Question

Given \(\vec{r}(t),\) find \(\vec{T}(t)\) and evaluate it at the indicated value of \(t\). $$ \vec{r}(t)=\langle\cos t, \sin t\rangle, \quad t=\pi $$

Short answer

\(\vec{T}(\pi) = \langle 0, -1 \rangle\)

Step-by-step solution

Differentiate the Vector Function

First, find the derivative of the vector function \( \vec{r}(t) = \langle \cos t, \sin t \rangle \). Differentiate each component:- The derivative of \( \cos t \) is \( -\sin t \).- The derivative of \( \sin t \) is \( \cos t \).Thus,\[ \vec{r}'(t) = \langle -\sin t, \cos t \rangle \]

Compute the Magnitude of the Derivative

Calculate the magnitude of \( \vec{r}'(t) = \langle -\sin t, \cos t \rangle \). The magnitude is calculated as:\[ \|\vec{r}'(t)\| = \sqrt{(-\sin t)^2 + (\cos t)^2} \]Simplify the expression:\[ \|\vec{r}'(t)\| = \sqrt{\sin^2 t + \cos^2 t} = \sqrt{1} = 1 \]

Find the Unit Tangent Vector

The unit tangent vector \( \vec{T}(t) \) is given by:\[ \vec{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|} = \frac{\langle -\sin t, \cos t \rangle}{1} = \langle -\sin t, \cos t \rangle \]

Evaluate \( \vec{T}(t) \) at \( t = \pi \)

Substitute \( t = \pi \) into \( \vec{T}(t) = \langle -\sin t, \cos t \rangle \):- Calculate \( -\sin(\pi) = 0 \)- Calculate \( \cos(\pi) = -1 \)So,\[ \vec{T}(\pi) = \langle 0, -1 \rangle \]

Key concepts

Vector Differentiation

Vector differentiation involves finding the derivative of a vector function, similar to differentiating a standard scalar function. In this context, differentiation identifies the rate of change of each component of the vector function. For example, given the vector function \( \vec{r}(t) = \langle \cos t, \sin t \rangle \), finding the derivative involves differentiating the \( x \)-component \( \cos t \) and the \( y \)-component \( \sin t \):

• The derivative of \( \cos t \) is \( -\sin t \).
• The derivative of \( \sin t \) is \( \cos t \).

The product is the derivative vector \( \vec{r}'(t) = \langle -\sin t, \cos t \rangle \).

This new vector describes not only the direction but also how these directions change as \( t \) changes. It's a powerful tool for understanding motion and curve properties in a two-dimensional plane.

Magnitude of a Vector

The magnitude of a vector measures its length or size and is derived from its components. To find it, we apply the formula for the Euclidean norm:

Given a vector \( \vec{a} = \langle a_1, a_2 \rangle \), its magnitude is:\[\| \vec{a} \| = \sqrt{a_1^2 + a_2^2}\]
In our example with \( \vec{r}'(t) = \langle -\sin t, \cos t \rangle \), we calculate the magnitude as:

\[\| \vec{r}'(t) \| = \sqrt{(-\sin t)^2 + (\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t}\]
Trigonometric identities simplify \( \sin^2 t + \cos^2 t \) to \( 1 \), resulting in \( \sqrt{1} = 1 \). This shows the vector has a constant magnitude, keeping its path along a unit circle.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental in the study of periodic phenomena and occur frequently in vector analysis. These functions have specific properties that make them the perfect tools for describing circular motion.

Let's look at the basic properties:

• \( \sin(t) \) gives the \( y \)-coordinate on the unit circle.
• \( \cos(t) \) gives the \( x \)-coordinate on the unit circle.
• The identity \( \sin^2(t) + \cos^2(t) = 1 \) is pivotal in proofs and simplifications within trigonometry.

Trigonometric functions not only provide coordinates for the unit circle but also help measure angles and distances between points on the circle. They form the basis for circular and oscillatory motion descriptions, making them essential for understanding vector calculus in the context of rotational dynamics.