Problem 39 - Calculus III

Question

Find the equation of the tangent line to the curve \(\mathbf{r}(t)=\left\langle e^{t}, e^{-t}, 0\right\rangle\) at \(t=0\).

Step-by-Step Solution

Answer

The tangent line is given by \( x = 1 + t, y = 1 - t, z = 0 \).

Step 1: Point on the Curve

To find the point on the curve where the tangent line is, evaluate the vector function \( \mathbf{r}(t) \) at \( t = 0 \). Since \( \mathbf{r}(t) = \langle e^{t}, e^{-t}, 0 \rangle \), substitute \( t = 0 \) in the function: \( \mathbf{r}(0) = \langle e^{0}, e^{-0}, 0 \rangle = \langle 1, 1, 0 \rangle \). The point is (1, 1, 0).

Step 2: Derivative of the Vector Function

To find the tangent line, calculate the derivative \( \mathbf{r}'(t) \) to obtain the tangent vector at any time \( t \). Differentiate each component of \( \mathbf{r}(t) \):\[ \mathbf{r}'(t) = \left\langle \frac{d}{dt}(e^{t}), \frac{d}{dt}(e^{-t}), \frac{d}{dt}(0) \right\rangle = \left\langle e^{t}, -e^{-t}, 0 \right\rangle. \]

Step 3: Evaluate Derivative at Specific Point

Evaluate \( \mathbf{r}'(t) \) at \( t = 0 \) to get the tangent vector at that point. Substitute \( t = 0 \): \( \mathbf{r}'(0) = \langle e^{0}, -e^{0}, 0 \rangle = \langle 1, -1, 0 \rangle \).

Step 4: Equation of the Tangent Line

The equation of the tangent line can be written using the point \((1, 1, 0)\) and the direction vector \(\langle 1, -1, 0 \rangle\). The parametric equations for the tangent line are: \[ x = 1 + t, \quad y = 1 - t, \quad z = 0. \] These represent the equation of the tangent line.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields, functions that assign a vector to every point in space. It is essential in the study of physical phenomena such as electromagnetism and fluid dynamics, where quantities like force and velocity are naturally represented as vectors.
In vector calculus, we often work with vector functions that describe curves or surfaces in three-dimensional space. These functions are expressed in terms of a parameter, typically denoted as \( t \). The vector function \( \mathbf{r}(t) \) gives a point on the curve as \( t \) varies, mapping it directly to the Cartesian coordinates using vectors.
For example, \( \mathbf{r}(t) = \langle e^{t}, e^{-t}, 0 \rangle \) provides a path in the xy-plane, with the components \( e^{t} \) and \( e^{-t} \) describing the x and y coordinates, respectively, and a fixed z-component. Notably, the curve remains in the xy-plane since its z-component is always zero.

Derivative

Derivatives are a fundamental concept in calculus, representing the rate at which a function is changing at any given point. In vector calculus, we can take derivatives of vector functions, which will yield a vector that gives us both a direction and a rate of change at each point.
For a vector function \( \mathbf{r}(t) = \langle e^{t}, e^{-t}, 0 \rangle \), the derivative \( \mathbf{r}'(t) \) is found by differentiating each component individually:

• \( \frac{d}{dt}(e^{t}) = e^{t} \)
• \( \frac{d}{dt}(e^{-t}) = -e^{-t} \)
• \( \frac{d}{dt}(0) = 0 \)

Thus, the derivative \( \mathbf{r}'(t) = \langle e^{t}, -e^{-t}, 0 \rangle \) encodes the tangent vector, indicating the direction in which the curve is proceeding at any particular \( t \).
Evaluating this derivative at a specific point, such as \( t = 0 \), allows us to determine the exact tangent line at that point. Here, \( \mathbf{r}'(0) = \langle 1, -1, 0 \rangle \) provides the direction of the tangent line, while the point \( (1, 1, 0) \) lies on the tangent line.

Parametric Equations

Parametric equations utilize one or more parameters to express a set of related quantities. They allow us to describe curves or surfaces in a more flexible and effective way compared to standard Cartesian equations.
For the tangent line, which passes through a specific point on the curve and extends in the direction defined by the tangent vector, parametric equations naturally emerge. Using the point \( (1, 1, 0) \) where the curve touches the tangent line and the direction vector \( \langle 1, -1, 0 \rangle \), the parametric equations become:

• \( x = 1 + t \)
• \( y = 1 - t \)
• \( z = 0 \)

These equations indicate how each coordinate of the line changes with the parameter \( t \). As \( t \) varies, it traces out the complete tangent line in space, providing a comprehensive description of the line's path starting from the initial point \((1, 1, 0)\).
Parametric equations are particularly useful because they allow for straight-forwardly representing curves that cannot be defined easily by implicit or explicit equations.