Question

Find the exact points on the circle \(x^{2}+y^{2}=2\) at which the field \(\mathbf{F}=\langle f, g\rangle=\left\langle x^{2}, y\right\rangle\) switches from pointing inward to pointing outward on the circle, or vice versa.

Short answer

Question: Determine the exact points on the circle with equation \(x^2+y^2=2\) where the vector field \(\mathbf{F}=\langle x^2,y\rangle\) changes direction from pointing inward to pointing outward, or vice versa. Answer: The exact points on the circle at which the field switches direction are \((\pm\sqrt{2}, 0)\), \((0,\pm\sqrt{2})\), and \(\left(\pm\frac{1}{\sqrt{2}}, \pm\frac{1}{\sqrt{2}}\right)\).

Step-by-step solution

Parametrize the circle and find its tangent vectors

To parametrize the given circle \(x^2+y^2=2\), we can use a trigonometric parametrization with radius \(\sqrt{2}\). We obtain the parametrization: \[x(\theta) = \sqrt{2}\cos\theta\] \[y(\theta) = \sqrt{2}\sin\theta\] Now, we can find the tangent vectors \(\mathbf{T}(\theta)\) by differentiating the parametric equations with respect to \(\theta\): \[\frac{dx}{d\theta} = -\sqrt{2}\sin\theta\] \[\frac{dy}{d\theta} = \sqrt{2}\cos\theta\] So, the tangent vector is: \[\mathbf{T}(\theta)=\left\langle -\sqrt{2}\sin\theta, \sqrt{2}\cos\theta\right\rangle\]

Find the directional derivative of the vector field F along the tangent vectors

To find the directional derivative of the vector field \(\mathbf{F}\) along the tangent vectors, we need to compute the dot product \(\mathbf{F}\cdot\mathbf{T}\) at point \((x(\theta), y(\theta))\). We have: \[\mathbf{F}(\theta)=\langle (x(\theta))^2, y(\theta)\rangle = \langle 2\cos^2\theta, \sqrt{2}\sin\theta\rangle\] So the dot product is: \[\mathbf{F}\cdot\mathbf{T} = 2\cos^2\theta(-\sqrt{2}\sin\theta) + (\sqrt{2}\sin\theta)(\sqrt{2}\cos\theta) = -2\sqrt{2}\sin\theta\cos^2\theta + 2\sin\theta\cos\theta\]

Determine the points where the directional derivative changes its sign

The directional derivative changes its sign when it becomes zero. Therefore, we need to find the value of \(\theta\) for which the expression for the dot product is equal to zero: \[-2\sqrt{2}\sin\theta\cos^2\theta + 2\sin\theta\cos\theta = 0\] We can factor out a term \(2\sin\theta\cos\theta\) from the expression: \[2\sin\theta\cos\theta(-\sqrt{2}\cos\theta + 1) = 0\] This equation is satisfied when either \(\sin\theta\cos\theta=0\), or \((-\sqrt{2}\cos\theta+1)=0\). We can solve these sub-equations for \(\theta\): Case 1: \(\sin\theta\cos\theta=0\) This is true when \(\sin\theta=0\) or \(\cos\theta=0\). The solutions are \(\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\). Case 2: \(-\sqrt{2}\cos\theta+1=0\) This gives us \(\cos\theta = \frac{1}{\sqrt{2}}\), and the solutions are \(\theta = \frac{\pi}{4}, \frac{5\pi}{4}\). Now, we can calculate the exact points on the circle where the field switches direction: 1. At \(\theta = 0, \pi\), we have the points \((\pm\sqrt{2}, 0)\). 2. At \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\), we have the points \((0, \pm\sqrt{2})\). 3. At \(\theta = \frac{\pi}{4}, \frac{5\pi}{4}\), we have the points \(\left(\pm\frac{1}{\sqrt{2}}, \pm\frac{1}{\sqrt{2}}\right)\). So, the exact points on the circle at which the field switches direction are \((\pm\sqrt{2}, 0)\), \((0,\pm\sqrt{2})\), and \(\left(\pm\frac{1}{\sqrt{2}}, \pm\frac{1}{\sqrt{2}}\right)\).

Key concepts

Parametrizing Curves

In the world of mathematics, especially in calculus and geometry, parametrizing curves is an essential concept. It involves expressing a curve in a coordinate plane or space using a parameter, typically denoted as \( \theta \) or \( t \). For instance, a circle can be conveniently parametrized using trigonometric functions.

To express the circle \( x^2 + y^2 = 2 \), we can use the radius, which is \( \sqrt{2} \), and trigonometric identities. The transformation is expressed as:

• \( x(\theta) = \sqrt{2}\cos\theta \)
• \( y(\theta) = \sqrt{2}\sin\theta \)

These are the parametric equations for the circle, allowing us to describe every point on the circle by simply varying \( \theta \). Parametrization simplifies the process of analyzing the geometric properties of curves and aids in performing calculus-based operations on these curves.

Vector Fields

Vector fields are crucial in both physics and mathematics because they specify a vector to every point in a subset of space. A vector field is denoted as \( \mathbf{F} = \langle f, g \rangle \), where \( f \) and \( g \) are functions describing the vectors' components in the field.

In the given exercise, the vector field \( \mathbf{F} = \langle x^2, y \rangle \) assigns a vector based on the coordinates \( (x, y) \). Vector fields help us understand the direction and magnitude of quantities, such as force or velocity, at different points in space.

By evaluating the vector field at points on the circle, we can determine how these vectors interact with the curve's orientation. Understanding vector fields is critical for analyzing systems affected by various forces or currents, both in natural and artificial environments.

Dot Product

The dot product is a fundamental operation in vector algebra. It measures how much one vector extends in the direction of another and is crucial for finding angles and projections between vectors. The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In this exercise, the dot product helps determine where the vector field \( \mathbf{F} \) changes direction along the circle's tangent.

By computing \( \mathbf{F} \cdot \mathbf{T} \), where \( \mathbf{T} \) is the circle's tangent vector, we find points of directional change. If the dot product equals zero, the vectors are perpendicular, indicating a switch from inward to outward direction (or vice versa). This is foundational in understanding vector interactions on curves.

Differential Calculus

Differential calculus deals with the study of how functions change using derivatives. It plays a pivotal role in understanding dynamics in mathematics and the physical sciences. The derivative of a function provides vital information about its rate of change and is depicted as \( \frac{df}{dt} \) or \( f'(x) \).

To analyze the circle \( x^2 + y^2 = 2 \), we first derive tangent vectors by differentiating parametric equations. For example, \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) unveil the directions where the curve changes with respect to \( \theta \). This technique allows us to explore the geometry and orientation of curves deeper.

These tools enable mathematicians and scientists to dissect small local changes in functions, aiding in constructing more complex, global understandings of various phenomena, from motion to growth processes.