}-y^{3}\right) \mathbf{i}+x y^{2} \mathbf{j} ; \quad C\) is the curve \(x=t^{2}\), \(y=t^{3},-1 \leq t \leq 0 .\)

Step-by-Step Solution

Answer

The work done by the force field is \( \frac{15}{44} \).

Step 1: Understand the Problem

We need to find the work done by the force field \( \mathbf{F}(x, y) = (x^3 - y^3) \mathbf{i} + xy^2 \mathbf{j} \) along the curve \( C \) given by the parametric equations \( x = t^2 \) and \( y = t^3 \) from \( t = -1 \) to \( t = 0 \). This involves calculating the line integral of \( \mathbf{F} \) along \( C \).

Step 2: Parametrize the Curve and Find Derivatives

With \( x = t^2 \) and \( y = t^3 \), differentiate to find \( \frac{dx}{dt} = 2t \) and \( \frac{dy}{dt} = 3t^2 \). The curve \( C \) is parametrized by \( t \) in the interval \([-1, 0]\).

Step 3: Substitute Parametric Equations into Force Field

Replace \( x \) and \( y \) in \( \mathbf{F}(x, y) \) with \( t^2 \) and \( t^3 \) respectively: \( \mathbf{F}(t) = ((t^2)^3 - (t^3)^3) \mathbf{i} + (t^2)(t^3)^2 \mathbf{j} \). Simplifying, \( \mathbf{F}(t) = (t^6 - t^9) \mathbf{i} + t^8 \mathbf{j} \).

Step 4: Calculate the Line Integral

The work done \( W \) is given by the integral \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \). With \( d\mathbf{r} = \frac{dx}{dt} \mathbf{i} + \frac{dy}{dt} \mathbf{j} \ dt = (2t \mathbf{i} + 3t^2 \mathbf{j}) dt \), the dot product \( \mathbf{F} \cdot d\mathbf{r} = (t^6 - t^9)(2t) + t^8(3t^2) \). Simplifying, \( \mathbf{F} \cdot d\mathbf{r} = 2t^7 - 2t^{10} + 3t^{10} \).

Step 5: Integrate over the Interval

Combine terms to get \( 2t^7 + t^{10} \). Integrate from \( t = -1 \) to \( t = 0 \): \( \int_{-1}^{0} (2t^7 + t^{10}) dt \). Perform the integration: \( \frac{2}{8}t^8 + \frac{1}{11}t^{11} \) evaluated from \(-1\) to \(0\). This results in \(0 - (-\frac{1}{4} - \frac{1}{11})\).

Step 6: Simplify and Compute the Work Done

Simplify the expression: \( \frac{1}{4} + \frac{1}{11} = \frac{11 + 4}{44} = \frac{15}{44} \). The work done by the force field is \( \frac{15}{44} \).

Key Concepts

Line Integral

In vector calculus, a line integral is a method used to find the work done by a force field along a given curve or path. To calculate a line integral, we need to evaluate the integral of a vector field along a certain path. This essentially summarizes how much the vector field \( \mathbf{F} \) "pushes" along the path \C\. Consider a vector field \( \mathbf{F}(x,y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \). The line integral along a path \C\ is given by:\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (P \, dx + Q \, dy) \]Where \(d\mathbf{r}\) is the differential path element along \C\. This expression forms a dot product typically integrating over parameter \(t\) if the curve is parametrized.Understanding line integrals is key to finding the work done in various physics applications, notably in electromagnetism or fluid dynamics. It quantifies how much of a vector field aligns with or acts along a path specified by a parameterization of the curve.

Parametric Equations

Parametric equations are crucial in expressing curves in a manner that simplifies calculations such as line integrals. Instead of describing a curve as a function of \(x\) and \(y\) (e.g., \(y=f(x)\)), parametric equations introduce a third variable, \(t\), called the parameter.For the given exercise, the curve \C\ is parametrized by:

• \(x = t^2\)
• \(y = t^3\)

Where \(t\) represents values ranging from \(-1\) to \(0\). This parameterization allows us to convert the expressions for \(dx\) and \(dy\) into terms of \(dt\), i.e., find derivatives \(\frac{dx}{dt} = 2t\) and \(\frac{dy}{dt} = 3t^2\). Thus enabling integration over a single variable, in this case, \(t\).Parameterizing a curve is a powerful approach as it can define complex curves that would be difficult or impossible to describe with simple \(y=f(x)\) relationships.

Vector Calculus

Vector calculus involves differentiation and integration of vector fields, and is critical for understanding physical fields involving quantities like velocity, force, and electromagnetism. It encompasses operations like divergence, gradient, and curl. However, in the context of work done by a force field, the focus is on understanding:

• Vector functions: Mathematical expressions defining vectors in space, e.g., \(\mathbf{F}(x, y)\).
• Vector fields: Representations of how vectors vary over a space, key in defining force fields influencing motion.
• Dot products: Measure how much of one vector aligns with another, crucial for concepts such as work done by forces.

In this exercise, vector calculus provides the framework for algebraically handling and manipulating vectors. It allows for describing the complete movement along the curve, evaluating how the force field influences or "works on" any underlying path parameterized.

Force Field Calculation

The calculation of the work done by a force field often stems from a vector field representation of that force, like \(\mathbf{F}(x, y) = (x^3 - y^3) \mathbf{i} + xy^2 \mathbf{j}\). Once the parametric equations for a path are determined, substitute these values into the force field equation to express the field as a function of parameter \(t\).In the example, substituting \(x = t^2\) and \(y = t^3\), the force field simplifies to:

• \( \mathbf{F}(t) = (t^6 - t^9) \mathbf{i} + t^8 \mathbf{j} \)

This representation permits us to determine how the force varies as the parameter \(t\) changes along the curve. Then, by integrating along the path, the total work done is discovered. This process is central in physics to calculate energy transfers and insights into dynamic systems driven by varying forces across a trajectory.