Question

A circle, centered at the origin with a radius of 2 units, lies in the \(x y\) plane. Determine the unit vector in rectangular components that lies in the \(x y\) plane, is tangent to the circle at \((-\sqrt{3}, 1,0)\), and is in the general direction of increasing values of \(y\).

Short answer

Question: Find a unit vector in the \(xy\) plane that is tangent to the given circle with center (0, 0, 0) and radius 2 at the point \((-\sqrt{3}, 1, 0)\) such that the tangent vector has an increasing \(y\) direction. Answer: A possible unit vector is \(\left(\frac{1}{\sqrt{1^2 + k^2}}, \frac{k}{\sqrt{1^2 + k^2}}\right)\), where \(k>0\).

Step-by-step solution

Find the direction vector of the tangent at \((-\sqrt{3}, 1, 0)\).

To find the direction vector of the tangent, we first need to determine a point on the circle that is very close to the given point \((-\sqrt{3}, 1, 0)\). Let it be \((-\sqrt{3}+ dx, 1+ dy, 0)\). As the point lies on the circle with a radius of 2 units, it must satisfy the equation \((x^2 + y^2 = 4)\). So, \(((- \sqrt{3} + dx)^2 + (1 + dy)^2 = 4 \) We can now find the direction vector, \(\overrightarrow{d}\), as: \(\overrightarrow{d} = ((-\sqrt{3} + dx ) - (-\sqrt{3}), (1 + dy) -1) = (dx, dy)\)

Use the equation of the circle for a point close to \((-\sqrt{3}, 1, 0)\).

As derived in Step 1: \(((- \sqrt{3} + dx)^2 + (1 + dy)^2 = 4\) Taking the derivative with respect to \(x\), we get: \(2(-\sqrt{3} + dx)\cdot1 + 2(1 + dy) \cdot \frac{dy}{dx} = 0\) This gives us the value of slope \(\frac{dy}{dx}\).

Calculate the slope and determine the direction vector.

Simplifying the equation from Step 2, we get: \(\frac{dy}{dx} = \frac{\sqrt{3} - dx}{1 + dy}\) Using the fact that the tangent vector has an increasing \(y\) direction, the slope \(\frac{dy}{dx}>0\). So the direction vector is \((1, \frac{dy}{dx}) = (1, k)\), where \(k = \frac{\sqrt{3} - dx}{1 + dy}\) and \(k > 0\).

Find the magnitude of the direction vector.

The magnitude of the direction vector, \(||(1,k)||\), is given by: \(||\overrightarrow{d}|| = \sqrt{1^2 + k^2}\)

Calculate the unit vector.

To find the unit vector, we must divide the direction vector by its magnitude: \(\hat{d} = \frac{\overrightarrow{d}}{||\overrightarrow{d}||} = \left(\frac{1}{\sqrt{1^2 + k^2}}, \frac{k}{\sqrt{1^2 + k^2}}\right)\) Since we are looking for a unit vector in the direction of increasing values of \(y\), we must choose a \(k>0\). \(\hat{d} = \left(\frac{1}{\sqrt{1^2 + k^2}}, \frac{k}{\sqrt{1^2 + k^2}}\right)\) for \(k>0\) There can be multiple values of \(k>0\) that give unit vectors tangent to the circle at the given point and in the increasing direction of \(y\).

Key concepts

Unit Vector

In electromagnetics, a unit vector is a vector with a magnitude of 1. It's used to express direction without regard to magnitude, making it a fundamental concept in physics and engineering. Unit vectors are particularly useful when dealing with vectors in 2D or 3D spaces, as they help simplify expressions and calculations.

To find a unit vector from a given vector, \(\overrightarrow{v}\), the formula is:

• Divide each component of the vector by the vector's magnitude, \(||\overrightarrow{v}||\).

A unit vector maintains the direction of the original vector but strips away its length, normalizing it:
\[\hat{v} = \frac{\overrightarrow{v}}{||\overrightarrow{v}||}\]

In our exercise example, after deriving the tangent direction vector, the process of normalization gives us the unit vector that is tangent to the circle and points in the desired direction. This process allows us to emphasize direction as opposed to quantity.

Rectangular Components

Rectangular components refer to the parts of a vector that are aligned with the axes of a rectangular coordinate system, typically the x and y axes in 2D space. These components allow for easy manipulation and understanding of vectors by breaking them down along orthogonal directions.

Each vector in 2D can be defined in terms of its x-component and y-component:

• For a vector \(\overrightarrow{v} = (v_x, v_y)\), the components are \(v_x\) for the x-axis and \(v_y\) for the y-axis.
• The overall magnitude of the vector can be calculated using Pythagoras: \(||\overrightarrow{v}|| = \sqrt{v_x^2 + v_y^2}\).

In the context of the circle, determining rectangular components includes finding these x and y directions of the tangent vector that satisfy the problem condition. Such knowledge is crucial for correct computation of the tangent direction and subsequently the unit vector.

Tangent Vector

A tangent vector to a path like a circle represents the direction in which the path proceeds at any given point. In our exercise, you're supposed to find the tangent to the circle at a specific point.

• The direction of the tangent is perpendicular to the radius of the circle at the point of tangency.
• At \(p(x, y) = (-\sqrt{3}, 1)\) on the circle, the tangent vector would lie flat against the circle and can be expressed as \(dx, dy\), a small change from the point tangent.

Using derivatives, the direction of the tangent was found in terms of its slope, which was part of establishing the condition for a tangent in the direction of increasing \(y\). Recognizing that a tangent vector is free of magnitude constraints emphasizes the need for normalization to yield a unit direction.

Circle Equation

The circle equation is a fundamental tool in geometry and physics used to describe a circle's position and size. The standard equation for a circle centered at the origin in the xy-plane is given by:
\[ x^2 + y^2 = r^2 \]
where \(r\) represents the radius of the circle.

In many problems involving circles, checking if a point lies on the circle entails verifying it satisfies this equation. This holds true for our exercise where point \((-\sqrt{3}, 1)\) is confirmed to lie on the circle since it satisfies:
\[ (-\sqrt{3})^2 + 1^2 = 4 \]

This foundation aids in defining other elements such as deriving tangent properties and identifying points near the circumference that comply with the given conditions. Utilizing the circle's equation simplifies many related perpendicular or tangential problems.