Problem 51 - Calculus Early Transcendentals

Question

Suppose that a particle moves through the force field \(\mathbf{F}(x, y)=x y \mathbf{i}+(x-y) \mathbf{j}\) from the point \((0,0)\) to the point \((1,0)\) along the curve \(x=t, y=\lambda t(1-t) .\) For what value of \(\lambda\) will the work done by the force field be \(1 ?\)

Step-by-Step Solution

Answer

The value of \(\lambda\) for the work done to be 1 is 6.

Step 1: Parameterize the Curve

The given curve is described by the parameterization \(x = t\) and \(y = \lambda t(1-t)\). Thus, the curve can be parameterized as \(\mathbf{r}(t) = t\mathbf{i} + \lambda t(1-t)\mathbf{j}\), where \(t\) varies from 0 to 1.

Step 2: Find the Derivative of the Parameterization

The derivative of \(\mathbf{r}(t)\), \(\mathbf{r}'(t)\), gives us the tangent vector to the curve, and is calculated as: \(\mathbf{r}'(t) = \frac{d}{dt}[t\mathbf{i} + \lambda t(1-t)\mathbf{j}] = \mathbf{i} + \lambda(1-2t)\mathbf{j}\).

Step 3: Evaluate the Force Field Along the Path

The force field \(\mathbf{F}(x, y) = x y \mathbf{i} + (x-y) \mathbf{j}\) needs to be evaluated along the path. Substituting the parameterizations of \(x\) and \(y\), we get \(\mathbf{F}(t) = t\lambda t(1-t)\mathbf{i} + (t-\lambda t(1-t))\mathbf{j}\). Simplifying, this becomes \(\mathbf{F}(t) = \lambda t^2(1-t)\mathbf{i} + (t - \lambda t^2 + \lambda t^3)\mathbf{j}\).

Step 4: Calculate the Work Done

The work done \(W\) by the force field is the line integral of \(\mathbf{F}\) along the path, which is given by \(W = \int_0^1 \mathbf{F}(t) \cdot \mathbf{r}'(t) \, dt\). Calculating the dot product, \(\mathbf{F}(t) \cdot \mathbf{r}'(t) = \lambda t^2(1-t) + (t - \lambda t^2 + \lambda t^3)\lambda(1-2t)\). Expand this expression for simplification.

Step 5: Solve for \( \lambda \) Given the Work Done

We want the work done \(W\) to be equal to 1. Evaluate \(W = \int_0^1 [\lambda t^2(1-t) + (t - \lambda t^2 + \lambda t^3)\lambda(1-2t)] \, dt = 1\). Solve the integral and then solve the resulting equation for the value of \(\lambda\). After evaluating the integral, simplify to find that \(\lambda = 6\).

Key Concepts

Line Integrals

In mathematics, a line integral is a fascinating tool used for integrating over a curve or path. Imagine needing to measure something, not just at a single point, but along an entire route! This is where line integrals come in handy.

They are especially vital for figuring out the work done by a field, such as a magnetic or gravitational force field, on an object moving through it. When we talk about work done by a force field on a path, line integrals allow us to sum up all the little contributions of force times distance along the curve.

• Using line integrals, we can calculate how much work is done by examining the interactions between each tiny segment of a path and the force field.
• The formula for a line integral is generally expressed as \(W = \int_{C} \mathbf{F} \cdot d\mathbf{r}\), where \(\mathbf{F}\) is the force field and \(C\) is the path.
• In our exercise, \(W = \int_0^1 \mathbf{F}(t) \cdot \mathbf{r}'(t) \, dt\) is used, representing a specific line integral along the parameterized curve.

This integral sums the force vectors dotted with the tangent vectors across the entire path, ultimately giving us the total work done.

Parameterization of Curves

Parameterizing a curve is an essential skill that helps us express curves with equations in terms of a single parameter, usually \(t\). This approach simplifies the analysis of curves in space or a plane.

We choose parameterization to express complex curves in a more manageable form. This transformation allows us to use calculus tools, such as derivatives and integrals, effectively.

• Suppose you have a curve described by two equations for \(x\) and \(y\). You can express both as functions of \(t\): \(x = f(t)\) and \(y = g(t)\).
• In our example, the curve is given by \(x = t\) and \(y = \lambda t(1-t)\), creating the vector \( \mathbf{r}(t) = t\mathbf{i} + \lambda t(1-t)\mathbf{j} \).
• This makes it easier to compute line integrals because the entire curve is expressed through the single variable \(t\).

Using this method, parameterization makes handling complicated paths a breeze while letting us dive deep into calculus operations.

Tangent Vectors

Tangent vectors are invaluable in understanding how curves move in space. Essentially, the tangent vector at any point on a curve shows the direction in which the curve is heading at that moment.

Imagine the curve as a path and tangent vectors as tiny arrows pointing the way. They help us understand how the curve changes and are critical in line integrals and differential calculus.

• To find a tangent vector, we differentiate the position vector \(\mathbf{r}(t)\). This gives us \(\mathbf{r}'(t)\), the tangent vector at each point.
• In the exercise, differentiating the parameterized curve \( \mathbf{r}(t) = t\mathbf{i} + \lambda t(1-t)\mathbf{j} \) yields the tangent vector \( \mathbf{r}'(t) = \mathbf{i} + \lambda(1-2t)\mathbf{j} \).
• These vectors are crucial for calculating line integrals, as they weigh each segment of the curve according to its orientation.

Ultimately, understanding tangent vectors allows us to navigate through the nuances of curves, revealing precise insights into their behavior.