Problem 109 - Precalculus

Question

There is a branch of calculus devoted to the study of vector valued functions; these are functions that map real numbers onto vectors. For example, \(v(t)=\langle t, 2 t\rangle\) Find the magnitude of the vector-valued function \(\mathbf{v}(t)=\langle\cos t, \sin t\rangle\)

Step-by-Step Solution

Answer

The magnitude of the vector-valued function is 1.

Step 1: Understanding the Function

The given vector-valued function is \(\mathbf{v}(t)=\langle\cos t, \sin t\rangle\). This function maps a real number \(t\) to a vector with components \(\cos t\) and \(\sin t\).

Step 2: Recall the Magnitude Formula

The magnitude of a vector \(\mathbf{v}(t) = \langle x(t), y(t) \rangle\) is calculated using the formula: \[\|\mathbf{v}(t)\| = \sqrt{x(t)^2 + y(t)^2}\].

Step 3: Apply the Magnitude Formula

Substitute \(x(t) = \cos t\) and \(y(t) = \sin t\) into the formula: \[\|\mathbf{v}(t)\| = \sqrt{(\cos t)^2 + (\sin t)^2}\].

Step 4: Simplify the Expression

Simplify \((\cos t)^2 + (\sin t)^2\). We know from the Pythagorean identity that \(\cos^2 t + \sin^2 t = 1\). Thus, \[\|\mathbf{v}(t)\| = \sqrt{1} = 1\].

Key Concepts

Calculus

Calculus is a branch of mathematics that is pivotal in understanding how things change. When studying vector-valued functions, such as the example \(v(t) = \langle \cos t, \sin t \rangle\), calculus helps analyze the rate and direction of change of vectors over time.
Vector-valued functions are unique because they map real numbers to vectors, not just single values. This means they are ideal for describing paths in space, motion, and other multi-dimensional scenarios.
In this example, the function maps each value of \(t\) to a point on the unit circle since the coordinates are \(\langle\cos t, \sin t\rangle\).
Understanding calculus in this context allows for analyzing these paths and determining how properties like arc length and curvature change.

Magnitude of a Vector

The magnitude of a vector measures its length or 'size' in a coordinate system. For a vector \(\mathbf{v}(t) = \langle x(t), y(t) \rangle\), the magnitude is calculated using the formula:

• \[\|\mathbf{v}(t)\| = \sqrt{x(t)^2 + y(t)^2}\]

This formula is derived from the distance theorem in geometry.
In our example, we calculate the magnitude of \(\mathbf{v}(t) = \langle \cos t, \sin t \rangle\). By substituting \(x(t) = \cos t\) and \(y(t) = \sin t\), we obtained:

• \[\|\mathbf{v}(t)\| = \sqrt{(\cos t)^2 + (\sin t)^2}\]

The magnitude tells us how far the vector is from the origin, providing insight into its length.

Pythagorean Identity

The Pythagorean identity is a fundamental principle in trigonometry, stating that for any angle \(t\):

• \[\cos^2 t + \sin^2 t = 1\]

This identity is crucial when calculating the magnitude of vectors involving trigonometric functions. It simplifies expressions by using known trigonometric truths, reducing complex equations to their simplest form.
In our specific problem, understanding the Pythagorean identity allows the expression \(\cos^2 t + \sin^2 t\) to simplify to 1, hence:

• \[\|\mathbf{v}(t)\| = \sqrt{1} = 1\]

Recognizing this identity helps rapidly simplify calculations in vector analysis, showing that any vector of the form \(\langle \cos t, \sin t \rangle\) lies on the unit circle with a constant magnitude of 1.