Question

Give the equation of the line tangent to the graph of \(\vec{r}(t)\) at the given \(t\) value. $$ \vec{r}(t)=\left\langle e^{t}, \tan t, t\right\rangle \text { at } t=0 $$

Short answer

The equation of the tangent line is \(\vec{L}(t) = \left\langle 1 + t, t, t \right\rangle\).

Step-by-step solution

Compute Derivative of \(\vec{r}(t)\)

The tangent line to the graph at a point can be found by taking the derivative of \(\vec{r}(t)\). Compute the derivative: \(\vec{r}'(t) = \left\langle\frac{d}{dt}e^t, \frac{d}{dt}\tan t, \frac{d}{dt}t\right\rangle = \left\langle e^t, \sec^2 t, 1\right\rangle\).

Evaluate Derivative at \(t=0\)

Substitute \(t = 0\) into the derivative \(\vec{r}'(t)\) to find the tangent vector at \(t=0\): \(\vec{r}'(0) = \left\langle e^0, \sec^2(0), 1\right\rangle = \left\langle 1, 1, 1\right\rangle\).

Compute \(\vec{r}(0)\)

Evaluate \(\vec{r}(t)\) at \(t=0\) to find the point of tangency: \(\vec{r}(0) = \left\langle e^0, \tan 0, 0\right\rangle = \left\langle 1, 0, 0\right\rangle\).

Formulate the Tangent Line Equation

The equation of the line tangent to \(\vec{r}(t)\) at \(t=0\) is given by the equation of the line in vector form: \(\vec{L}(t) = \vec{r}(0) + t\vec{r}'(0)\). Substitute the known values: \(\vec{L}(t) = \left\langle 1, 0, 0\right\rangle + t\left\langle 1, 1, 1\right\rangle = \left\langle 1 + t, t, t\right\rangle\).

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics focused on differentiation and integration of vector fields. It plays a crucial role in understanding phenomena in physics and engineering. In the context of finding a tangent line, vector calculus allows us to differentiate a vector-valued function, providing us essential tools to analyze motion in space.

When dealing with vector-valued functions like \( \vec{r}(t) = \langle e^t, \tan t, t \rangle \), we consider each component of the vector separately. The derivative represents a vector itself and can be thought of as the velocity of a point moving in space, describing the direction and speed of change for each component.

By computing the derivative \( \vec{r}'(t) \) and evaluating it at a specific point \( t = 0 \), one finds the tangent vector. This vector gives not only the slope but also the direction of the tangent line at that point, which is critical in understanding the geometric structure of the graph of \( \vec{r}(t) \).

Derivative

The derivative is a foundational tool in calculus that helps us comprehend how a function changes at any given point. For a vector function like \( \vec{r}(t) = \langle e^t, \tan t, t \rangle \), taking the derivative means differentiating each component function separately. This results in another vector that tells us the rate of change of the original vector function with respect to \( t \).

In our problem, the derivative \( \vec{r}'(t) = \langle e^t, \sec^2 t, 1 \rangle \) was computed. Here, \( e^t \) gives the rate of change of the exponential component, \( \sec^2 t \) provides the rate of change of the tangent component, which isn't defined everywhere, and \( 1 \) is a constant rate of change in the linear component.

• \( e^t \): Derivative of the exponential function \( e^t \) remains \( e^t \)
• \( \sec^2 t \): Derivative of \( \tan t \) is given by \( \sec^2 t \)
• \( 1 \): Derivative of \( t \) is simply 1

By evaluating this derivative at \( t = 0 \), we find \( \vec{r}'(0) = \langle 1, 1, 1 \rangle \), giving us the direction vector of the tangent line at that point.

Parametric Equations

Parametric equations are equations that express the coordinates of the points of a curve as functions of a parameter, usually \( t \). They offer great flexibility and creativity in describing complex curves and shapes in multidimensional spaces. With vector calculus, parametric equations become immensely useful for describing paths and motions.

In our exercise, \( \vec{r}(t) = \langle e^t, \tan t, t \rangle \) is a parameterized curve where \( t \) can be thought of as time, controlling the position of a particle in 3D space. The values \( e^t, \tan t, \) and \( t \) define its path along the x, y, and z axes respectively.

• \( e^t \): Describes the path's progression along the x-axis, growing exponentially.
• \( \tan t \): A periodic function affecting the y-axis, with undefined points, resulting in interesting behavior.
• \( t \): A straightforward linear increase on the z-axis.

By evaluating the position \( \vec{r}(0) = \langle 1, 0, 0 \rangle \), it reveals the specific point through which the tangent line passes. This point, combined with the tangent vector \( \vec{r}'(0) \), provides the line's equation, \( \vec{L}(t) = \langle 1 + t, t, t \rangle \), illustrating its parametric nature and confirming its role in understanding the dynamics of curves.