Problem 28 - Thomas Calculus

Question

Work Find the work done by the gradient of \(f(x, y)=(x+y)^{2}\) counterclockwise around the circle \(x^{2}+y^{2}=4\) from \((2,0)\) to itself.

Step-by-Step Solution

Answer

The work done is zero.

Step 1: Identify the vector field

The gradient of a scalar function is a vector field. Here, we need to find the gradient of the function \(f(x, y) = (x+y)^2\). The gradient \(abla f\) can be calculated as the vector \(\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle\).

Step 2: Calculate the gradient

Calculate the partial derivative of \(f(x, y) = (x+y)^2\) with respect to \(x\): \(\frac{\partial f}{\partial x} = 2(x+y)\). Calculate the partial derivative with respect to \(y\): \(\frac{\partial f}{\partial y} = 2(x+y)\). Hence, the gradient is \(abla f = \langle 2(x+y), 2(x+y) \rangle\).

Step 3: Parameterize the circle

The path of integration is around the circle defined by \(x^2 + y^2 = 4\). We can parameterize this circle in terms of \(t\) from \(0\) to \(2\pi\): \(x = 2\cos(t)\), \(y = 2\sin(t)\).

Step 4: Compute the line integral of the gradient

The work done by the gradient field is given by the line integral \(\int_C abla f \cdot d\mathbf{r}\), where \(d\mathbf{r} = \langle -2\sin(t), 2\cos(t) \rangle dt\). Thus, we compute: \(\int_0^{2\pi} \langle 2(x+y), 2(x+y) \rangle \cdot \langle -2\sin(t), 2\cos(t) \rangle dt\).

Step 5: Evaluate the line integral

Substitute \(x = 2\cos(t)\) and \(y = 2\sin(t)\) into the gradient: \(2(x+y) = 4(\cos(t)+\sin(t))\). The dot product becomes: \(8(\cos(t)+\sin(t))(-2\sin(t), 2\cos(t))\). Integrate this expression over \(t\) from \(0\) to \(2\pi\).

Step 6: Solve the integral

Evaluate \(\int_0^{2\pi} 8(-2\sin(t)\cos(t) + 2\cos(t)\sin(t)) \, dt\). The integrand simplifies to zero since \(-2\sin(t)\cos(t) + 2\cos(t)\sin(t) = 0\). Thus, the integral is zero.

Key Concepts

Gradient

In vector calculus, the term "gradient" is crucial to understanding how functions can change. The gradient of a real-valued function measures the rate and direction of change.
For a function like \(f(x, y) = (x+y)^2\), the gradient is represented as a vector field. This field gives us insight into the steepness and direction of the function's increase at any point.
The components of the gradient vector are the partial derivatives of the function with respect to its variables. In our case,

• \( \frac{\partial f}{\partial x} = 2(x+y) \)
• \( \frac{\partial f}{\partial y} = 2(x+y) \)

This shows that the gradient is \( abla f = \langle 2(x+y), 2(x+y) \rangle \). The gradient points in the direction where the function increases fastest, and its magnitude is the rate of increase.

Line Integral

The concept of the line integral extends the idea of integration to functions along curves. It is widely used in physics to compute work done by a force along a path.
In this exercise, we're determining the work done by a gradient field along a circle. The line integral has the form: \[ \int_C abla f \cdot d\mathbf{r} \] where \( abla f \) is the gradient of the function and \( d\mathbf{r} \) is an infinitesimal piece of the path.
This integral essentially sums up all the little bits of work done as you traverse along the curve \(C\). If the vector field is conservative, as is the case with gradient fields, the work done is path-independent and solely relies on the initial and final points, often resulting in zero if the path is closed.

Parameterization

Parameterization is a technique in vector calculus used to express a curve as a set of equations dependent on a parameter, often simplifying calculations, especially for curves like circles.
Here, we parameterized the circle described by the equation \(x^2 + y^2 = 4\) to make integration manageable. By letting

• \(x = 2\cos(t)\)
• \(y = 2\sin(t)\)

where \(t\) ranges from 0 to \(2\pi\), we can describe every point on the circle as \(t\) varies.
This parameterization helps in expressing both the position \( \mathbf{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \) and the differential element \( d\mathbf{r} = \langle -2\sin(t), 2\cos(t) \rangle \, dt \), aiding in the line integral computation.

Circle

The circle is a fundamental shape in geometry that can be defined as all points equidistant from a center point.
For this exercise, the circle is given by the equation \(x^2 + y^2 = 4\), which represents a circle of radius 2 centered at the origin \((0,0)\).
When dealing with vector calculus and line integrals, understanding the geometry of the path, like circles, is crucial. These paths can often be parameterized, allowing functions to be evaluated over the entire shape. In this instance, we use the parameterization to simplify calculations with trigonometric functions

• \(x = 2\cos(t)\)
• \(y = 2\sin(t)\)

This not only assists with integrating over the curve but also visualizing the motion around the circle, starting and returning to the same point \((2,0)\).