Question

Give the equation of the line tangent to the graph of \(\vec{r}(t)\) at the given \(t\) value. $$ \vec{r}(t)=\langle 3 \cos t, \sin t\rangle \text { at } t=\pi / 4 $$

Short answer

The tangent line is \( \langle x, y \rangle = \langle \frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle + s\langle -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \).

Step-by-step solution

Find the Derivative of the Vector Function

To find the equation of the tangent line, we first need the derivative of the vector function \( \vec{r}(t)=\langle 3 \cos t, \sin t\rangle \). The derivative with respect to \( t \) is the vector \( \vec{r}'(t) = \langle -3 \sin t, \cos t \rangle \).

Evaluate the Derivative at the Given Value

We evaluate \( \vec{r}'(t) = \langle -3 \sin t, \cos t \rangle \) at \( t = \pi/4 \). - \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \) and \( \cos(\pi/4) = \frac{\sqrt{2}}{2} \), so - \( \vec{r}'(\pi/4) = \langle -3 \cdot \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle = \langle -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \).

Evaluate the Original Function at the Given Value

Now, evaluate the original function \( \vec{r}(t) \) at \( t = \pi/4 \).- \( \vec{r}(\pi/4) = \langle 3 \cos(\pi/4), \sin(\pi/4) \rangle = \langle 3 \cdot \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle = \langle \frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \).

Set Up the Equation of the Tangent Line

The equation of a line through a point \( (x_0, y_0) \) with direction \( \langle a, b \rangle \) is\[\vec{r} = \vec{r}_0 + t\vec{r}'(t).\]So, the tangent line at \( t = \pi/4 \) is \[\langle x, y \rangle = \langle \frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle + s\langle -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle,\]where \( s \) is the parameter for the line.

Key concepts

Vector Functions

Vector functions are an exciting realm in calculus that help us describe motion in space using a multi-component approach. Imagine trying to illustrate not just where something is but also how it's moving. This is what vector functions do well.

A vector function - Often represented as \( \vec{r}(t) = \langle f(t), g(t) \rangle \) in two dimensions or \( \vec{r}(t) = \langle f(t), g(t), h(t) \rangle \) in three dimensions.- Uses components, like \( f(t), g(t) \), to define the x and y positions over time.Through this, vector functions help in visualizing trajectories or paths that are traced out by moving particles or objects. For example, our function \( \vec{r}(t) = \langle 3 \cos t, \sin t \rangle \) gives both the x and y coordinates at any time \( t \). This mapping is particularly useful in applications like physics or engineering, where predicting the motion of objects is key.

Understanding vector functions sets the foundation for navigating complex motions and geometrical shapes in higher dimensions.

Derivative of Vector Functions

Derivatives of vector functions enable us to find additional properties, like velocities, for the trajectories we describe using our vector functions. This is akin to discovering the speed at which a certain object is moving at any given moment.

To find a derivative - Take the derivative of each component separately.For example, for our function \( \vec{r}(t) = \langle 3 \cos t, \sin t \rangle \), we find the derivative \( \vec{r}'(t) = \langle -3 \sin t, \cos t \rangle \).This derivative gives us the direction and speed of movement along the curve at any point \( t \). It's the backbone for determining tangent lines as well as analyzing accelerated motion.

By mastering the derivative of vector functions, we gain potent tools to decipher dynamic systems represented by mathematical curves.

Evaluating Derivatives

Evaluating derivatives at specific points equips us with precise insights about what happens at those exact moments in a path. This step bridges the gap between general functionalities and specific instances.

For example, evaluating the derivative \( \vec{r}'(t) = \langle -3 \sin t, \cos t \rangle \) at \( t = \pi/4 \) involves plugging \( \pi/4 \) into the derivative components:

• Determine \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \) and \( \cos(\pi/4) = \frac{\sqrt{2}}{2} \).
• Substitute these into \( \vec{r}'(\pi/4) \) to find \( \langle -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \).

This evaluated derivative provides the direction of the tangent at a particular point, facilitating the creation of tangent lines and understanding instantaneous velocity at that moment.

Recognizing how to evaluate derivatives effectively enables one to transition smoothly from broad math concepts to precise, real-world applications.

Parametric Equations

Parametric equations provide an enriched way to express curves, allowing us to formulate the coordinates of points on a path using a parameter, often denoted by \( t \). They reveal the connection between mathematical expressions and the geometric shapes they produce.

In the context of our exercise,- Each component of the vector function is defined separately as a function of \( t \), namely \( x(t) = 3 \cos t \) and \( y(t) = \sin t \). This can be particularly versatile in describing motion, where the interplay between different variables is crucial. They differ from traditional forms by offering flexibility in dealing with curves and paths that may loop, twist, or cross themselves.

Understanding parametric equations allows for more expressive and visually intuitive representations of objects in space. With these equations, one can effortlessly trace curves, making them indispensable in both theoretical explorations and practical modeling.