Problem 85 - Calculus Early Transcendentals: Pearson New International Edition

Question

A fly is crawling along a wire helix so that its position vector is \(\mathbf{r}(t)=6 \cos \pi t \mathbf{i}+6 \sin \pi t \mathbf{j}+2 t \mathbf{k}, t \geq 0\). At what point will the fly hit the sphere \(x^{2}+y^{2}+z^{2}=100\), and how far did it travel in getting there (assuming that it started when \(t=0\) )?

Step-by-Step Solution

Answer

The fly hits the sphere at point (6, 0, 8) and travels approximately 28.65 units to get there.

Step 1: Understand the Problem

We have a fly moving along a helix, represented by the position vector \( \mathbf{r}(t)=6 \cos \pi t \mathbf{i}+6 \sin \pi t \mathbf{j}+2 t \mathbf{k} \). The fly will hit a sphere described by \( x^{2}+y^{2}+z^{2}=100 \). We need to determine when the coordinates of the helix satisfy the sphere equation and the distance traveled by the fly until it hits the sphere.

Step 2: Set Sphere Equation with Position Vector

The sphere equation is \( x^{2}+y^{2}+z^{2}=100 \). Plug in \( x=6\cos(\pi t) \), \( y=6\sin(\pi t) \), and \( z=2t \) from the position vector into the sphere equation:\[ (6\cos(\pi t))^2 + (6\sin(\pi t))^2 + (2t)^2 = 100 \]

Step 3: Simplify and Solve for t

Simplify the equation:\[ 36\cos^2(\pi t) + 36\sin^2(\pi t) + 4t^2 = 100 \]Using the identity \( \cos^2(\pi t) + \sin^2(\pi t) = 1 \), this reduces to:\[ 36 + 4t^2 = 100 \]Now, solve for \( t \):\[ 4t^2 = 64 \]\[ t^2 = 16 \]\[ t = 4 \] since \( t \geq 0 \).

Step 4: Calculate Position at t = 4

When \( t = 4 \), calculate \( \mathbf{r}(4) \):\[ x = 6 \cos(4 \pi) = 6 \cdot 1 = 6 \]\[ y = 6 \sin(4 \pi) = 6 \cdot 0 = 0 \]\[ z = 2 \cdot 4 = 8 \]So, the point is \( (6, 0, 8) \).

Step 5: Calculate Distance Travelled

The arc length formula for vector functions can be used to find the traveled distance up to \( t = 4 \):\[ \int_{0}^{4} \| \mathbf{r}'(t) \| \, dt \]First, find \( \mathbf{r}'(t) \):\[ \mathbf{r}'(t) = \langle -6\pi\sin(\pi t), 6\pi\cos(\pi t), 2 \rangle \]Calculate \( \| \mathbf{r}'(t) \| \):\[ \sqrt{(6\pi)^2 + 2^2} = \sqrt{36\pi^2 + 4} = \sqrt{36\pi^2 + 4} \]Integrate:\[ \int_{0}^{4} \sqrt{36\pi^2 + 4} \, dt = \sqrt{36\pi^2 + 4} \times 4 \]\[ = 4\sqrt{36\pi^2 + 4} \]

Step 6: Evaluate the Integral

Substituting the constant \( \sqrt{36\pi^2 + 4} \) into the expression from Step 5 gives:\[ 4 \sqrt{36\pi^2 + 4} = 4 \sqrt{36\cdot9.87 + 4} \approx 28.65 \]The distance the fly travels is approximately 28.65 units.

Key Concepts

Helix

A helix is a type of curve in three-dimensional space, characterized by a constant radius and a linear climb in the vertical axis. It resembles a spiral staircase or a coiled spring. Each point on the helix can be described using a position vector, which provides information about its location along this path.

In our problem, the fly moves along a helical path as defined by the position vector \( \mathbf{r}(t) = 6 \cos \pi t \mathbf{i} + 6 \sin \pi t \mathbf{j} + 2 t \mathbf{k} \).
This vector expression shows the fly's position in terms of three components:

• \(6 \cos(\pi t)\) for the \(x\)-component.
• \(6 \sin(\pi t)\) for the \(y\)-component.
• \(2t\) for the \(z\)-component.

As \(t\) increases, the fly not only rotates around the \(z\)-axis but also moves upwards, forming the typical spiral shape of a helix. Understanding this motion is crucial for solving problems involving particles or objects moving in helical paths, as it involves both circular and linear components.

Arc Length

The arc length is the distance traveled along a curved path, and in this context, it refers to the total length of the helix that the fly traverses before hitting the sphere. Calculating the arc length for a vector function involves integrating the magnitude of its derivative over the interval, from the initial to the final position.

For the position vector \( \mathbf{r}(t) \), the arc length \(s\) is given by the integral:
\[ s = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \]
This integral calculates the sum of "infinitesimal distances" over the chosen interval, generating the total length traveled.

To find \( \mathbf{r}'(t) \), we differentiate \( \mathbf{r}(t) \) term by term:

• The derivative of \(6 \cos(\pi t)\) is \(-6\pi \sin(\pi t)\).
• The derivative of \(6 \sin(\pi t)\) is \(6\pi \cos(\pi t)\).
• The derivative of \(2t\) is \(2\).

Thus, \(\mathbf{r}'(t) = \langle -6\pi \sin(\pi t), 6\pi \cos(\pi t), 2 \rangle\). Finding \( \| \mathbf{r}'(t) \| \) by these components, and integrating from \(0\) to \(4\), accounts for the traveled path length, yielding approximately 28.65 units. This calculation assists in understanding how different trajectories can affect the overall distance an object covers in space.

Sphere Equation

The sphere equation describes a set of points in three-dimensional space that all lie at the same distance from a central point. The general form is \(x^2 + y^2 + z^2 = r^2\), where \(r\) is the sphere's radius.

In the exercise, the sphere is defined by the equation \(x^2 + y^2 + z^2 = 100\), indicating a sphere with a radius of \(10\), since \(\sqrt{100} = 10\).
This equation is fundamental in determining when the fly, traveling along the helix, reaches the sphere's surface. By substituting the parametric equations for \(x, y,\) and \(z\) from the helix into the sphere's equation,
we can pinpoint when the fly is exactly \(10\) units away from the origin, marking its first contact with the sphere.

• The terms \(x = 6 \cos(\pi t)\) and \(y = 6 \sin(\pi t)\) cover the circle within the \(xy\)-plane.
• Adding \(z = 2t\) expands the search to three dimensions, confirming when the helix intersects the sphere.

Understanding these interactions enables us to grasp how paths and surfaces can coincide, a key component in solving geometrical intersection problems in physics and mathematics.