Question

Find the arc length of \(\vec{r}(t)\) on the indicated interval. $$ \vec{r}(t)=\left\langle e^{-t} \cos t, e^{-t} \sin t\right\rangle \text { on }[0,1] $$

Short answer

The arc length is \( \sqrt{2} (1 - \frac{1}{e}) \).

Step-by-step solution

Find the Derivative of \( \vec{r}(t) \)

First, we need to find the derivative of the vector function \( \vec{r}(t) = \langle e^{-t} \cos t, e^{-t} \sin t \rangle \). To do this, differentiate each component separately. The derivative of the first component \( \frac{d}{dt}[e^{-t} \cos t] = -e^{-t} \cos t - e^{-t} \sin t \). The derivative of the second component \( \frac{d}{dt}[e^{-t} \sin t] = -e^{-t} \sin t + e^{-t} \cos t \). So, \( \vec{r}'(t) = \langle -e^{-t} \cos t - e^{-t} \sin t, -e^{-t} \sin t + e^{-t} \cos t \rangle \).

Compute the Magnitude of \( \vec{r}'(t) \)

Next, compute the magnitude \( |\vec{r}'(t)| \). The magnitude of a vector \( \langle a, b \rangle \) is given by \( \sqrt{a^2 + b^2} \).For our vector:\[ |\vec{r}'(t)| = \sqrt{(-e^{-t} \cos t - e^{-t} \sin t)^2 + (-e^{-t} \sin t + e^{-t} \cos t)^2} \]Simplify:\[ = \sqrt{e^{-2t} (\cos^2 t + 2 \cos t \sin t + \sin^2 t) + e^{-2t} (\sin^2 t - 2 \cos t \sin t + \cos^2 t)} \]Notice the simplification:\[ = \sqrt{2e^{-2t} (\cos^2 t + \sin^2 t)} = \sqrt{2e^{-2t}} = \sqrt{2} e^{-t} \].

Integrate to Find the Arc Length

The formula for arc length \( L \) is given by:\[ L = \int_a^b |\vec{r}'(t)| \, dt \]Substitute in our values:\[ L = \int_0^1 \sqrt{2} e^{-t} \, dt \]The antiderivative of \( \sqrt{2} e^{-t} \) is \( -\sqrt{2} e^{-t} \).Evaluate from 0 to 1:\[ \left[ -\sqrt{2} e^{-t} \right]_0^1 = -\sqrt{2} e^{-1} + \sqrt{2} = \sqrt{2} (1 - \frac{1}{e}) \].

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and differential operations over them. It relies on vectors, which are quantities having both magnitude and direction. In essence, vector calculus is crucial in describing physical quantities in physics and engineering, such as velocity fields and forces.

A vector function, like the one in the given exercise, is a function that takes one or more variables (like time, in this case) and returns a vector. This allows modeling phenomena where direction and magnitude change over time. Here, the function \( \vec{r}(t) = \langle e^{-t} \cos t, e^{-t} \sin t \rangle \) lives in the plane and traces out a path as \( t \) varies. Understanding these vector paths becomes critical in fields such as fluid dynamics, electromagnetics, and computer graphics.

Derivative of a Vector Function

Finding the derivative of a vector function is akin to finding the instantaneous rate of change of the vector with respect to a variable, such as time. In simpler terms, it tells us how the vector itself is changing at any given moment.

Here’s how it’s done for the given function: the derivative \( \vec{r}'(t) \) is determined by differentiating each component of the vector function individually. Given \( \vec{r}(t) = \langle e^{-t} \cos t, e^{-t} \sin t \rangle \), its derivative is \( \vec{r}'(t) = \langle -e^{-t} \cos t - e^{-t} \sin t, -e^{-t} \sin t + e^{-t} \cos t \rangle \).

• This operation is performed similarly to how scalar functions are differentiated, following the rules of differentiation, such as the product rule and chain rule.
• The resulting vector gives the velocity of a point on the curve traced by \( \vec{r}(t) \).

Magnitude of a Vector

The magnitude of a vector represents the vector's length, which is calculated using the components of the vector. In our exercise, it’s essential to determine how long the tangent vector, \( \vec{r}'(t) \), is, because it’s part of finding the arc length.

Here's how it works: For a 2D vector \( \langle a, b \rangle \), the magnitude is calculated as \( \sqrt{a^2 + b^2} \). For our derivative vector, \( \vec{r}'(t) = \langle -e^{-t} \cos t - e^{-t} \sin t, -e^{-t} \sin t + e^{-t} \cos t \rangle, \) the magnitude simplifies to \( \sqrt{2} e^{-t}. \)

• This simplifies the work substantially, thanks to trigonometric identities such as \( \cos^2 t + \sin^2 t = 1. \)
• Understanding vector magnitude is pivotal when measuring distances, speeds, or any scalar quantity associated with vectors.

Integral Calculation

The last step in finding the arc length involves performing an integral calculation over the interval of interest. The integral essentially accumulates all the calculated magnitudes of the tangent vectors along the path traced by \( \vec{r}(t) \).

The formula used is:
\[ L = \int_a^b |\vec{r}'(t)| \, dt \] For this function, with the limits 0 and 1, our arc length calculation is \( L = \int_0^1 \sqrt{2} e^{-t} \, dt. \)

• The antiderivative we find is \( -\sqrt{2} e^{-t}, \) which we evaluate at the bounds to find \( \sqrt{2} (1 - \frac{1}{e}). \)
• This concept is essential for deriving many physical properties, such as work done along a path in physics.

Engaging with integrals gives us the capacity to accumulate quantities over a continuous range, crucial in much of the work in calculus and related fields.