Problem 41 - Advanced Engineering Mathematics

Question

In Problems, find the length of the curve traced by the given vector function on the indicated interval. $$ \mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}+\text { ct } \mathbf{k} ; 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

The length of the curve is \( 2\pi \sqrt{a^2 + c^2} \).

Step 1: Understand the Problem

We need to find the length of the curve traced by the vector function \( \mathbf{r}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} + ct \mathbf{k} \) over the interval \( 0 \leq t \leq 2\pi \). This problem requires applying the formula for the arc length of a vector function.

Step 2: Recall the Arc Length Formula

The arc length \( L \) of a vector function \( \mathbf{r}(t) \) over an interval \([a, b]\) is given by the integral: \[ L = \int_{a}^{b} \left| \frac{d\mathbf{r}}{dt} \right| \, dt \]where \( \left| \frac{d\mathbf{r}}{dt} \right| \) is the magnitude of the derivative of \( \mathbf{r}(t) \).

Step 3: Find the Derivative of the Vector Function

Calculate the derivative of the vector function \( \mathbf{r}(t) \) with respect to \( t \):\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(a \cos t \mathbf{i} + a \sin t \mathbf{j} + ct \mathbf{k}) = -a \sin t \mathbf{i} + a \cos t \mathbf{j} + c \mathbf{k} \]

Step 4: Calculate the Magnitude of the Derivative

The magnitude of the derivative \( \frac{d\mathbf{r}}{dt} = -a \sin t \mathbf{i} + a \cos t \mathbf{j} + c \mathbf{k} \) is found using:\[ \left| \frac{d\mathbf{r}}{dt} \right| = \sqrt{(-a \sin t)^2 + (a \cos t)^2 + c^2} = \sqrt{a^2 \sin^2 t + a^2 \cos^2 t + c^2} \]Since \( \sin^2 t + \cos^2 t = 1 \), this simplifies to:\[ \left| \frac{d\mathbf{r}}{dt} \right| = \sqrt{a^2 + c^2} \]

Step 5: Integrate to Find the Arc Length

Substitute \( \left| \frac{d\mathbf{r}}{dt} \right| = \sqrt{a^2 + c^2} \) into the arc length formula:\[ L = \int_{0}^{2\pi} \sqrt{a^2 + c^2} \, dt \]Since \( \sqrt{a^2 + c^2} \) is a constant, the integral simplifies to:\[ L = \sqrt{a^2 + c^2} \int_{0}^{2\pi} 1 \, dt = \sqrt{a^2 + c^2} [t]_{0}^{2\pi} = \sqrt{a^2 + c^2} (2\pi - 0) \]Thus, the arc length is:\[ L = 2\pi \sqrt{a^2 + c^2} \]

Step 6: Present the Final Answer

The length of the curve traced by the vector function \( \mathbf{r}(t) \) is \( 2\pi \sqrt{a^2 + c^2} \).

Key Concepts

Vector Calculus

Vector calculus is a powerful tool in mathematics, particularly in physics and engineering, which allows us to analyze and compute various properties of vector functions. A vector function assigns a vector to each point in its domain, forming a curve in three-dimensional space. The example provided, \( \mathbf{r}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} + ct \mathbf{k} \), is a typical vector function describing a helix.
Understanding how vector calculus works allows us to dissect this curve to compute properties like arc length, which is the total 'distance' traversed along the curve.
Fundamental to vector calculus are operations like differentiation and integration, which help in finding the rate of change and cumulative attributes of these functions.

Derivative of Vector Function

To solve problems involving vector functions, taking their derivative with respect to their parameter, such as time \( t \), is crucial. The derivative gives a new vector function, often representing the tangent to the curve at any given point. For the function \( \mathbf{r}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} + ct \mathbf{k} \), the derivative is computed as:

• \( \frac{d}{dt} (a \cos t \mathbf{i} + a \sin t \mathbf{j} + ct \mathbf{k}) \)
• This results in \( -a \sin t \mathbf{i} + a \cos t \mathbf{j} + c \mathbf{k} \).

This new function \( \frac{d\mathbf{r}(t)}{dt} \) shows how the curve changes its direction and length per unit time. Differentiating vector functions is just like differentiating scalar functions, treating each component separately but keeping the vector nature intact.

Magnitude of Vector

Once we have the derivative of a vector function, we need to determine its magnitude, or length, to help find the arc length. The magnitude is essentially the "size" of the vector, calculated by taking the square root of the sum of the squares of each component.
For example, if \( \frac{d\mathbf{r}}{dt} = -a \sin t \mathbf{i} + a \cos t \mathbf{j} + c \mathbf{k} \), its magnitude is:
\[\left| \frac{d\mathbf{r}}{dt} \right| = \sqrt{(-a \sin t)^2 + (a \cos t)^2 + c^2}\]
The trigonometric identity \( \sin^2 t + \cos^2 t = 1 \) simplifies the expression to:
\[\left| \frac{d\mathbf{r}}{dt} \right| = \sqrt{a^2 + c^2}\]Calculating this magnitude helps us understand the rate at which the curve is traced out.

Integral Calculus

Integral calculus comes to play when finding the total length of the curve described by the vector function. After determining the magnitude of the derivative, we integrate this constant over the given interval \( [0, 2\pi] \).
The integral needed is:
\[L = \int_{0}^{2\pi} \sqrt{a^2 + c^2} \, dt\]
Since \( \sqrt{a^2 + c^2} \) is constant, the integral simplifies to:
\[L = \sqrt{a^2 + c^2} \int_{0}^{2\pi} 1 \, dt = \sqrt{a^2 + c^2} [t]_{0}^{2\pi} = \sqrt{a^2 + c^2} (2\pi - 0)\]
This yields the arc length:
\[ L = 2\pi \sqrt{a^2 + c^2} \]
Integral calculus allows for accumulation, measuring the extent of curves and areas, proving essential in physics and engineering where analysis over intervals is frequent.