Problem 65 - Calculus III

Question

Find the length for the following curves. $$ \mathbf{r}(t)=\langle 3(t), 4 \cos (t), 4 \sin (t)\rangle \text { for } 1 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The length of the curve is 15 units.

Step 1: Find the velocity vector

First, find the derivative of the position vector \( \mathbf{r}(t) \) to obtain the velocity vector. Differentiate each component of the vector \( \mathbf{r}(t) = \langle 3t, 4\cos(t), 4\sin(t) \rangle \) with respect to \( t \):\[ \mathbf{r}'(t) = \left\langle 3, -4\sin(t), 4\cos(t) \right\rangle \]

Step 2: Find the magnitude of the velocity vector

Compute the magnitude of \( \mathbf{r}'(t) \) to find the speed at any time \( t \). The magnitude is given by the square root of the sum of the squares of its components:\[ \| \mathbf{r}'(t) \| = \sqrt{3^2 + (-4\sin(t))^2 + (4\cos(t))^2} \]Calculate:\[ = \sqrt{9 + 16\sin^2(t) + 16\cos^2(t)} \]Using the identity \( \sin^2(t) + \cos^2(t) = 1 \), we simplify to:\[ \| \mathbf{r}'(t) \| = \sqrt{25} = 5 \]

Step 3: Set up the integral for arc length

The arc length \( L \) of the curve from \( t = 1 \) to \( t = 4 \) is given by the integral of the magnitude of the velocity. Set up the integral:\[ L = \int_{1}^{4} \| \mathbf{r}'(t) \| \, dt = \int_{1}^{4} 5 \, dt \]

Step 4: Evaluate the integral

Integrate the constant 5 with respect to \( t \):\[ L = 5 \int_{1}^{4} \, dt = 5[t]_{1}^{4} \]Compute the definite integral:\[ L = 5(4 - 1) = 5 \cdot 3 = 15 \]

Key Concepts

Arc Length

In Calculus III, finding the arc length of a curve is akin to measuring out the distance a path covers. We work with a vector function \( \mathbf{r}(t) \), a smooth vector line on a graph, which traces out our path. The arc length is found by integrating the magnitude of the velocity vector over the interval of interest. This tells us how far along the curve we travel between specific points.

For instance, if you want to find how long a section of a roller coaster track is, you calculate the arc length over that section. In our initial example, \(\mathbf{r}(t)=\langle 3t, 4 \cos (t), 4 \sin (t)\rangle\) represents a helical path, and we found the arc length from \( t = 1\) to \( t = 4\) as 15 units.

• Start by finding the velocity vector and its magnitude, which represents the path's speed.
• Integrate this magnitude over your interval to find the arc length.
• The integration part turns a complex 3D curve calculation into something manageable, giving an accumulated distance value.

Velocity Vector

The velocity vector is the first derivative of the position vector \( \mathbf{r}(t) \). It's like the speedometer reading at any point along a path, only in multiple directions. It shows how fast and in what direction the position on the path is changing.

To find it, differentiate each component of \( \mathbf{r}(t) \). For \( \mathbf{r}(t) = \langle 3t, 4\cos(t), 4\sin(t) \rangle \), the derivative \( \mathbf{r}'(t) = \langle 3, -4\sin(t), 4\cos(t) \rangle \).

This step is crucial because understanding how the position changes tells us everything we need to calculate arc length, curvature, and more.

In practice, it helps:

• Reveal the dynamics of a moving point on a curve.
• Help calculate arc length through its magnitude.
• Contribute to understanding forces and accelerations in physics and engineering problems.

Integral of Vector Functions

Integrals, especially in calculus, help accumulate quantities. When dealing with vector functions, such as velocities, integrals accumulate the instantaneous values over a time span to provide overall quantities like distance or displacement.

In the original problem, the integral \( \int_{1}^{4} \| \mathbf{r}'(t) \| \, dt \) calculates the total arc length from \( t=1 \) to \( t=4 \). Solving it simply requires evaluating a standard integral once the velocity vector's magnitude is determined. We saw this integral result in 15, showing the total length of the path.

When handling such integrals, keep these in mind:

• They transform velocity information into spatial conclusions (distances, areas).
• They demand understanding of underlying functions and their magnitudes.
• The more insights you have into a function's behavior, the more accurate your accumulated result will be.

Integrals are powerful, converting a collection of instantaneous data into a bigger picture of what a vector function like \( \mathbf{r}(t) \) actually does.