Problem 56 - Essential Calculus Early Transcendentals

Question

(a) Find the point of intersection of the tangent lines to the curve \(\mathbf{r}(t)=\langle\sin \pi t, 2 \sin \pi t, \cos \pi t\rangle\) at the points where \( t=0\) and \(t=0.5\) . (b) Illustrate by graphing the curve and both tangent lines.

Step-by-Step Solution

Answer

The tangent lines intersect at the point \( \langle \pi/2, \pi, 1 \rangle \).

Step 1: Find the Position Vectors

First, find the position vectors at the given points by substituting the values of \( t \). For \( t = 0 \), \( \mathbf{r}(0) = \langle \sin(0), 2\sin(0), \cos(0) \rangle = \langle 0, 0, 1 \rangle \). For \( t = 0.5 \), \( \mathbf{r}(0.5) = \langle \sin(\pi/2), 2\sin(\pi/2), \cos(\pi/2) \rangle = \langle 1, 2, 0 \rangle \).

Step 2: Compute the Derivative

The tangent vector to the curve can be found by differentiating \( \mathbf{r}(t) \) with respect to \( t \). The derivative is \( \mathbf{r}'(t) = \langle \pi \cos(\pi t), 2\pi \cos(\pi t), -\pi \sin(\pi t) \rangle \).

Step 3: Find Tangent Vectors at Specific Points

Substitute \( t = 0 \) and \( t = 0.5 \) into the derivative to find the tangent vectors. \( \mathbf{r}'(0) = \langle \pi, 2\pi, 0 \rangle \) and \( \mathbf{r}'(0.5) = \langle 0, 0, -\pi \rangle \).

Step 4: Write Equations for Tangent Lines

Using the position and tangent vectors, the equations of the tangent lines can be constructed. At \( t = 0 \), \( \mathbf{T}_1(s) = \langle 0, 0, 1 \rangle + s \langle \pi, 2\pi, 0 \rangle \). At \( t = 0.5 \), \( \mathbf{T}_2(s) = \langle 1, 2, 0 \rangle + s \langle 0, 0, -\pi \rangle \).

Step 5: Solve for Intersection Point

To find the point of intersection, set \( \mathbf{T}_1(s) \) and \( \mathbf{T}_2(s) \) equal and solve for \( s \). \( \langle \pi s, 2\pi s, 1 \rangle = \langle 1, 2, -\pi s \rangle \). Solving gives \( s = 0.5 \), substituting in either line equation gives intersection at \( \langle \pi/2, \pi, 1 \rangle \).

Step 6: Graph the Curve and Tangent Lines

Use graphing software to plot the curve \( \mathbf{r}(t) = \langle \sin(\pi t), 2 \sin(\pi t), \cos(\pi t) \rangle \) over a suitable range, and add the tangent lines using their parametric equations, highlighting the intersection point \( \langle \pi/2, \pi, 1 \rangle \).

Key Concepts

Parametric Curves

Parametric curves are a wonderful topic in the world of vector calculus, providing a unique way to represent curves in space. Unlike a traditional function that uses a single equation with a dependent and an independent variable, parametric curves employ parameters, typically represented as \( t \), to define coordinates separately. This allows for a more flexible and powerful representation, especially useful in 3D space or for modeling complex movements.

• In our initial exercise, the curve is represented by \( \mathbf{r}(t) = \langle \sin(\pi t), 2 \sin(\pi t), \cos(\pi t) \rangle \).
• This representation allows us to trace out a path in 3-dimensional space as \( t \) varies from one value to another. Think of it as drawing a line in the air by moving an invisible pen that records its motion as \( t \) changes.

Parametric curves are particularly advantageous for visualizing and analyzing the motion of points through different spaces, providing insights into their behavior that traditional methods might obscure. They also play a crucial role in computer graphics and robotic path planning.

Tangent Lines

Tangent lines are lines that just touch a curve at a given point, moving in the exact direction the curve is going at that point. In the context of parametric curves, we determine tangent lines by calculating derivatives. This derivative is essentially a vector that tells us the direction and speed at the particular moment, akin to a velocity vector in physics.

• For example, by differentiating \( \mathbf{r}(t) \), we obtained the tangent vector \( \mathbf{r}'(t) = \langle \pi \cos(\pi t), 2\pi \cos(\pi t), -\pi \sin(\pi t) \rangle \).
• This helps us find the tangent vectors at specific points, such as \( \langle \pi, 2\pi, 0 \rangle \) at \( t = 0 \), indicating the direction of the curve at that point.

The full tangent line equation combines the position vector at the point with the tangent vector. This allows one to extend a line from any point on the curve, exploring where it might head if the curve continued in its tangent path. Tangential analysis is crucial for understanding properties like instantaneous rate of change.

Intersection Point

Finding the intersection point of curves or lines is a fundamental task in vector calculus. It involves discovering a common point that two or more geometric entities share. This can be especially interesting and challenging in three-dimensional space.

• In the example problem, we determined the point where the tangent lines at \( t=0 \) and \( t=0.5 \) intersected, using parametric equations for each tangent line.
• This was done by setting the equations of the tangent lines equal and solving for the parameter \( s \). This led us to the common point \( \langle \pi/2, \pi, 1 \rangle \).

The intersection provides insights into how different paths cross each other and can be especially valuable in physics, computer graphics, and robotics. Understanding intersections aids in collision detection or determining meeting points in various applications.