Question

Given the force field \(\mathbf{F},\) find the work required to move an object on the given oriented curve. \(\mathbf{F}=\langle y, x\rangle\) on the parabola \(y=2 x^{2}\) from (0,0) to (2,8)

Short answer

The work required to move an object along the curve \(y = 2x^2\) from point (0,0) to point (2,8) under the force field \(\mathbf{F} = \langle y, x \rangle\) is 16 Joules.

Step-by-step solution

Parametrize the curve

We are given a curve defined by the equation \(y = 2x^2\). We can parameterize this curve using a parameter \(t\) as follows: $$ \mathbf{r}(t) = \langle t, 2t^2 \rangle $$ The range of \(t\) should be such that it covers the interval from (0,0) to (2,8). For that, we need \(t\) to range from 0 to 2.

Compute the differential of the position vector

Next, we need to compute the differential of the position vector \(\mathbf{r}(t)\). To do that, we need to differentiate \(\mathbf{r}(t)\) w.r.t \(t\): $$ \frac{d\mathbf{r}}{dt} = \langle 1, 4t \rangle $$ Let's denote the differential vector as $$ d\mathbf{r}= \langle 1, 4t\rangle dt $$

Substitute the vector values into the force field

The force field is given by \(\mathbf{F} = \langle y, x \rangle\). We can substitute the values of \(x\) and \(y\) from the parametrization, \(\mathbf{r}(t) = \langle t, 2t^2 \rangle\), as follows: $$ \mathbf{F} = \langle 2t^2, t \rangle $$

Compute the dot product of the force field and the differential vector

Now, we need to compute the dot product of the force field \(\mathbf{F}\) an dthe differential vector \(d\mathbf{r}\). This will give us the integrand we need to evaluate the line integral: $$ \mathbf{F} \cdot d\mathbf{r} = \langle 2t^2, t \rangle \cdot \langle 1, 4t \rangle dt = (2t^2)(1)+(t)(4t)dt = 2t^2+4t^2 dt = 6t^2 dt $$

Evaluate the line integral to find the work done

Finally, we need to evaluate the line integral to find the work done by the force field \(\mathbf{F}\) as it moves the object along the curve: $$ W = \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{t=0}^{t=2} 6t^2 dt $$ We can compute this integral using the fundamental theorem of calculus: $$ W = \left[2t^3\right]_0^2 = 2(2^3)-2(0^3) = 16 J $$ So, the work required to move an object on the given oriented curve is 16 J (Joules).

Key concepts

Line Integrals

Line integrals allow us to compute the total effect of a vector field along a curve path. They are essential in physics to calculate the work done by a force field when moving an object along a specific pathway.
In the context of force fields, a line integral sums up the effects of the field at each point along a curve. To compute a line integral, you integrate the dot product of the vector field and the curve's differential vector over the curve's path.
Key steps in evaluating a line integral include:

• Parameterizing the curve to determine a suitable path.
• Calculating the dot product between the vector field and the differential vector.
• Integrating the resultant expression over the defined parameter range.

Each of these steps is crucial to ensure that the computation accurately reflects the work done by the force along the curve.

Parameterization of Curves

The parameterization of curves is a technique used to express a curve using one or more parameters. In our example, the parabola given by the equation \( y = 2x^2 \) is parameterized using the variable \( t \).
The goal of parameterization is to represent each point on the curve as a function of a parameter. For our parabola, we set:

• \( x = t \)
• \( y = 2t^2 \)

This creates the vector function \( \mathbf{r}(t) = \langle t, 2t^2 \rangle \), with \( t \) ranging from 0 to 2.}
Parameterizing a curve simplifies the process of calculating integrals because it transforms the curve into a single variable expression, making it easier to evaluate moving forward.

Dot Product

The dot product (or scalar product) is a mathematical operation that takes two equal-length sequences of numbers and returns a single number. It's crucial when calculating line integrals in vector fields.
In the step-by-step solution, the dot product is calculated between the force field \( \mathbf{F} = \langle 2t^2, t \rangle \) and the differential vector \( d\mathbf{r} = \langle 1, 4t \rangle dt \).
The dot product is computed as follows:

• Multiply corresponding components together, \((2t^2 \cdot 1)\) and \((t \cdot 4t)\).
• Sum the results, giving \(2t^2 + 4t^2 = 6t^2\).

This product tells us how much of the force field's effect is aligned with the curve's path, which is essential in calculating the total work done over the curve.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation with integration, providing a manageable way to evaluate definite integrals. It's particularly useful in calculating line integrals.
In the given solution, once the dot product is integrated over the range of \( t \) from 0 to 2, the Fundamental Theorem of Calculus is applied:

• \( W = \int_{t=0}^{t=2} 6t^2 dt \).
• This integral simplifies to \( \left[2t^3\right]_0^2 \).
• Evaluating this expression at the bounds gives \( 16 J \) (Joules).

The theorem simplifies calculating the exact work done by the force field, as it moves the object along the designated path. It transforms the integration task into straightforward arithmetic, facilitating the process of determining the work involved.