Problem 217 - Advanced Calculus

Question

a) Find the integral of the vector field \(\mathrm{F}^{-}(\mathrm{x}, \mathrm{y})=\left[\left(\mathrm{x} / \mathrm{r}^{3}\right),\left(\mathrm{y} / \mathrm{r}^{3}\right)\right.\), where \(\mathrm{r}=\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{1 / 2}\), along the curve \(\mathrm{C}^{\rightarrow}(\mathrm{t})=\left(\mathrm{e}^{\mathrm{t}} \cos \mathrm{t}\right.\), \(\mathrm{e}^{\mathrm{t}}\) sin \(\mathrm{t}\) ) from the point \((1,0)\) to the point \(\left(\mathrm{e}^{2 \pi}, 0\right)\). b) Does \(\mathrm{F}^{-}\) admit a potential function? If so, compute the integral in a) by this method.

Step-by-Step Solution

Answer

In summary, the integral of the given vector field along the curve is \( -(e^{-2\pi} - 1)\). The vector field F admits a potential function \( \varphi(x, y) =-\frac{1}{2r^2} + c\), where c is a constant. Using this potential function, we also find the same result for the integral.

Step 1: Compute the derivative of the curve C(t)

The curve C(t) is given by the parametric equation: \(C(t) = \left( e^t \cos t, e^t \sin t \right)\) Let's compute the derivative of C(t) with respect to t: \(C'(t) = \left( \frac{d}{dt}(e^t \cos t), \frac{d}{dt}(e^t \sin t) \right) = \left( e^t(\cos t - \sin t), e^t( \cos t + \sin t) \right)\)

Step 2: Plug in the parameterization of C(t) into F(x, y)

The vector field F is given by F(x, y) = \(\left(\frac{x}{r^3}, \frac{y}{r^3}\right),\) where r = \(\sqrt{x^2 + y^2}\) Now, let's plug in the parameterization of C(t) into F(x, y): \(F(C(t)) = \left( \frac{e^t \cos t}{(e^t)^3}, \frac{e^t \sin t}{(e^t)^3} \right) = \left( \frac{\cos t}{e^{2t}}, \frac{\sin t}{e^{2t}} \right)\)

Step 3: Integrate the dot product of F(C(t)) and C'(t)

Now we need to compute the integral of the dot product of F(C(t)) and C'(t) with respect to t from 0 to 2π: \(\int_0^{2\pi} \left\langle \frac{\cos t}{e^{2t}}, \frac{\sin t}{e^{2t}} \right\rangle \cdot \left\langle e^t(\cos t - \sin t), e^t( \cos t + \sin t) \right\rangle dt\) =\(\int_0^{2\pi} \left( \frac{\cos^2 t}{e^t} - \frac{\sin t \cos t}{e^t} + \frac{\sin t \cos t}{e^t} + \frac{\sin^2 t}{e^t}\right) dt\) =\(\int_0^{2\pi} \frac{1}{e^t}dt\) \(= \left[-\frac{1}{e^t}\right]_0^{2\pi} = -(e^{-2\pi} - 1)\) #b) Find the potential function and compute the integral using this method #

Step 1: Determine if the vector field F admits a potential function

To see whether F admits a potential function, we check if the partial derivatives of F satisfy the following condition: \(\frac{\partial}{\partial y} \left( \frac{x}{r^3} \right) = \frac{\partial}{\partial x} \left( \frac{y}{r^3} \right)\) If this condition holds, then F is conservative and admits a potential function. Following computations: \(\frac{\partial}{\partial y} \left( \frac{x}{r^3} \right) = \frac{-3x^2y}{r^7}\) \(\frac{\partial}{\partial x} \left( \frac{y}{r^3} \right) = \frac{-3x^2y}{r^7}\) Both partial derivatives are equal, thus F admits a potential function.

Step 2: Find the potential function

Since F is conservative and admits a potential function, we can find such a function ϕ: \(\nabla \varphi(x,y) = \left( \frac{x}{r^3}, \frac{y}{r^3} \right)\) From this gradient, we can compute the potential function as follows: \(\varphi(x, y) = \int \frac{x}{r^3} dx = -\frac{1}{2r^2} + \varphi_1(y)\) \(\varphi(x, y) = \int \frac{y}{r^3} dy = -\frac{1}{2r^2} + \varphi_2(x)\) Therefore, the potential function is ϕ(x, y) = \( -\frac{1}{2r^2} + c\), where c is a constant.

Step 3: Compute the integral using the potential function

Now we can compute the integral in part a) using the potential function: \(\int_C \bold{F} \cdot d \bold{r} = \varphi(\mathrm{e}^{2 \pi}, 0) - \varphi(1, 0)\) \(= -\frac{1}{2\left((\mathrm{e}^{2 \pi})^2\right)} + c - \left(-\frac{1}{2((1)^2)} + c \right)\) \(= -(e^{-2\pi} - 1)\) This is the same result obtained in part a).

Key Concepts

Line Integral of Vector Fields

The line integral of a vector field is a fundamental tool in multivariable calculus, particularly useful when trying to measure the cumulative effect of a field's force along a curve. A vector field assigns a vector to every point in space, representing something like force, velocity, or acceleration. When calculating the line integral, one essentially sums up the components of the vector field that are directed along a particular path or curve.

Imagine walking through a hilly terrain with varying wind speeds and directions. The line integral can tell us how much the wind aids or resists our progress along a chosen path. Mathematically, we compute this by taking a dot product of the vector field with the derivative of the curve, which gives us a scalar function. We then integrate this function over the interval corresponding to our path.

In our exercise, we consider the vector field \( \mathbf{F} \) and a curve \( \mathbf{C}(t) \) parametrized by \( t \). The line integral becomes a matter of calculating the integral of \( \mathbf{F}(\mathbf{C}(t)) \) dotted with \( \mathbf{C}'(t) \) from \(t = 0\) to \(t = 2\pi\), yielding the total effect of the field along the curve. This calculation involves trigonometric identities and the properties of the exponential function to simplify and evaluate the integral.

Potential Function of Vector Fields

A potential function for a vector field is akin to finding a 'height map' or 'energy landscape' from which the vector field itself could be seen as the slopes at each point. If a vector field \( \mathbf{F} \) has a potential function \( \varphi \), this implies that \( \mathbf{F} \) can be expressed as the gradient of \( \varphi \). This is a powerful notion as it helps simplify many calculations, especially line integrals, by moving from a path-dependent process to a much easier evaluation at endpoints.

For our vector field, a potential function is found by integrating the components of \( \mathbf{F} \) with respect to their respective variables, ensuring that the resulting functions are compatible — their mixed second-order partial derivatives should match. In the solution provided, careful integration of \( \mathbf{F} \) yields a potential function \( \varphi(x, y) = -\frac{1}{2r^2} + c \), where \( c \) is a constant and \( r \) is the radius function of \( x \) and \( y \). This discovery offers an alternative method for computing line integrals, as shown in the solution.

Conservative Vector Fields

Conservative vector fields are special because knowing them at one point allows you to predict their behavior throughout the space. Much like conserving energy in physics, these vector fields have a 'memoryless' quality — the path taken between two points doesn't affect the work done or energy transferred.

To establish whether a vector field is conservative, one checks the equality of its mixed partial derivatives. If these are equal, we're given a green light; the vector field is indeed conservative, and this unlocks the potential function method for simplifying the cumbersome line integral calculation. In particular, it lets us evaluate the line integral merely by taking the difference of the potential function at the endpoints of our curve.

The exercise implements this condition and shows that the given vector field is conservative. Conveniently, the path independence of conservative fields confirms our previous line integral result, emphasizing that the outcome only depends on the start and endpoints — not on the particularities of the journey we undertake within this mathematical landscape.