Problem 167 - Calculus Volume 3

Question

A person on a hang glider is spiraling upward as a result of the rapidly rising air on a path having position vector \(\mathbf{r}(t)=(3 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+t^{2} \mathbf{k}\). The path is similar to that of a helix, although it is not a helix. The graph is shown here: Find the following quantities: The glider's speed at any time

Step-by-Step Solution

Answer

The speed of the glider is \( \sqrt{9 + 4t^2} \).

Step 1: Understand the Problem

We are given a position vector \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + t^2 \mathbf{k} \). The task is to find the speed of the glider, which is the magnitude of the velocity vector.

Step 2: Calculate Velocity

The velocity vector is the derivative of the position vector with respect to time \( t \). Therefore, compute \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \). Differentiate each component: \( \begin{align*}\frac{d}{dt}(3 \cos t) &= -3 \sin t, \\frac{d}{dt}(3 \sin t) &= 3 \cos t, \\frac{d}{dt}(t^2) &= 2t. \\end{align*}\)Thus, the velocity vector is \( \mathbf{v}(t) = (-3 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 2t \mathbf{k} \).

Step 3: Calculate Speed

The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Use the formula for the magnitude of a vector:\( \| \mathbf{v}(t) \| = \sqrt{(-3 \sin t)^2 + (3 \cos t)^2 + (2t)^2} \).

Step 4: Simplify the Speed Expression

Simplify the expression for speed:\(\begin{align*}\| \mathbf{v}(t) \| &= \sqrt{9 \sin^2 t + 9 \cos^2 t + 4t^2} \&= \sqrt{9(\sin^2 t + \cos^2 t) + 4t^2} \&= \sqrt{9 \cdot 1 + 4t^2} \&= \sqrt{9 + 4t^2}.\end{align*}\)

Step 5: Conclusion

The formula \( \sqrt{9 + 4t^2} \) represents the speed of the glider at any time \( t \).

Key Concepts

Velocity Vector

In physics and mathematics, the velocity vector is a fundamental concept, representing both the direction and speed of an object in motion. It is derived from the position vector, which defines the location of an object in space at any given time.

• The velocity vector, \( \mathbf{v}(t) \), is calculated by taking the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \).
• For the hang glider scenario, the position vector is \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + t^2 \mathbf{k} \).
• When we differentiate each component separately, the resulting velocity vector is \( \mathbf{v}(t) = (-3 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 2t \mathbf{k} \).

The velocity vector is crucial as it provides a complete description of the glider's motion, showing both how fast and in which direction it is moving.

Speed Calculation

Speed is the scalar measure of how fast an object is moving, irrespective of its direction. It is determined by computing the magnitude of the velocity vector.

• Given the velocity vector \( \mathbf{v}(t) = (-3 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 2t \mathbf{k} \), the speed is found using the formula for vector magnitude.
• The calculation involves squaring each component of \( \mathbf{v}(t) \), summing them up, and finding the square root of that sum.

This method yields the formula for speed: \( \| \mathbf{v}(t) \| = \sqrt{9 \sin^2 t + 9 \cos^2 t + 4t^2} \). Simplifying using trigonometric identities, the expression becomes \( \sqrt{9 + 4t^2} \), indicating how the glider's speed changes concerning time \( t \).

Vector Magnitude

The vector magnitude is a measure of the length of a vector, providing insight into its size without considering its direction. For any vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), the magnitude \( \| \mathbf{v} \| \) is calculated by this formula:\[\| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2}\]

• This calculation involves squaring each component of the vector, adding the squares, and then taking the square root of the result.
• For the glider's velocity vector, it results in \( \sqrt{(-3 \sin t)^2 + (3 \cos t)^2 + (2t)^2} \).
• Using trigonometric identity \( \sin^2 t + \cos^2 t = 1 \), further simplifies to \( \sqrt{9 + 4t^2} \).

The magnitude calculation is essential in determining the speed, offering a clear snapshot of the motion intensity at any instant.

Derivative of Position Vector

The derivative of a position vector is central to determining an object's velocity and, subsequently, its speed. Differentiating the position vector with respect to time provides the velocity vector.

• Each component of the vector needs to be differentiated independently.
• For \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + t^2 \mathbf{k} \), differentiating yields each part: \(-3 \sin t \), \(3 \cos t \), and \(2t\).

The product, \( \mathbf{v}(t) = (-3 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 2t \mathbf{k} \), is the velocity vector. This differentiation is critical, as it translates the positional change into motion, showing how quickly and in which direction the glider's path evolves. It is the first step in finding both the velocity and speed of an object.