Problem 104 - Calculus 3

Question

Find the arc length of the curve on the given interval. \(\mathbf{r}(t)=\langle 2 \sin t, 5 t, 2 \cos t\rangle, 0 \leq t \leq \pi .\) This portion of the graph is shown here:

Step-by-Step Solution

Answer

The arc length is \( \pi \sqrt{29} \).

Step 1: Formula for Arc Length

Recall that the arc length of a vector function \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) from \( t = a \) to \( t = b \) is given by the integral \( L = \int_a^b \| \mathbf{r}'(t) \| \, dt \). Here, \( \| \mathbf{r}'(t) \| \) represents the magnitude of the derivative of \( \mathbf{r}(t) \).

Step 2: Find the Derivative

Compute the derivative of \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \). The derivative is \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \).

Step 3: Magnitude of the Derivative

Calculate the magnitude of \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \). Using the magnitude formula, \( \| \mathbf{r}'(t) \| = \sqrt{(2 \cos t)^2 + 5^2 + (-2 \sin t)^2} = \sqrt{4 \cos^2 t + 25 + 4 \sin^2 t} \).

Step 4: Simplify the Expression

Simplify the expression inside the square root: \( 4 \cos^2 t + 4 \sin^2 t = 4(\cos^2 t + \sin^2 t) = 4 \times 1 = 4 \). Thus, the magnitude becomes \( \| \mathbf{r}'(t) \| = \sqrt{4 + 25} = \sqrt{29} \).

Step 5: Calculate the Arc Length

Substitute the simplified magnitude into the arc length formula: \( L = \int_0^\pi \sqrt{29} \, dt \). Since \( \sqrt{29} \) is a constant, the integral can be simplified to \( L = \sqrt{29} \int_0^\pi dt = \sqrt{29} [t]_0^\pi = \sqrt{29} (\pi - 0) = \pi \sqrt{29} \).

Key Concepts

Calculus 3

Calculus 3 is an advanced level of calculus that expands on the foundational concepts taught in earlier calculus courses. It introduces three-dimensional space and adds more complexity to mathematical calculations. Students learn to work with vector functions, which describe curves in this three-dimensional space.

• Understanding vector functions is crucial as they describe the motion along a path, offering a comprehensive way to model real-life scenarios, such as the flight path of an airplane or the trajectory of a satellite.
• Students also become familiar with surfaces and curves, partial derivatives, multiple integrals, and vector calculus, which broadens the scope of problems they can solve.

These sophisticated tools are a step up from Calculus 1 and 2 and offer valuable skills for tackling more challenging mathematical problems.

Arc Length Formula

The arc length formula provides a method to determine the length of a curve defined by a vector function. In calculus, this is particularly useful for vector functions defined in multiple dimensions. The arc length of a vector function \[ L = \int_a^b \| \mathbf{r}'(t) \| \, dt \]is calculated by integrating the magnitude of the function's derivative over a specified interval.

• \(L\) represents the total distance covered by the curve from \(t = a\) to \(t = b\).
• This method helps in finding the actual curve length in multi-dimensional space, rather than just a straight-line distance.

The process requires finding the derivative of the vector function and then calculating its magnitude, which will be used in the integration to find the arc length.

Derivative of a Vector Function

To find the arc length of a curve, understanding the derivative of a vector function is essential. A vector function such as \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\) describes a curve in space. Its derivative, \(\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle\), provides us with the velocity vector at any point \(t\).

• This derivative indicates how rapidly and in what direction the curve is changing at each instant.
• By computing \(\mathbf{r}'(t)\), we lay the groundwork for determining the magnitude, another requisite component of the arc length calculation.

The derivative's role is fundamental; without it, you cannot proceed to the integration step needed for the arc length.

Magnitude of a Vector

The magnitude of a vector is a measure of its length or size. For a derivative vector function \(\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle\), its magnitude is
\[ \| \mathbf{r}'(t) \| = \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2} \]This magnitude is a scalar quantity that represents the speed or rate of change of the position with respect to the parameter \(t\).

• Calculating the magnitude involves applying the square root to the sum of the squares of the components of the derivative.
• For arc length problems, it's vital as it becomes the integrand in the arc length formula.

This integral of the magnitude over an interval yields the length of the path traversed by the vector function. Thus, understanding and calculating vector magnitude is pivotal to solving problems involving curves and distances in vector calculus.