Problem 41 - Calculus Early Transcendentals

Question

Find the work done by the force field \(\mathbf{F}\) on a particle that moves along the curve \(C\). $$ \begin{array}{l}{\mathbf{F}(x, y)=x y \mathbf{i}+x^{2} \mathbf{j}} \\ {C: x=y^{2} \text { from }(0,0) \text { to }(1,1)}\end{array} $$

Step-by-Step Solution

Answer

The work done is \( \frac{3}{5} \).

Step 1: Parameterize the curve

The curve is given by \( x = y^2 \). To parameterize it, let's use \( t \) as the parameter. We can set \( y = t \), which implies that \( x = t^2 \). Therefore, the parameterization of the curve \( C \) becomes \( \mathbf{r}(t) = t^2 \mathbf{i} + t \mathbf{j} \) with \( t \) ranging from 0 to 1.

Step 2: Express the differential of the parameterization

To find the work done, we need \( d\mathbf{r} \), the differential of the position vector. Differentiating \( \mathbf{r}(t) = t^2 \mathbf{i} + t \mathbf{j} \), we obtain \( d\mathbf{r} = (2t \mathbf{i} + \mathbf{j}) dt \). This differential will be used in the line integral.

Step 3: Substitute the parameterized values into the force field

Substitute \( x = t^2 \) and \( y = t \) into the force field \( \mathbf{F}(x, y) = xy \mathbf{i} + x^2 \mathbf{j} \). This gives \( \mathbf{F}(t) = (t^2 \cdot t) \mathbf{i} + (t^2)^2 \mathbf{j} = t^3 \mathbf{i} + t^4 \mathbf{j} \).

Step 4: Set up the line integral

The work done by the force field is given by the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Substitute the expressions we found earlier: \( \int_0^1 (t^3 \mathbf{i} + t^4 \mathbf{j}) \cdot (2t \mathbf{i} + \mathbf{j}) dt \).

Step 5: Compute the dot product inside the integral

Calculate the dot product: \((t^3 \mathbf{i} + t^4 \mathbf{j}) \cdot (2t \mathbf{i} + \mathbf{j}) = t^3 \cdot 2t + t^4 \cdot 1 = 2t^4 + t^4 = 3t^4\). Thus, the integral becomes \( \int_0^1 3t^4 \ dt \).

Step 6: Evaluate the integral

To evaluate \( \int_0^1 3t^4 \ dt \), calculate the antiderivative: \( \int 3t^4 \ dt = \left[ \frac{3}{5}t^5 \right]_0^1 \). This evaluates to \( \frac{3}{5}(1)^5 - \frac{3}{5}(0)^5 = \frac{3}{5} \).

Key Concepts

Parameterization of Curves

Imagine you are on a thrilling roller coaster ride, but instead of being locked on a fixed path, you could define the path using mathematical expressions. This is precisely the magic of parameterization! When dealing with vector fields, parameterizing curves helps us describe the path of a moving object or point on a curve using one or more parameters, typically denoted by the letter \( t \). This method simplifies complex curves by expressing them as functions of \( t \), allowing us to easily manipulate and analyze these equations.

In essence, parameterization converts a potentially tricky path into something smooth and manageable. For instance, consider the curve given by \( x = y^2 \). By setting \( y = t \) and therefore \( x = t^2 \), we have crafted a lovely parameterization: \( \mathbf{r}(t) = t^2 \mathbf{i} + t \mathbf{j} \). This representation empowers us to further utilize this parameterization in subsequent calculations, like finding work in vector fields.

Line Integrals

Think of a line integral like strolling along a path and measuring how the journey affects certain values, such as the accumulation of work. In vector calculus, line integrals involve integrating a vector field along a curve to find values like work done by a force. This method helps us calculate how the field acts along the path taken by an object.

The curve ##\( C \)## in our example extends from point (0,0) to (1,1) with the equation \( x = y^2 \). By parameterizing this path and calculating its differential, we gain the needed components for setting up a line integral: \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Keeping in mind the force field \( \mathbf{F}(x, y) = xy \mathbf{i} + x^2 \mathbf{j} \), we replace \( x \) and \( y \) with their parameterized counterparts. Then, we compute the integral by evaluating the dot product and integrating it over the specified range. Each of these steps helps us aggregate the impact of the force along the entire path of the curve.

Dot Product in Vector Calculus

The dot product, a fundamental concept in vector calculus, is like finding common ground between two sets of directions. It reveals how much one vector influences the direction of another, producing a scalar (a single number). In the context of line integrals, we calculate the dot product between the force field's vector and the differential path vector to assess how they interact along a curve.

In our example problem, once we replace \( x \) and \( y \) with their parameterized values, we have the vector field \( \mathbf{F}(t) = t^3 \mathbf{i} + t^4 \mathbf{j} \) and the path differential \( d\mathbf{r} = (2t \mathbf{i} + \mathbf{j}) dt \). The dot product of these expressions \((t^3 \mathbf{i} + t^4 \mathbf{j}) \cdot (2t \mathbf{i} + \mathbf{j})\) simplifies to \( 3t^4 \). By integrating this dot product over the interval from 0 to 1, we effectively calculate the total work done by the force field along the path.