Problem 1 - Essential Calculus Early Transcendentals

Question

Find the length of the curve. $$\mathbf{r}(t)=\langle t, 3 \cos t, 3 \sin t\rangle, \quad-5 \leqslant t \leqslant 5$$

Step-by-Step Solution

Answer

The length of the curve is \(10\sqrt{10}\).

Step 1: Understand the Problem

We are given a vector function \( \mathbf{r}(t) = \langle t, 3\cos t, 3\sin t \rangle \) and need to find the length of the curve over the interval \(-5 \leq t \leq 5\).

Step 2: Recall the Formula for Arc Length

The arc length of a curve defined by a vector function \( \mathbf{r}(t) \) from \( a \) to \( b \) is given by \[L = \int_a^b \left\| \frac{d\mathbf{r}}{dt} \right\| \, dt\] We need to find the derivative of \( \mathbf{r}(t) \) and then its magnitude.

Step 3: Differentiate the Vector Function

Calculate \( \frac{d\mathbf{r}}{dt} \). We differentiate each component of \( \mathbf{r}(t) = \langle t, 3\cos t, 3\sin t \rangle \): - \( \frac{d}{dt}(t) = 1 \)- \( \frac{d}{dt}(3\cos t) = -3\sin t \)- \( \frac{d}{dt}(3\sin t) = 3\cos t \)Thus, \( \frac{d\mathbf{r}}{dt} = \langle 1, -3\sin t, 3\cos t \rangle \).

Step 4: Compute the Magnitude of the Derivative

Find \( \left\| \frac{d\mathbf{r}}{dt} \right\| \):\[\left\| \frac{d\mathbf{r}}{dt} \right\| = \sqrt{1^2 + (-3\sin t)^2 + (3\cos t)^2} = \sqrt{1 + 9\sin^2 t + 9\cos^2 t}\]Using the identity \( \sin^2 t + \cos^2 t = 1 \), the expression simplifies to \[\sqrt{1 + 9} = \sqrt{10}\]

Step 5: Integrate to Find the Arc Length

Substitute \( \left\| \frac{d\mathbf{r}}{dt} \right\| = \sqrt{10} \) into the arc length formula:\[L = \int_{-5}^{5} \sqrt{10} \, dt\]Since \( \sqrt{10} \) is a constant, integrate it over \(-5\) to \(5\):\[L = \sqrt{10} \cdot (5 - (-5)) = \sqrt{10} \times 10 = 10\sqrt{10}\]

Step 6: Confirm the Final Answer

We have calculated the length of the curve using the arc length formula, finding it to be \(10\sqrt{10}\).

Key Concepts

Arc Length Formula

The arc length formula is a key concept in vector calculus and is used to determine the length of a curve defined by a vector function. When you have a curve represented by a vector function \( \mathbf{r}(t) \), the arc length from \( t = a \) to \( t = b \) is found by integrating the magnitude of the derivative of the vector function, \( \frac{d\mathbf{r}}{dt} \). This means:

• Step 1: Differentiate the vector function to get \( \frac{d\mathbf{r}}{dt} \).
• Step 2: Compute the magnitude of this derivative.
• Step 3: Integrate this magnitude over the given interval \([a, b]\).

This process allows us to "add up" all the little segments of the curve to find its total length. The formula is expressed as:\[L = \int_a^b \left\| \frac{d\mathbf{r}}{dt} \right\| \, dt\]where \( \left\| \frac{d\mathbf{r}}{dt} \right\| \) is the magnitude of the derivative of the vector function. Thus, this formula gives an exact measure of how long the curve is between the specified bounds.

Vector Functions

Vector functions are the building blocks for describing curves in vector calculus. A vector function combines components that each represent a dimension. For instance, the function \( \mathbf{r}(t) = \langle t, 3\cos t, 3\sin t \rangle \) provides a way to express curves in three-dimensional space through its components:

• \( t \) is the component along the x-axis, which linearly changes with \( t \).
• \( 3\cos t \) is the component along the y-axis, which varies with the trigonometric function cosine.
• \( 3\sin t \) is the component along the z-axis, which varies with the trigonometric sine function.

Each component is a function of \( t \), the parameter that typically describes time or progression along a curve. By evaluating each component over a range of \( t \), the vector function maps out a continuous path or curve in space. Understanding how these components relate to each other is crucial to analyzing the behavior and properties of the curve they define.

Curve Length Calculation

To find the length of a curve described by a vector function, we begin by calculating the derivative of the vector function. Let's use our example \( \mathbf{r}(t) = \langle t, 3\cos t, 3\sin t \rangle \). We derive each component separately:

• The derivative of \( t \) with respect to \( t \) is 1.
• The derivative of \( 3\cos t \) is \(-3\sin t \).
• The derivative of \( 3\sin t \) is \( 3\cos t \).

Thus, \( \frac{d\mathbf{r}}{dt} = \langle 1, -3\sin t, 3\cos t \rangle \).Next, we find the magnitude of this derivative:\[ \left\| \frac{d\mathbf{r}}{dt} \right\| = \sqrt{1^2 + (-3\sin t)^2 + (3\cos t)^2} \]This simplifies to \( \sqrt{1 + 9\sin^2 t + 9\cos^2 t} \), and using the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), it becomes \( \sqrt{10} \).Finally, integrate this magnitude over the desired interval \( [-5, 5] \):\[ L = \int_{-5}^{5} \sqrt{10} \, dt = \sqrt{10} \times 10 = 10\sqrt{10} \]In these steps, we effectively add up the arc lengths of tiny segments along the curve, yielding the total curve length.