Question

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

Short answer

The equation of the tangent line is: \( x = -3, \ y = -3t, \ z = \pi + t \).

Step-by-step solution

Find the Derivative of the Position Vector

To find the tangent line, we first need to find the derivative of the position vector \( \vec{r}(t) \), which will give us the velocity vector \( \vec{r}'(t) \).\[\vec{r}'(t) = \langle -3 \sin t, 3 \cos t, 1 \rangle\]

Evaluate the Derivative at the given t-value

Next, evaluate the derivative \( \vec{r}'(t) \) at the given value \( t = \pi \).\[\vec{r}'(\pi) = \langle -3 \sin \pi, 3 \cos \pi, 1 \rangle = \langle 0, -3, 1 \rangle\]

Find the Point on the Curve at t = π

Substitute \( t = \pi \) into the original position vector \( \vec{r}(t) \) to find the coordinates of the point.\[\vec{r}(\pi) = \langle 3 \cos \pi, 3 \sin \pi, \pi \rangle = \langle -3, 0, \pi \rangle\]

Write the Equation of the Tangent Line

Use the point \( \vec{r}(\pi) = \langle -3, 0, \pi \rangle \) and the direction vector \( \vec{r}'(\pi) = \langle 0, -3, 1 \rangle \) to write the equation of the tangent line.The parametric equations of the line are given by:\[\begin{align*}x(t) &= -3 + 0t = -3, \y(t) &= 0 - 3t, \z(t) &= \pi + t\end{align*}\]

Key concepts

Position Vector

A position vector is a mathematical tool that gives you the position of a point in space relative to a fixed origin. It's often denoted by \( \vec{r}(t) \) and is expressed as a combination of its components along the coordinate axes. For example, in three dimensions, you might write it as \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \). This vector is essential for describing the motion of an object over time because it shows how the object's position changes.In the given exercise, the position vector is \( \vec{r}(t) = \langle 3 \cos t, 3 \sin t, t \rangle \). This describes a point moving along a path defined by these components:

• The \( x \)-component \( 3 \cos t \) determines the horizontal position.
• The \( y \)-component \( 3 \sin t \) gives the vertical position.
• The \( z \)-component \( t \) indicates how the point's height changes.

Understanding the position vector is key to analyzing the object's trajectory and allows you to track the motion throughout its journey.

Parametric Equations

Parametric equations are a way to represent curves and surfaces in the coordinate plane or space. They express each coordinate (like \( x \), \( y \), and \( z \)) as a function of one or more parameters, usually denoted by \( t \). This approach is particularly useful for modeling paths of moving objects.In the context of the tangent line, parametric equations help describe the line's path using its position and direction vectors. Once you know the tangent point — here, \( \vec{r}(\pi) = \langle -3, 0, \pi \rangle \) — and the velocity vector \( \vec{r}'(\pi) = \langle 0, -3, 1 \rangle \), you can write parametric equations for the tangent line as:

• \( x(t) = -3 + 0t = -3 \)
• \( y(t) = 0 - 3t \)
• \( z(t) = \pi + t \)

These equations describe how to get from the starting point of the tangent line, moving with the velocity vector's direction, to represent its path in space.

Velocity Vector

The velocity vector is derived from the position vector by computing its derivative with respect to time. This vector indicates the direction and speed of an object's travel at any given moment. It's a fundamental concept in understanding how the object's position changes over time.In the exercise, the position vector \( \vec{r}(t) \) was differentiated to find the velocity vector \( \vec{r}'(t) \). The result, \( \vec{r}'(t) = \langle -3 \sin t, 3 \cos t, 1 \rangle \), gives:

• The \( x \)-component's rate of change as \( -3 \sin t \).
• The \( y \)-component's rate of change as \( 3 \cos t \).
• The constant \( 1 \) for the \( z \)-component, indicating uniform movement upward.

When evaluated at \( t = \pi \), the velocity vector becomes \( \langle 0, -3, 1 \rangle \). This shows that there's no change in the \( x \)-direction, a downward move in the \( y \)-direction, and an upward move in \( z \). The velocity vector is crucial for defining how the tangent line behaves, offering insight into both direction and speed.