Problem 59 - Precalculus

Question

Assume that the head of \(\mathbf{u}\) is restricted so that its tail is at the origin and its head is on the unit circle in quadrant II or quadrant III. A vector \(\mathbf{v}\) has its tail at the origin and its head must lie on the line \(y=2-x\) in quadrant I. Find the least value of \(\mathbf{u} \cdot \mathbf{v}\).

Step-by-Step Solution

Answer

The least value of \( \mathbf{u} \cdot \mathbf{v} \) is \(-2\).

Step 1: Identify Constraints

Vector \( \mathbf{u} \) lies on the unit circle in quadrant II or III; hence, \( \mathbf{u} = (\cos \theta, \sin \theta) \) with \( \pi/2 < \theta < 3\pi/2 \). Vector \( \mathbf{v} \) lies on the line \( y = 2 - x \) in quadrant I, so \( \mathbf{v} = (x, 2-x) \) with \( 0 < x < 2 \).

Step 2: Express Dot Product

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is given by \( (\cos \theta, \sin \theta) \cdot (x, 2-x) = x \cos \theta + (2-x) \sin \theta \).

Step 3: Simplify Expression

The expression for the dot product can be simplified to \( \cos \theta \cdot x + 2\sin \theta - x \sin \theta = x (\cos \theta - \sin \theta) + 2\sin \theta \).

Step 4: Find Critical Points

To minimize the dot product, consider the derivative with respect to \( x \): \( f'(x) = \cos \theta - \sin \theta \). This implies \( x \) doesn't affect the minimum directly, so we consider the endpoints of the interval for \( x \).

Step 5: Consider Boundary Conditions

Evaluate the dot product at the boundaries \( x = 0 \) and \( x = 2 \): - For \( x = 0 \), the dot product is \( 2\sin \theta \).- For \( x = 2 \), the dot product is \( 2\cos \theta - 2\sin \theta \).

Step 6: Determine Optimal Configuration

Since we require the minimum dot product and \( \cos \theta \) and \( \sin \theta \) affect quadrants, consider \( \theta = \frac{3\pi}{4} \) (ideal for \( 2\sin \theta \) least when \( \cos \theta = \sin \theta \)): - Substitute \( \theta = \frac{3\pi}{4} \) to find the least value at either endpoint.

Step 7: Calculate Minimum Value

Calculate for \( \theta = \frac{3\pi}{4} \):- At \( x = 0 \), dot product is \( 2\sin\left(\frac{3\pi}{4}\right) = 2\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2} \).- At \( x = 2 \), dot product is \(-2 \), using \( \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).

Step 8: Conclude the Minimum Value

The least value of \( \mathbf{u} \cdot \mathbf{v} \) is \(-2\) at \( x = 2 \) and \( \theta = \frac{3\pi}{4} \).

Key Concepts

Understanding Vectors

Vectors are mathematical objects that have both magnitude and direction. In this context, vectors are visualized as arrows originating from the origin. They extend to a specific point in the coordinate plane.
Vectors can be used in many fields such as physics and engineering to represent quantities like force or velocity.
A vector \( \mathbf{u} \) can be denoted by its components along the x-axis and y-axis, for example, \( (x, y) \).
This makes it versatile for calculations, including the dot product - an operation discussed later. It's crucial to note how these components change based on their restrictions, like only lying in certain quadrants.
In exercises like this, constraints dictate where the head of the vector can be, such as on a specific line or shape like a unit circle.

The Unit Circle

The unit circle is an essential concept in trigonometry and vector analysis. It is a circle with a radius of one unit, centered at the origin of a coordinate plane.
In this problem, the unit circle is particularly useful for determining directions of vectors. When a vector's head lies on this circle, it means that the vector's magnitude is always one.
The position on the unit circle can be expressed in terms of sine and cosine functions: \( \mathbf{u} = (\cos \theta, \sin \theta) \), where \( \theta \) is the angle formed with the positive x-axis.
The unit circle also helps in understanding angles belonging to different quadrants, which affect the signs and values of \( \cos \theta \) and \( \sin \theta \). This is crucial for calculating the dot product of vectors restricted to specific quadrants.

Exploration of Quadrants

The Cartesian plane is divided into four quadrants. Each quadrant affects the sign of \( x \) and \( y \) components of vectors.
Understanding these quadrants is essential to rightly manipulate vector components for calculations.

• Quadrant I: Both x and y are positive.
• Quadrant II: x is negative, y is positive.
• Quadrant III: Both x and y are negative.
• Quadrant IV: x is positive, y is negative.

Vectors in quadrant II have a negative cosine and positive sine value, while vectors in quadrant III have both negative cosine and sine values.
These properties influence the calculations of dot products because they determine whether the contributions from cosine and sine will add or subtract to the sum.

Minimization in Vector Calculations

Minimization is aimed at finding the smallest possible value of an expression. In our context, it involves the dot product of two vectors.
The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated using the formula \( x \cos \theta + y \sin \theta \).
We often differentiate the resulting expression to find critical points that give us minimum values. This involves calculating the derivative and analyzing conditions at endpoints.
Boundary conditions, as discussed, play an important role if the expression depends only on one variable while others are constant.
In problems like this, determining specific angles, such as \( \theta = \frac{3\pi}{4} \), can be more effective to achieve minimization when endpoints themselves give extreme values that can be tested directly without further manipulation.