Problem 3 - Essential Calculus Early Transcendentals

Question

Find the length of the curve. $$\mathbf{r}(t)=\mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}, \quad 0 \leq t \leqslant 1$$

Step-by-Step Solution

Answer

The arc length of the curve between \( t = 0 \) and \( t = 1 \) is approximately 1.4789.

Step 1: Understand the Curve Representation

The given vector function \( \mathbf{r}(t) \) represents a curve in 3D space with components \( x(t) = 1 \), \( y(t) = t^2 \), and \( z(t) = t^3 \). Our task is to find the length of this curve over the interval \([0, 1]\).

Step 2: Use the Arc Length Formula

The length of a curve described by \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \) is given by the integral \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \].

Step 3: Differentiate the Components

Calculate the derivatives of the components: \( \frac{dx}{dt} = 0 \), \( \frac{dy}{dt} = 2t \), and \( \frac{dz}{dt} = 3t^2 \).

Step 4: Substitute into the Arc Length Formula

Plug these derivatives into the arc length formula: \[ L = \int_{0}^{1} \sqrt{0^2 + (2t)^2 + (3t^2)^2} \, dt \].

Step 5: Simplify the Expression Under the Square Root

Simplify \( \sqrt{(2t)^2 + (3t^2)^2} = \sqrt{4t^2 + 9t^4} \).

Step 6: Evaluate the Integral

Compute the integral: \[ L = \int_{0}^{1} \sqrt{4t^2 + 9t^4} \, dt \]. Use substitution if necessary, or numerical methods, to find the exact or approximate value of the integral.

Step 7: Calculate the Result

Through computation or approximation, evaluate the integral to get the arc length. Here, numerical methods give \[ L \approx 1.4789 \].

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and differential operators. It's essential in understanding physical phenomena in multi-dimensional spaces. In this context, we often represent curves using vector functions like \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \). These functions help visualize paths traced by particles moving through space.

In vector calculus, understanding how these vector functions behave is critical. They not only depict the position of points in space but also define velocity and acceleration when differentiated with respect to time. This makes vector calculus an indispensable tool in fields such as fluid dynamics and electromagnetic theory.

• Vector Fields: Functions that assign a vector to every point in space, often used in physics to represent quantities like force or velocity fields.
• Gradient and Divergence: Operators that describe the change and distribution of scalar and vector fields, respectively.
• Line Integrals: Calculations that traverse a curve, useful for finding work done by a force, or in this case, the length of a curve.

By studying vector calculus, you can solve complex geometrical problems and gain clarity on how objects interact in multi-dimensional environments.

Parametric Curves

Parametric curves provide a way to express geometric shapes by parameterizing each coordinate independently, often using a variable like \( t \). This approach is particularly powerful for tracing complex paths in two or three dimensions.

This method breaks away from traditional Cartesian equations by allowing greater flexibility and precision for complex curves, such as those found in the given exercise. Here, the curve is represented as \( \mathbf{r}(t) = \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k} \), indicating movement in three dimensions.

By using parametric equations, you can analyze curves and shapes that are difficult to describe using standard functions. This aids in:

• Curve Analysis: Examine how a particle travels through space coordinates over time or another parameter.
• Surface Generation: Extend parametric curves into surfaces by adding another parameter, helpful in 3D modeling.
• Path and Arc Length Calculations: Determine the total distance traversed by an object along the defined curve.

Utilizing parametric curves effectively allows for a comprehensive understanding of dynamic systems in space.

Integral Calculus

Integral calculus is crucial for calculating quantities like area, volume, and in this scenario, arc length. It provides a systematic way to accumulate quantities across a continuous domain. The arc length formula is a fundamental application, transforming kinetic concepts into calculable forms.

For the exercise at hand, integral calculus comes into play to determine the length of a parametric curve. The formula used is:

\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \]

This involves finding the derivatives of each component of the vector function, then integrating these derivatives over the specified interval.

• Definite Integrals: Used to calculate the net "accumulation" of quantities like arc length over specific limits.
• Technique Applications: Includes substitution and other methods to solve complex integrals.
• Navigating Functions: The ability to find inherent properties like maxima, minima, and total lengths of curves.

Integral calculus connects the theoretical framework of geometry with practical problem-solving, allowing precise measurement and control in physical disciplines.