Problem 28 - Calculus Early Transcendentals

Question

At what points does the helix \(\mathbf{r}(t)=\langle\sin t, \cos t, t\rangle\) intersect the sphere \( x^{2}+y^{2}+z^{2}=5 ?\)

Step-by-Step Solution

Answer

The helix intersects the sphere at \( (\sin(2), \cos(2), 2) \) and \( (\sin(-2), \cos(-2), -2) \).

Step 1: Understand the Problem

We need to determine the points of intersection between a helix given by \( \mathbf{r}(t)=\langle\sin t, \cos t, t\rangle \) and a sphere described by the equation \( x^{2}+y^{2}+z^{2}=5 \). This requires finding values of \( t \) where the parametric coordinates satisfy the sphere's equation.

Step 2: Substitute Helix into Sphere Equation

Substitute the helix's parametric expressions into the sphere's equation: \( x = \sin t \), \( y = \cos t \), and \( z = t \). The sphere equation becomes \( (\sin t)^{2} + (\cos t)^{2} + t^{2} = 5 \).

Step 3: Simplify the Equation

By the Pythagorean identity, \( (\sin t)^{2} + (\cos t)^{2} = 1 \). This simplifies our equation to \( 1 + t^{2} = 5 \).

Step 4: Solve for t

Subtract 1 from both sides: \( t^{2} = 4 \). Take the square root of both sides to find \( t = \pm2 \).

Step 5: Find Intersection Points

Substitute \( t = 2 \) into \( \mathbf{r}(t) \) to find the point: \( \mathbf{r}(2) = \langle \sin 2, \cos 2, 2 \rangle \). Similarly, substitute \( t = -2 \) to find the point \( \mathbf{r}(-2) = \langle \sin(-2), \cos(-2), -2 \rangle \).

Step 6: Calculate Trigonometric Values

Use the trigonometric values: \( \sin(2), \cos(2)\), \(\sin(-2),\cos(-2)\) to find the exact points. Thus, the intersection points are approximately \( ( \sin(2), \cos(2), 2 ) \) and \( ( \sin(-2), \cos(-2), -2 ) \).

Step 7: Verify Your Points

Verify that the points \( \langle \sin(2), \cos(2), 2 \rangle \) and \( \langle \sin(-2), \cos(-2), -2 \rangle \) satisfy the sphere equation \( x^{2} + y^{2} + z^{2} = 5 \).

Key Concepts

Intersection of Curves

The concept of intersection of curves is vital when dealing with questions involving a parameter-based path like a helix meeting another geometric object such as a sphere.

When two curves intersect, they share one or more common points. In this exercise, we are asked to find where the helix intersects the sphere with equation \[x^2 + y^2 + z^2 = 5\].

This involves

• Substituting the parametric equations for the helix into the sphere's equation.
• Solving the resulting expressions to find the parameter values at which the curves meet.
• Using these parameter values to compute the actual spatial points of intersections.

Finding the intersections involves a blend of algebra and geometry, ensuring that conditions set by one shape are satisfied by the path traced by another.

Spherical Coordinates

Spherical coordinates offer a unique way of representing points in three-dimensional space, especially for objects like spheres. Though we don't directly use spherical coordinates in this exercise, understanding them helps visualize the sphere's shape described by \[x^2 + y^2 + z^2 = 5\].

Typically, in spherical coordinates, a point in space is expressed with

• radial distance \(r\), which serves as the distance from a central origin,
• polar angle \( \theta \), the angle from the positive z-axis, and
• azimuthal angle \( \phi \), which measures the angle within the xy-plane from the positive x-axis.

While our problem uses Cartesian coordinates, recognizing the sphere's radius as \( \sqrt{5} \) reinforces its symmetrical properties in three dimensions.

Trigonometry

Trigonometry is a mathematical tool that is invaluable in this problem, thanks to the helix's reliance on sine and cosine functions.

The helix expression \[ extbf{r}(t) = \langle \sin t, \cos t, t \rangle \] makes use of trigonometric functions to define motion in the xy-plane.

Here is how trigonometry plays a role:

• Trigonometric identities like \(\sin^2 t + \cos^2 t = 1\) simplify calculations, as seen when replacing this in the sphere's equation.
• Calculations with the specific angles, \(t = 2\) and \(t = -2\), require understanding sine and cosine functions at non-standard angles.

Understanding these functions helps locate the exact points of intersection when placed in the sphere's three-dimensional context.

Vector Calculus

Vector calculus is a perspective that assists in analyzing paths and interactions in high-dimensional spaces, like the helix meeting a sphere.

The vector function \[ extbf{r}(t) = \langle \sin t, \cos t, t \rangle \] describes the helix and involves the concepts of position vectors and parametric equations.

Key points to consider:

• The component vectors \(\langle \sin t, \cos t \rangle\) concentrate on circular motion in the xy-plane.
• The third component \(t\) introduces linear growth along the z-axis, effectively twisting the helix upwards or downwards.
• Using vector functions, we can analyze the overall curvature, angles, and intersections in a unified way.

By understanding both vector paths and their interactions, we gain powerful insights into the geometry involved.