Problem 111 - Calculus 3

Question

Find the length of the curve \(\mathbf{r}(t)=\left\langle\sqrt{2} t, e^{t}, e^{-t}\right\rangle\) over the interval \( 0 \leq t \leq 1 .\) The graph is shown here:

Step-by-Step Solution

Answer

The length of the curve is \( e - \frac{1}{e} \).

Step 1: Understand the Problem

We need to find the length of the given vector function \( \mathbf{r}(t) = \langle \sqrt{2}t, e^t, e^{-t} \rangle \) from \( t = 0 \) to \( t = 1 \). The length of a curve described by a vector function from \( a \) to \( b \) is given by the integral \( L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \).

Step 2: Find the Derivative of the Vector Function

Compute the derivative of \( \mathbf{r}(t) = \langle \sqrt{2}t, e^t, e^{-t} \rangle \). The derivative is \( \mathbf{r}'(t) = \langle \sqrt{2}, e^t, -e^{-t} \rangle \).

Step 3: Calculate the Magnitude of the Derivative

To find \( \| \mathbf{r}'(t) \| \), compute the magnitude: \( \| \mathbf{r}'(t) \| = \sqrt{(\sqrt{2})^2 + (e^t)^2 + (-e^{-t})^2} = \sqrt{2 + e^{2t} + e^{-2t}} \).

Step 4: Set Up the Integral for Curve Length

The length of the curve \( L \) is the integral from 0 to 1 of the magnitude found in Step 3: \( L = \int_{0}^{1} \sqrt{2 + e^{2t} + e^{-2t}} \, dt \).

Step 5: Simplify the Integrand

Recognize \( e^{2t} + e^{-2t} \) can be rewritten as \( \left(e^t + e^{-t}\right)^2 - 2 \). Therefore, the integrand becomes \( \sqrt{\left(e^t + e^{-t}\right)^2} \). This simplifies to \( e^t + e^{-t} \).

Step 6: Evaluate the Integral

Evaluate \( \int_{0}^{1} (e^t + e^{-t}) \, dt \) by finding the antiderivative: \( \int e^t \, dt = e^t \) and \( \int e^{-t} \, dt = -e^{-t} \). Thus, the integral becomes \([e^t - e^{-t}]_{0}^{1} = \left(e^1 - e^{-1}\right) - \left(e^0 - e^0\right) = e - \frac{1}{e} \).

Step 7: Calculate the Final Answer

Substitute the limits: \( (e - \frac{1}{e}) - (1 - 1) = e - \frac{1}{e} \). Thus, the length of the curve is \( e - \frac{1}{e} \).

Key Concepts

Curve Length

The concept of curve length is a fascinating topic in vector calculus. It refers to the total distance traveled along a curve in a vector field. When we have a vector function that describes a curve, we can calculate its length using integral calculus.

To find the curve length, we integrate the magnitude of the derivative of the vector function over a specified interval. For a vector function \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\), the length L of the curve from \( t = a \) to \( t = b \) is given by:

• \( L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \)

By integrating the magnitude of the derivative, which gives the curve's instantaneous rate of change, we determine the total path length of the curve. Understanding how to compute the length of curves is essential for applications in physics, engineering, and computer graphics.

Vector Functions

Vector functions describe objects moving through space, and they are essential for modeling curves and trajectories in three-dimensional space. They use a parameter, often denoted by \( t \), to provide positions, called vectors, that change over time.

For the example exercise, the vector function \( \mathbf{r}(t) = \langle \sqrt{2} t, e^{t}, e^{-t} \rangle \) gives the position of a point as it moves along a curve. Each component of the vector function corresponds to a coordinate: \( x(t) = \sqrt{2} t \), \( y(t) = e^t \), and \( z(t) = e^{-t} \).

• Derivative of a Vector Function: Calculating \( \mathbf{r}'(t) \) finds how each component changes with respect to \( t \). It's vital for finding tangent vectors and determining curve length.
• In the exercise, the derivative was \( \mathbf{r}'(t) = \langle \sqrt{2}, e^t, -e^{-t} \rangle \).

Vector functions are used in various fields to track position and movement along curves.

Integral Calculus

Integral calculus plays a crucial role in determining quantities like areas, volumes, and lengths. When finding the length of a curve, we use integration to accumulate small elements of arc length to a total distance.

The process involves integrating the magnitude of the velocity vector, \( \| \mathbf{r}'(t) \| \), which represents the speed of moving along the curve: \( L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \).

Importance of Integral Calculus:

• Allows calculation of cumulative quantities like curve length by summing infinitesimal contributions.
• Transforms problems of motion and fields into manageable calculations.

In the original exercise, using integral calculus involved first simplifying the integrand from \( \sqrt{2 + e^{2t} + e^{-2t}} \) to \( e^t + e^{-t} \), and then evaluating the integral to find the curve's exact length, resulting in \( e - \frac{1}{e} \). This method exemplifies how integration helps solve complex geometric and physical problems.