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 point \((0,-12,0)\) in the direction opposite to the direction of increasing arc length.

Step-by-Step Solution

Answer

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

Step 1: Understand the Parameterization

The curve is given by the vector function \( \mathbf{r}(t) = (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \). This implies that it is a helix with a radius of 12 in the xy-plane and rises linearly in the z direction. The parameter \( t \) represents the angle of rotation around the circle in the xy-plane, and the linear rise is at a rate of 5 units per unit increase of \( t \).

Step 2: Identify the Initial Point

We are given the initial point \((0,-12,0)\) on the curve. Set this point equal to the expression \( \mathbf{r}(t) \) to find the initial parameter \( t = 0 \). At \( t = 0, \; \sin(0) = 0, \cos(0) = 1, \) and the formula becomes \( \mathbf{r}(0) = 0 \cdot \mathbf{i} - 12 \cdot \mathbf{j} + 0 \cdot \mathbf{k} \), which matches the given point. Thus, \( t = 0 \) is the correct starting value for the parameter.

Step 3: Calculate the Arc Length Differential

The arc length \( ds \) along the curve is given by the magnitude of the derivative of \( \mathbf{r}(t) \): \( ds = \| \mathbf{r}'(t) \| \, dt \). Compute \( \mathbf{r}'(t) = (12 \cos t) \mathbf{i} + (12 \sin t) \mathbf{j} + 5 \mathbf{k} \). The magnitude is \( \| \mathbf{r}'(t) \| = \sqrt{(12 \cos t)^2 + (12 \sin t)^2 + 5^2} = \sqrt{144 + 25} = 13 \). Hence, \( ds = 13 \, dt \).

Step 4: Set Up the Arc Length Equation

To find a point at a distance \( 13\pi \) in the opposite direction, set up the equation \(-13\pi = \int_0^t 13 \, dt\). Integrating gives \(-13\pi = 13t\).

Step 5: Solve for t

Solve \(-13\pi = 13t\) for \( t \). Divide both sides by 13: \(-\pi = t\).

Step 6: Substitute t into the Curve Equation

Substitute \( t = -\pi \) into \( \mathbf{r}(t) = (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \) to find the point on the curve. Calculate \( \mathbf{r}(-\pi) = (12 \sin(-\pi)) \mathbf{i} - (12 \cos(-\pi)) \mathbf{j} + 5(-\pi) \mathbf{k} \).

Step 7: Compute the Coordinates

Compute the components: \( \sin(-\pi) = 0 \), \( \cos(-\pi) = -1 \). Therefore, \( \mathbf{r}(-\pi) = 0 \cdot \mathbf{i} + 12 \cdot \mathbf{j} - 5\pi \cdot \mathbf{k} = (0, 12, -5\pi) \).

Key Concepts

Helix

A helix is a three-dimensional spiral-like shape. Imagine a spring or the coil of a spiral staircase. The key characteristic of a helix is its tendency to progress along a circular path while simultaneously moving upward (or downward) along a straight line.In terms of mathematics, a helix can be described using vector functions. For example, the vector function \( \mathbf{r}(t) = (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \) represents a helix with:

• A radius of 12 in the xy-plane.
• A rise in the z direction at a rate of 5 units per unit increase in \( t \).

Understanding helices is essential in vector calculus as they often model real-world phenomena like the path of a particle in a magnetic field or the DNA helical structure.

Arc Length

The arc length refers to the distance along a curve. To calculate it, especially for curves defined by vector functions, a specific formula is used.The arc length \( ds \) for a curve given by a vector function \( \mathbf{r}(t) \) is calculated using the derivative of the vector function's magnitude:\[ ds = \| \mathbf{r}'(t) \| \ dt \]Here, \( \mathbf{r}'(t) \) is the derivative of the vector function with respect to the parameter \( t \).

• First, compute \( \mathbf{r}'(t) \).
• Then, calculate its magnitude \( \| \mathbf{r}'(t) \| \).

In the given example, the magnitude was found to be 13, thus \( ds = 13 \, dt \). This means for each small increase in \( t \), the curve stretches by 13 units, allowing us to determine distances along the curve accurately.

Parametric Equations

Parametric equations express a set of related quantities as explicit functions of a common parameter. They are particularly useful for describing curves in space.In the context of the exercise, the curve is defined by the parametric equations:

• \( x(t) = 12 \sin t \)
• \( y(t) = -12 \cos t \)
• \( z(t) = 5t \)

These equations associate each \( t \) value with a specific point \( (x, y, z) \) in three-dimensional space. This representation is fundamental in vector calculus as it allows for the study of dynamic points on curves, like finding arc lengths or tangents.

Vector Functions

Vector functions are essential tools in calculus used to represent multi-dimensional quantities that vary with respect to one or more parameters.A vector function combines multiple functions representing different dimensions into one entity. The function \( \mathbf{r}(t) = (12 \sin t) \mathbf{i} - (12 \cos t) \mathbf{j} + 5t \mathbf{k} \) is an example, describing a curve in three-dimensional space:

• \( 12 \sin t \cdot \mathbf{i} \) represents the x-component, moving in a sinusoidal manner.
• \( -12 \cos t \cdot \mathbf{j} \) represents the y-component, also following a sinusoidal path.
• \( 5t \cdot \mathbf{k} \) represents the z-component, linearly increasing as \( t \) grows.

Understanding vector functions helps in describing and analyzing complex curves and surfaces in vector calculus.