Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each line integral. \(\int_{C}\left(x^{2}+y^{2}+z^{2}\right) d s ; \quad C\) is the curve \(x=4 \cos t\), \(y=4 \sin t, z=3 t, 0 \leq t \leq 2 \pi .\)

Step-by-Step Solution

Answer

The line integral evaluates to \( \sqrt{41} (32\pi + 24\pi^3) \).

Step 1: Parameterize the Curve

The given parameterization of the curve is \( \mathbf{r}(t) = \langle x, y, z \rangle = \langle 4 \cos t, 4 \sin t, 3t \rangle \). This will serve as the parameters of the curve \( C \) for the line integral.

Step 2: Compute the Derivative of the Curve

Calculate the derivative \( \frac{d \mathbf{r}}{dt} \) with respect to \( t \):\[ \frac{d \mathbf{r}}{dt} = \left\langle \frac{d}{dt}(4 \cos t), \frac{d}{dt}(4 \sin t), \frac{d}{dt}(3t) \right\rangle = \langle -4 \sin t, 4 \cos t, 3 \rangle \]

Step 3: Find the Magnitude of the Derivative

Compute the magnitude of \( \frac{d \mathbf{r}}{dt} \):\[ \left\| \frac{d \mathbf{r}}{dt} \right\| = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2 + (3)^2} = \sqrt{16 + 16 \cos^2 t + 16 \sin^2 t} = \sqrt{32 + 9} = \sqrt{41} \]

Step 4: Substitute into the Integral

Substitute \( x(t), y(t), z(t) \) into the integrand \( x^2 + y^2 + z^2 \):\[ (4 \cos t)^2 + (4 \sin t)^2 + (3t)^2 = 16(\cos^2 t + \sin^2 t) + 9t^2 \]Since \( \cos^2 t + \sin^2 t = 1 \),\[ x^2 + y^2 + z^2 = 16 + 9t^2 \]

Step 5: Set Up the Integral

The line integral is given by:\[ \int_{C} (x^2 + y^2 + z^2) \, ds = \int_{0}^{2\pi} (16 + 9t^2) \sqrt{41} \, dt \]

Step 6: Evaluate the Integral

Solve the integral:\[ \sqrt{41} \int_{0}^{2\pi} (16 + 9t^2) \, dt = \sqrt{41} \left[ 16t + 3t^3 \right]_{0}^{2\pi} \]\[ = \sqrt{41} \left( 16(2\pi) + 3(2\pi)^3 \right) = \sqrt{41} \left( 32\pi + 24\pi^3 \right) \]\[ = \sqrt{41} (32\pi + 24\pi^3) \]

Step 7: Confirm the Final Answer

The final result of the line integral is \( \sqrt{41} (32\pi + 24\pi^3) \).

Key Concepts

Parameterization of Curves

Parameterization of a curve involves expressing a curve as a set of equations in terms of a single parameter, usually denoted as \( t \). This parameter helps to trace the path of the curve as it changes value. In our original exercise, the curve was parameterized using \( t \) with:

• \( x = 4 \cos t \)
• \( y = 4 \sin t \)
• \( z = 3t \)

These parametric equations break down three-dimensional motion into simple terms.
It provides a straightforward way to define the coordinates of each point on the curve as \( t \) varies over a specified interval, \( 0 \leq t \leq 2\pi \). In essence, parameterization converts complex geometric paths into manageable pieces with clearly defined expressions.

Derivative of Vector Functions

The derivative of a vector function is analogous to taking the derivative of a regular function, applied to each component of the vector. In our problem, given \[ \mathbf{r}(t) = \langle 4 \cos t, 4 \sin t, 3t \rangle \]we compute the derivative with respect to \( t \) as follows:

• The derivative of \( 4 \cos t \) results in \( -4 \sin t \).
• The derivative of \( 4 \sin t \) results in \( 4 \cos t \).
• The derivative of \( 3t \) results in \( 3 \).

Thus, the derivative of the vector function is:\[ \frac{d \mathbf{r}}{dt} = \langle -4 \sin t, 4 \cos t, 3 \rangle \]
This derivative provides the tangent vector at any point on the curve
and is essential for computing the integral along the path. Remember, the derivative captures how the curve's position changes as \( t \) changes, serving as a velocity vector for the parameterized curve.

Magnitude of a Vector

The magnitude of a vector provides its length, which is a crucial step in evaluating line integrals. For a vector \( \mathbf{v} = \langle a, b, c \rangle \), the magnitude is calculated as:\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \]In our example:\[ \frac{d \mathbf{r}}{dt} = \langle -4 \sin t, 4 \cos t, 3 \rangle \]the magnitude is:\[ \left\| \frac{d \mathbf{r}}{dt} \right\| = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2 + (3)^2} = \sqrt{16 \sin^2 t + 16 \cos^2 t + 9} \]Utilizing the identity \( \cos^2 t + \sin^2 t = 1 \), this simplifies to:\[ \sqrt{32 + 9} = \sqrt{41} \]The magnitude reflects the rate of change in the curve's path, converting parametric motion into a scalar that contributes to calculating the integrated path length.

Trigonometric Identities

Trigonometric identities are mathematical equalities involving trigonometric functions that prove incredibly useful in simplifying expressions. One key identity, \( \cos^2 t + \sin^2 t = 1 \), frequently appears in problems involving circular motion or parameterized curves.
In our line integral, this identity helped simplify the magnitude calculation:\[ 16(\cos^2 t + \sin^2 t) + 9t^2 \rightarrow 16 + 9t^2 \]Understanding trigonometric identities allows for straightforward manipulation of complex expressions, reducing redundancy and easing computations.
This particular identity emphasizes the inherent relationship between sine and cosine, further strengthening the treatment of parametric curves typically involving these functions. Know these identities well as they can drastically streamline solving problems involving circular paths or periodic functions.