Problem 89 - Precalculus

Question

There is a branch of calculus devoted to the study of vectorvalued functions; these are functions that map real numbers onto vectors. For example, \(v(t)=\langle t, 2 t\rangle\). Find the angle between the vector-valued functions \(\mathbf{u}(t)=\langle\sin t, \cos t\rangle\) and \(\mathbf{v}(t)=\langle\csc t,-\cos t\rangle\) when \(t=\frac{\pi}{6}\).

Step-by-Step Solution

Answer

The angle is \( \cos^{-1}\left(\frac{1}{2\sqrt{19}}\right) \) radians at \( t = \frac{\pi}{6} \).

Step 1: Evaluate the Vectors at t = π/6

Substitute \( t = \frac{\pi}{6} \) into the vector functions. For \( \mathbf{u}(t) = \langle \sin t, \cos t \rangle \), calculate: \[ \mathbf{u}(\frac{\pi}{6}) = \langle \sin\frac{\pi}{6}, \cos\frac{\pi}{6} \rangle = \langle \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \]For \( \mathbf{v}(t) = \langle \csc t,-\cos t \rangle \), calculate: \[ \mathbf{v}(\frac{\pi}{6}) = \langle \csc\frac{\pi}{6}, -\cos\frac{\pi}{6} \rangle = \langle 2, -\frac{\sqrt{3}}{2} \rangle \]

Step 2: Calculate Dot Product of the Vectors

The dot product of two vectors \( \mathbf{u} = \langle a_1, a_2 \rangle \) and \( \mathbf{v} = \langle b_1, b_2 \rangle \) is given by:\[ \mathbf{u} \cdot \mathbf{v} = a_1 b_1 + a_2 b_2 \]Plug in the values from Step 1:\[ \mathbf{u}(\frac{\pi}{6}) \cdot \mathbf{v}(\frac{\pi}{6}) = \left(\frac{1}{2}\right)(2) + \left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) \]Calculate:\[ 1 - \frac{3}{4} = \frac{1}{4} \]

Step 3: Calculate Magnitudes of the Vectors

The magnitude of a vector \( \mathbf{u} = \langle a_1, a_2 \rangle \) is:\[ \| \mathbf{u} \| = \sqrt{a_1^2 + a_2^2} \]Calculate for both vectors:\[ \| \mathbf{u}(\frac{\pi}{6}) \| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = 1 \]\[ \| \mathbf{v}(\frac{\pi}{6}) \| = \sqrt{2^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{4 + \frac{3}{4}} = \sqrt{\frac{19}{4}} = \frac{\sqrt{19}}{2} \]

Step 4: Use the Dot Product Formula to Find the Angle

The angle \( \theta \) between two vectors can be found using:\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \cdot \| \mathbf{v} \|} \]Substitute the values from Step 2 and Step 3:\[ \cos \theta = \frac{\frac{1}{4}}{1 \cdot \frac{\sqrt{19}}{2}} = \frac{1}{2\sqrt{19}} \]Therefore, the angle is:\[ \theta = \cos^{-1}\left(\frac{1}{2\sqrt{19}}\right) \]

Step 5: Conclusion: Angle Between the Vectors

The angle between the vector-valued functions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \) at \( t = \frac{\pi}{6} \) is given by:\[ \theta = \cos^{-1}\left(\frac{1}{2\sqrt{19}}\right) \]

Key Concepts

Vector-valued functions

Vector-valued functions are a fascinating aspect of calculus that involve mapping real numbers onto vectors rather than individual scalar quantities. These types of functions play a crucial role in modeling physical phenomena in physics and engineering, among other fields. A vector-valued function like \( \mathbf{v}(t)=\langle t, 2t \rangle \) maps a real number \( t \) to a vector consisting of its components in the plane.

Each component of the vector is itself a real-valued function and, together, they provide a geometric representation of the vector as a changing entity over time or another variable. In our exercise, \( \mathbf{u}(t)=\langle \sin t, \cos t \rangle \) and \( \mathbf{v}(t)=\langle \csc t, -\cos t \rangle \) are examples of such functions. They describe paths in a 2-dimensional space based on the input variable \( t \). Through this type of function, you can define positions, directions, or velocities in space over specific intervals of \( t \), useful for simulating movements and other dynamic systems.

Dot product

The dot product, also known as the scalar product, is a crucial operation in vector calculus that connects vectors with geometry and algebra. It provides a way to multiply two vectors to produce a single scalar quantity. Given two vectors \( \mathbf{u} = \langle a_1, a_2 \rangle \) and \( \mathbf{v} = \langle b_1, b_2 \rangle \), the dot product is calculated as:

\[ \mathbf{u} \cdot \mathbf{v} = a_1 b_1 + a_2 b_2 \]

This is a fundamental operation because it reveals important information about the vectors, such as whether they are perpendicular. When the dot product is zero, the vectors are orthogonal.

In the context of our exercise, the dot product helps in calculating the angle between two vector-valued functions. By computing \( \mathbf{u}(\frac{\pi}{6}) \cdot \mathbf{v}(\frac{\pi}{6}) \), we get a scalar that serves as part of the input for determining the angle through trigonometric relations.

Magnitude of a vector

The magnitude of a vector, often referred to as its length or norm, is a measure of how long the vector is in a geometric space. It is calculated using the Pythagorean theorem in the context of vector components. For a vector \( \mathbf{u} = \langle a_1, a_2 \rangle \), its magnitude is given by:

\[ \| \mathbf{u} \| = \sqrt{a_1^2 + a_2^2} \]

The magnitude is always a non-negative scalar value and gives an intuitive sense of the size of the vector.

In finding angles between vectors as in our example, the magnitude of each vector plays a critical role. We know that for \( \mathbf{u}(\frac{\pi}{6}) \), the magnitude is calculated to ensure accuracy in other parts of our problem, such as evaluating the cosine of the angle between vectors. This magnitude is also vital in physical interpretations where it might represent quantities like speed or force.

Angle between vectors

Determining the angle between two vectors is a fundamental task in vector calculus, which can reveal how aligned or opposing two directions are. It helps in understanding the relationship between vector directions in various fields, including computer graphics, physics, and navigation. The angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is found using the dot product and magnitudes of the vectors:

\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \cdot \| \mathbf{v} \|} \]

This relation is derived from the cosine law and allows us to compute the actual angle by taking the inverse cosine (\( \cos^{-1} \)) of the calculated expression.

In our example exercise, substituting the computed dot product and magnitudes of \( \mathbf{u}(\frac{\pi}{6}) \) and \( \mathbf{v}(\frac{\pi}{6}) \) into this formula gives the angle. This solution, \( \theta = \cos^{-1}(\frac{1}{2\sqrt{19}}) \), tells us precisely how these two direction vectors orient with respect to each other at that particular \( t \). This understanding is crucial for controlling and predicting movements in applications like robotics, animation, and mechanical design.