Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the symmetric equations of the tangent line to the curve with equation $$ \mathbf{r}(t)=2 \cos t \mathbf{i}+6 \sin t \mathbf{j}+t \mathbf{k} $$ at \(t=\pi / 3\).

Step-by-Step Solution

Answer

\( \frac{x - 1}{-\sqrt{3}} = \frac{y - 3\sqrt{3}}{3} = \frac{z - \pi/3}{1} \)

Step 1: Differentiate the Vector Equation

To find the direction of the tangent line, we need the derivative of the vector-valued function \( \mathbf{r}(t) \). This represents the velocity vector, which is tangent to the curve. Differentiate each component:- The derivative of \( 2 \cos t \) is \( -2 \sin t \).- The derivative of \( 6 \sin t \) is \( 6 \cos t \).- The derivative of \( t \) is \( 1 \).Thus, \( \mathbf{r}'(t) = -2 \sin t\, \mathbf{i} + 6 \cos t\, \mathbf{j} + \mathbf{k} \).

Step 2: Evaluate the Derivative at \( t = \pi/3 \)

Substitute \( t = \pi/3 \) into \( \mathbf{r}'(t) \):- \( \sin(\pi/3) = \sqrt{3}/2 \) and \( \cos(\pi/3) = 1/2 \).After substitution, \[ \mathbf{r}'(\pi/3) = -2 \left( \frac{\sqrt{3}}{2} \right) \mathbf{i} + 6 \left( \frac{1}{2} \right) \mathbf{j} + \mathbf{k} = -\sqrt{3}\, \mathbf{i} + 3\, \mathbf{j} + \mathbf{k} \].

Step 3: Find the Point on the Curve at \( t = \pi/3 \)

Substitute \( t = \pi/3 \) into \( \mathbf{r}(t) \) to find the specific point:- \( 2 \cos(\pi/3) = 1 \),- \( 6 \sin(\pi/3) = 3\sqrt{3} \),- \( t = \pi/3 \).Thus, the point is \( (1, 3\sqrt{3}, \pi/3) \).

Step 4: Form the Symmetric Equations of the Tangent Line

The symmetric equations of a line that passes through point \( (x_0, y_0, z_0) \) with direction vector \( (a, b, c) \) are given by:\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \].For this problem:- Point \( (1, 3\sqrt{3}, \pi/3) \),- Direction \( (-\sqrt{3}, 3, 1) \).Thus, the symmetric equations are \[ \frac{x - 1}{-\sqrt{3}} = \frac{y - 3\sqrt{3}}{3} = \frac{z - \pi/3}{1} \].

Key Concepts

Vector-Valued Function

A vector-valued function is a mathematical representation where each input, usually denoted as time \(t\), corresponds to a vector output. The vector output can be in the form of three-dimensional vectors, as in the exercise we are considering. This is generally written as:\[ \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \]where \(x(t)\), \(y(t)\), and \(z(t)\) are functions of \(t\) corresponding to the coordinates in space, and \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are unit vectors in the x, y, and z directions.

• The given function \( \mathbf{r}(t)=2 \cos t \mathbf{i}+6 \sin t \mathbf{j}+t \mathbf{k} \) describes a curve in 3D space.
• The role of \(t\) is like a parameter that indicates a position along the curve.

This function allows us to observe how each of the spatial components evolves with \(t\). It serves as a critical part of finding tangent lines or other geometrical properties.

Derivative

The derivative of a vector-valued function provides a vector that represents the rate at which the function changes with respect to its parameter, often time \(t\). In the context of the exercise, the derivative is essential for finding the direction of the tangent line.The derivative can be expressed as:\[ \mathbf{r}'(t) = \frac{d\mathbf{r}}{dt} = \frac{dx}{dt} \mathbf{i} + \frac{dy}{dt} \mathbf{j} + \frac{dz}{dt} \mathbf{k} \]

• By differentiating each component of our vector function \(\mathbf{r}(t)\), we identify how each spatial coordinate changes.
• In our example, the differentiation yielded \( \mathbf{r}'(t) = -2 \sin t \mathbf{i} + 6 \cos t \mathbf{j} + \mathbf{k} \).

The derivative not only indicates the curve's direction at a given point but also helps compute other rotational or linear details about the motion or structure described by the vector.

Tangent Line

The tangent line is a straight line that touches a curve at a single point and matches the curve's direction at that point. It represents the instantaneous direction of the curve's path.In this exercise:

• The point on the curve is determined by evaluating the vector function at a specific \(t\), here \(t = \pi/3\).
• The direction is provided by the derivative of the vector function evaluated at the same \(t\).

These components combine to form the tangent line, characterized in vector form by:\[ \mathbf{r}(t_0) + \lambda \mathbf{r}'(t_0) \]where \(\lambda\) is a scalar parameter. In the given problem, the symmetric equations transform the differential form into:\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \]where \((x_0, y_0, z_0)\) is the point on the curve and \((a, b, c)\) is the direction vector. This represents a concise way to describe the line's geometry in space.

Curve

A curve in mathematical terms is a one-dimensional continuous path in space. It can be linear or nonlinear, varying in shape and curvature.

• In vector calculus, curves are often defined using vector-valued functions where the parameter \(t\) represents a point along the path.
• The curve in our exercise is described by \( \mathbf{r}(t) = 2 \cos t \mathbf{i} + 6 \sin t \mathbf{j} + t \mathbf{k} \), which maps \(t\) to a path in three-dimensional space.

The curve’s geometry changes with \(t\), providing a dynamic representation of its form. Every point on a curve has a specific direction, defined by its tangent, and forming understanding through the curve involves examining how different values of \(t\) affect the position and trajectory. This curvature and directionality are imperative for analyzing motion dynamics or any geometrical property. Utilizing vector-valued functions allows thorough exploration of how objects move and interact spatially.