Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider the curve \(\mathbf{r}(t)=\sin t \cos t \mathbf{i}+\sin ^{2} t \mathbf{j}+\cos t \mathbf{k}\), \(0 \leq t \leq 2 \pi\). (a) Show that the curve lies on a sphere centered at the origin. (b) Where does the tangent line at \(t=\pi / 6\) intersect the \(x y\)-plane?

Step-by-Step Solution

Answer

(a) The curve lies on a unit sphere centered at the origin. (b) Tangent line intersects the XY-plane at \( (\frac{3\sqrt{3}}{4}, \frac{7}{4}, 0) \).

Step 1: Calculate the Position Vector Magnitude

To show that the curve lies on a sphere centered at the origin, we need to find the magnitude of the position vector \(\mathbf{r}(t)\). The formula for the magnitude is \[||\mathbf{r}(t)|| = \sqrt{(\sin t \cos t)^2 + (\sin^2 t)^2 + (\cos t)^2}\]

Step 2: Simplify the Magnitude Expression

Calculate each component separately: - \((\sin t \cos t)^2 = \sin^2 t \cos^2 t\)- \((\sin^2 t)^2 = \sin^4 t\)- \((\cos t)^2 = \cos^2 t\)Now substitute these into the magnitude expression: \[||\mathbf{r}(t)|| = \sqrt{\sin^2 t \cos^2 t + \sin^4 t + \cos^2 t}\]

Step 3: Use Trigonometric Identities

Use the identity \(\sin^2 t + \cos^2 t = 1\) to simplify the expression:\[||\mathbf{r}(t)|| = \sqrt{\sin^2 t (\cos^2 t + \sin^2 t) + \cos^2 t} = \sqrt{1}\]Thus, \(||\mathbf{r}(t)|| = 1\), confirming the curve lies on a sphere of radius 1 centered at the origin.

Step 4: Find the Tangent Vector at \(t=\pi/6\)

First, calculate the derivative \(\mathbf{r}'(t)\):\[\mathbf{r}'(t) = \frac{d}{dt}(\sin t \cos t) \mathbf{i} + \frac{d}{dt}(\sin^2 t) \mathbf{j} + \frac{d}{dt}(\cos t) \mathbf{k}\]Calculate each component separately:- \(\frac{d}{dt}(\sin t \cos t) = \cos 2t\)- \(\frac{d}{dt}(\sin^2 t) = 2 \sin t \cos t\)- \(\frac{d}{dt}(\cos t) = -\sin t\)Combine these to get:\[\mathbf{r}'(t) = \cos 2t \mathbf{i} + 2 \sin t \cos t \mathbf{j} - \sin t \mathbf{k}\]

Step 5: Evaluate the Tangent Vector at \(t=\pi/6\)

Substitute \(t=\pi/6\) into the tangent vector:- \(\cos 2(\pi/6) = \cos(\pi/3) = 1/2\)- \(\sin(\pi/6) = 1/2\), so \(2 \sin(\pi/6) \cos(\pi/6) = 2 \cdot 1/2 \cdot \sqrt{3}/2 = \sqrt{3}/2\)- \(-\sin(\pi/6) = -1/2\)Thus,\[\mathbf{r}'(\pi/6) = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\]

Step 6: Find the Position Vector at \(t=\pi/6\)

Substitute \(t=\pi/6\) into the original position vector:- \(\sin(\pi/6) \cos(\pi/6) = 1/2 \cdot \sqrt{3}/2 = \sqrt{3}/4\)- \(\sin^2(\pi/6) = (1/2)^2 = 1/4\)- \(\cos(\pi/6) = \sqrt{3}/2\)Thus,\[\mathbf{r}(\pi/6) = \frac{\sqrt{3}}{4} \mathbf{i} + \frac{1}{4} \mathbf{j} + \frac{\sqrt{3}}{2} \mathbf{k}\]

Step 7: Determine the Equation of the Tangent Line

The equation of the tangent line can be expressed as:\[\mathbf{R}(t) = \mathbf{r}(\pi/6) + s \cdot \mathbf{r}'(\pi/6)\]Expressing in vector components:\[(\frac{\sqrt{3}}{4} + \frac{1}{2} s)\mathbf{i} + (\frac{1}{4} + \frac{\sqrt{3}}{2} s)\mathbf{j} + (\frac{\sqrt{3}}{2} - \frac{1}{2} s)\mathbf{k}\]

Step 8: Solve for Intersection with the XY-Plane

To find where the tangent line intersects the \(xy\)-plane, set the \(z\)-component to zero:\[\frac{\sqrt{3}}{2} - \frac{1}{2} s = 0\]Solve for \(s\):\[s = \sqrt{3}\]Substitute \(s = \sqrt{3}\) back into the \(\mathbf{i}\) and \(\mathbf{j}\) components:- \(x = \frac{\sqrt{3}}{4} + \frac{1}{2}(\sqrt{3}) = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}\)- \(y = \frac{1}{4} + \frac{\sqrt{3}}{2}(\sqrt{3}) = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}\)Thus, the intersection point is \( (\frac{3\sqrt{3}}{4}, \frac{7}{4}, 0) \).

Key Concepts

Parametric Curves

Parametric curves are a fascinating concept in vector calculus. These curves are defined by a set of parametric equations, usually in terms of a parameter, often denoted as \( t \). Unlike traditional functions where \( y \) is solely a function of \( x \), parametric curves allow us to control both \( x \) and \( y \) independently through \( t \).

This approach gives us flexibility and power in describing curves that cannot be easily represented by standard functional forms. An example is the position vector \( \mathbf{r}(t) \) given as \( \sin t \cos t \mathbf{i} + \sin^2 t \mathbf{j} + \cos t \mathbf{k} \). Here, three equations control the components in the \( x \), \( y \), and \( z \) directions:

• \( x(t) = \sin t \cos t \)
• \( y(t) = \sin^2 t \)
• \( z(t) = \cos t \)

Such curves provide an insightful view of motion and space, showcasing how a point moves as \( t \) changes from 0 to \( 2\pi \). As we see in exercises, understanding parametric curves is crucial for solving problems involving complex shapes and motions.

Tangent Vector

The concept of a tangent vector is crucial when studying the behavior of parametric curves. A tangent vector provides not only the direction but also the instantaneous rate of change of a path described by a parametric curve. To find the tangent vector at any point on the curve, we differentiate the position vector with respect to the parameter \( t \).

In our exercise, the derivative \( \mathbf{r}'(t) \) was calculated to find the tangent vector. For the specific curve \( \mathbf{r}(t) = \sin t \cos t \mathbf{i} + \sin^2 t \mathbf{j} + \cos t \mathbf{k} \), and calculating at \( t = \pi/6 \), we find:

• \( \mathbf{r}'\left(\pi/6\right) = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \)

This tangent vector points in the direction the curve is heading at \( t = \pi/6 \). It is analogous to finding the slope of a line at a point for traditional functions but has additional dimensions in parametric contexts. This helps us determine where the tangent would intersect planes such as the \( xy \)-plane, providing vital geometric insights.

Sphere

A sphere in mathematics is defined as the set of all points that are equidistant from a central point. In this exercise, we are dealing with showing that a given parametric curve lies on a sphere centered at the origin.

To achieve this, we calculate the magnitude of the position vector \( \mathbf{r}(t) \). The square root of the sum of the squares of its components gives us the radius from the origin at any point \( t \):
\[||\mathbf{r}(t)|| = \sqrt{(\sin t \cos t)^2 + (\sin^2 t)^2 + (\cos t)^2}\]
By simplifying using trigonometric identities, particularly \( \sin^2 t + \cos^2 t = 1 \), this expression reduces to 1, leading us to confirm that the curve indeed lies on a sphere of radius 1. The reassuring part is recognizing the circle's intrinsic connection to our fundamental trigonometric identity, allowing us to grasp how this complex curve holds the spherical property throughout its entire domain.

Trigonometric Identities

Trigonometric identities are essential tools in calculus, especially when working with parametric curves. These identities allow us to simplify complex expressions and uncover hidden relationships. The most fundamental identity is \( \sin^2 t + \cos^2 t = 1 \), reflecting the Pythagorean theorem in a unit circle context.

In our exercise, this identity simplifies the calculation of the magnitude of the position vector, confirming the path's spherical nature. Additional identities such as \( 2 \sin t \cos t = \sin 2t \) and \( \cos^2 t - \sin^2 t = \cos 2t \) assist in derivative calculations, offering shortcuts to otherwise cumbersome algebra.

• They serve as foundational tools for reducing expressions.
• They enable direct simplification paths during derivative and integral computations.
• They deepen our understanding of the geometric properties of shapes and motion.

Understanding these identities allows students to navigate through complex vector calculus problems with greater ease. They not only aid in calculation but also provide insights into the harmonic nature of trigonometric forms in mathematics.