Problem 10 - Thomas Calculus

Question

Find the point on the curve $$ \mathbf{r}(t)=(12 \sin t) \mathbf{i}-(12 \cos t) \mathbf{j}+5 t \mathbf{k} $$ at a distance 13\(\pi\) units along the curve from the origin in the direction opposite to the direction of increasing are length.

Step-by-Step Solution

Answer

The point is \( (0, 12, -5\pi) \).

Step 1: Compute the Derivative

Find the derivative of the vector function \( \mathbf{r}(t) = (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \). This derivative is \( \mathbf{r}'(t) = (12 \cos t) \mathbf{i} + (12 \sin t) \mathbf{j} + 5 \mathbf{k} \).

Step 2: Calculate the Magnitude of the Derivative

The magnitude of \( \mathbf{r}'(t) \) is:\[ \| \mathbf{r}'(t) \| = \sqrt{(12 \cos t)^2 + (12 \sin t)^2 + 5^2} = \sqrt{144 + 25} = 13. \]

Step 3: Determine the Arc Length Formula

The arc length from \( t_0 \) to \( t \) is \( s(t) = \int_{t_0}^{t} \| \mathbf{r}'(u) \| \, du = \int_{t_0}^{t} 13 \, du = 13(t - t_0) \).

Step 4: Solve for Negative Distance along the Arc Length

Given that \( s(t) = -13\pi \), we set up the equation:\[ 13(t - t_0) = -13\pi. \]Solving for \( t \), we have \( t - t_0 = -\pi \) hence \( t = t_0 - \pi \).

Step 5: Substitute to Find Point on Curve

If we assume that the starting point (the origin) is at \( t_0 = 0 \), find the point on the curve at \( t = 0 - \pi = -\pi \).Substitute into \( \mathbf{r}(t) \):\[ \mathbf{r}(-\pi) = (12 \sin(-\pi)) \mathbf{i} - (12 \cos(-\pi)) \mathbf{j} + 5(-\pi) \mathbf{k}. \]

Step 6: Simplify to Determine the Point

Calculate each component:- \(12 \sin(-\pi) = 0 \),- \(-12 \cos(-\pi) = -12(-1) = 12 \),- \(5(-\pi) = -5\pi.\)Thus, the point is \( 0\mathbf{i} + 12\mathbf{j} -5\pi \mathbf{k} \).

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations on these fields. It has many practical applications in physics and engineering. In this exercise, the vector function \( \mathbf{r}(t) = (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \) represents a curve in three-dimensional space. The vector \( \mathbf{r}(t) \) can be thought of as assigning a vector to every point of \( t \), guiding the position of points along a curve.
To explore this curve, we compute its derivative \( \mathbf{r}'(t) \), which is \( (12 \cos t) \mathbf{i} + (12 \sin t) \mathbf{j} + 5 \mathbf{k} \). This derivative represents the tangent vector to the curve at any point \( t \). It points in the direction of the steepest increase along the curve. By taking the magnitude of \( \mathbf{r}'(t) \), we measure the rate of change of the position vector, which helps in calculating the arc length along the curve.
Understanding how these derivatives and magnitudes work is local to the core of vector calculus. They help illustrate how curves twist and turn in space, allowing for detailed navigation and measurement of paths.

Closed Curves

Closed curves are loops with no beginning or end, like circles or ellipses. However, in this exercise, the curve is a helix rather than a closed loop, as it spirals upward without closure. The curve defined by \( \mathbf{r}(t) = (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \) extends indefinitely as \( t \) increases.
In the problem, we are interested in finding a specific point on the curve at a certain arc length from the origin. This involves a unique approach to traveling backwards on the spiral. By calculating the arc length and using the derived equations, we determine a point that's precisely 13\( \pi \) units backward. Even when a curve is not closed, similar mathematical tools are used to explore properties and features of the curve, such as lengths and positions.

Parametric Equations

Parametric equations allow us to represent complex curves more intuitively by using a parameter, often \( t \), which can be adjusted to trace the curve. In this example, parametric equations \( (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \) break down into simpler components along the \( x \), \( y \), and \( z \) axes.
This gives us a clear view of how each coordinate evolves with \( t \):

• The \( x \)-coordinate \( 12 \sin t \) fluctuates smoothly between -12 and 12 as a sine wave, representing a circular motion in the \( xy \)-plane.
• The \( y \)-coordinate \( -12 \cos t \) completes this circle perpendicular to the \( x \)-axis.
• The \( z \)-coordinate steadily advances by 5 units per \( t \), meaning the curve ascends uniformly, like a spiral or helix.

When working with parametric equations, we look at each piece to understand the path, visualize it, and solve tasks like computing specific points or lengths. They help us manage motion and changes linearly, crucial for understanding motion in physics and modeling pathways in engineering.