Problem 55 - Essential Calculus Early Transcendentals

Question

Find parametric equations for the tangent line to the curve \(x=t \cos t, y=t, z=t \sin t\) at the point \((-\pi, \pi, 0) .\) Illustrate by graphing both the curve and the tangent line on a common screen.

Step-by-Step Solution

Answer

The parametric equations are \(x = -\pi - t, y = \pi + t, z = -\pi t\).

Step 1: Check the Curve Equation

Given the curve: \[x = t \cos t, \quad y = t, \quad z = t \sin t\]We need to determine the parameter \(t\) at the point \((-\pi, \pi, 0)\). By matching the coordinates, we see that \(y = t = \pi\), which suggests \( t = \pi \).

Step 2: Evaluate the Point on the Curve

Substitute \(t = \pi\) into the curve equations to verify the point:\[x = \pi \cos \pi = -\pi, \quad y = \pi, \quad z = \pi \sin \pi = 0\]This confirms the point \((-\pi, \pi, 0)\) lies on the curve.

Step 3: Find the Derivative

To find the tangent line, compute the derivative vector \(\mathbf{r}'(t)\), where:\[\mathbf{r}(t) = (t \cos t, t, t \sin t)\]Calculate each component:\[\frac{d}{dt}(t \cos t) = \cos t - t \sin t\]\[\frac{d}{dt}(t) = 1\]\[\frac{d}{dt}(t \sin t) = \sin t + t \cos t\]

Step 4: Evaluate the Derivative at the Point

Evaluate \(\mathbf{r}'(t)\) at \(t = \pi\):\[\mathbf{r}'(\pi) = (-1 - \pi \sin \pi, 1, \pi \cos \pi) = (-1, 1, -\pi)\]This gives the direction vector for the tangent line.

Step 5: Formulate Parametric Equations for the Tangent Line

Use the point \((-\pi, \pi, 0)\) and the direction vector \((-1, 1, -\pi)\) to create the equations for the tangent line:\[x = -\pi - t \]\[y = \pi + t \]\[z = -\pi t\]

Step 6: Graphing the Curve and Tangent Line

To graph the curve and tangent line, plot the curve \(( t \cos t, t, t \sin t )\) and the tangent line represented by the parametric equations \(x = -\pi - t, y = \pi + t, z = -\pi t\) on the same graph.

Key Concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. This line has the same direction as the curve at the contact point. For parametric equations, a tangent line can be represented by a parametric linear equation using a point on the line and a direction vector. In our exercise, the curve is given as \[(x, y, z) = (t \cos t, t, t \sin t)\] At the point \((-\pi, \pi, 0)\), we began by finding the parameter \(t\), which is equal to \(\pi\). The tangent line is calculated using the derivative at \(t = \pi\). Once the direction vector is found, it provides the slope of the tangent line. By using the point \((-\pi, \pi, 0)\), and the direction vector \((-1, 1, -\pi)\), we constructed the parametric equations for the tangent line.

• For the x-coordinate: \( x = -\pi - t \)
• For the y-coordinate: \( y = \pi + t \)
• For the z-coordinate: \( z = -\pi t \)

Each represents a straight line path in the 3D space. This way, the tangent line provides a linear approximation of the curve at the point of tangency.

3D Curve

In a three-dimensional space, a curve can be defined by parametric equations. Each parameter equation gives the coordinates as a function of a parameter, often denoted as \(t\). In our exercise, the curve is represented by the equations \(x = t \cos t\), \(y = t\), and \(z = t \sin t\). These expressions plot into a coil-like shape that moves through the 3D space as \(t\) changes. Curves in 3D can have varying complexities, and unlike simple 2D graphs, they stretch out into the third dimension making visualization a bit more challenging.

• The x-coordinate moves back and forth, influenced by the cosine oscillation.
• The y-coordinate increases linearly with \(t\).
• The z-coordinate shifts up and down because of the sine function.

These changes showcase how a single parameter can map out an entire path in three-dimensional space, forming what is essentially a detailed trajectory or path traced by a point as \(t\) changes.

Vector Calculus

Vector calculus is a branch of mathematics focusing on the differentiation and integration of vector fields, primarily in 3D spaces. This powerful tool enables us to explore and analyze curves and surfaces with more precision. In our exercise, we used vector calculus to find the tangent line to a 3D curve. Initially, the given parametric equations can be interpreted as vector functions: \( \mathbf{r}(t) = (t \cos t, t, t \sin t) \). The derivative of this vector function, \( \mathbf{r'}(t) \), provides the tangent vector of the curve at any point \(t\). Essentially, the derivative vector indicates the direction in which the curve is heading.

• If the curve is viewed as a trajectory, the derivative can be thought of as the curve's velocity vector.
• This directional component is crucial in forming the parametric equations of a tangent line.

Ultimately, using vector calculus allowed us to compute this derivative and identify both the shape and direction at the point of interest, providing a clear picture of where and how the curve shifts in the three-dimensional space.

Parametric Derivatives

Parametric derivatives represent the rate of change of each component of a vector function with respect to the parameter \(t\). In our exercise, they were essential in finding the tangent line to the given 3D curve. For a parametric curve represented as \(x(t), y(t), z(t)\), derivatives \(x'(t)\), \(y'(t)\), and \(z'(t)\) are calculated individually to yield the tangent vector \( \mathbf{r'}(t) \).

• In our case, the derivatives were: \( \frac{d}{dt}(t \cos t) = \cos t - t \sin t \)
• \( \frac{d}{dt}(t) = 1 \)
• \( \frac{d}{dt}(t \sin t) = \sin t + t \cos t \)

By evaluating these derivatives at \(t = \pi\), we obtained the tangent line's direction vector: \((-1, 1, -\pi)\). This yielded our tangent line direction and equations. Understanding and applying parametric derivatives helps connect the dynamic changes in the 3D curve with its static representation at a particular point. This seamless transition from motion to form is what makes parametric derivatives a cornerstone in vector calculus.