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Find the length of the curve with the given vector equation. $$ \mathbf{r}(t)=\sqrt{6} t^{2} \mathbf{i}+\frac{2}{3} t^{3} \mathbf{j}+6 t \mathbf{k} ; 3 \leq t \leq 6 $$

Short Answer

Expert verified
The length of the curve is 117.

Step by step solution

01

Understand the problem

We need to find the length of the parametric curve defined by the vector function \( \mathbf{r}(t) = \sqrt{6} t^2 \mathbf{i} + \frac{2}{3} t^3 \mathbf{j} + 6t \mathbf{k} \) for \( t \) ranging from 3 to 6.
02

Identify the formula for arc length

The arc length \( L \) of a curve defined by a vector function \( \mathbf{r}(t) \) from \( t = a \) to \( t = b \) is given by the integral \( L = \int_a^b \| \mathbf{r}'(t) \| \, dt \), where \( \mathbf{r}'(t) \) is the derivative of \( \mathbf{r}(t) \).
03

Find the derivative \( \mathbf{r}'(t) \)

Calculate the derivative of each component of the vector function: \( \frac{d}{dt}(\sqrt{6}t^2) = 2\sqrt{6}t \), \( \frac{d}{dt}(\frac{2}{3}t^3) = 2t^2 \), and \( \frac{d}{dt}(6t) = 6 \). So, \( \mathbf{r}'(t) = 2\sqrt{6}t \mathbf{i} + 2t^2 \mathbf{j} + 6 \mathbf{k} \).
04

Compute the magnitude \( \| \mathbf{r}'(t) \| \)

The magnitude of \( \mathbf{r}'(t) \) is calculated as \( \| \mathbf{r}'(t) \| = \sqrt{(2\sqrt{6}t)^2 + (2t^2)^2 + 6^2} \). After simplifying, this becomes \( \| \mathbf{r}'(t) \| = \sqrt{24t^2 + 4t^4 + 36} \).
05

Simplify under the square root

Combine and simplify the terms under the square root: \( 4t^4 + 24t^2 + 36 \) can be rewritten as \( (2t^2 + 6)^2 \). Thus, \( \| \mathbf{r}'(t) \| = \sqrt{(2t^2 + 6)^2} = 2t^2 + 6 \) because the expression is squared.
06

Integrate the magnitude over the interval

Integrate \( 2t^2 + 6 \) from 3 to 6: \[ L = \int_3^6 (2t^2 + 6) \, dt = \left[ \frac{2}{3}t^3 + 6t \right]_3^6 \].
07

Evaluate the integral

Calculate the definite integral: First find \( \left. \frac{2}{3}t^3 + 6t \right|_3^6 = \left( \frac{2}{3}(6)^3 + 6(6) \right) - \left( \frac{2}{3}(3)^3 + 6(3) \right) \). This simplifies to \( 156 - 39 = 117 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Calculus
Vector calculus is an essential field of mathematics that deals with vector fields and operations on vectors, usually in three-dimensional space. It combines the concepts of calculus and vectors to aid in solving geometric and physical problems.
Vector calculus has several applications, including physics, engineering, and computer graphics. In the exercise given above, vector calculus is used to calculate the arc length of a vector function. This involves understanding the structure and behavior of the vector field defined by the given equation.
The core components often include:
  • Vector operations like addition, subtraction, and scalar multiplication.
  • Differentiation and integration of vector functions - both crucial for arc length calculations.
  • Understanding vector magnitude and direction.

By leveraging these concepts, the exercise allows the calculation of the curve's length, a common problem type when examining paths through space in vector calculus.
Parametric Curves
Parametric curves provide a means to describe a path using a set of equations called parameterized equations. Typically, we use a parameter, like \( t \), to express both the \( x \) and \( y \) coordinates of a point on a curve. This approach makes it easy to explore complex curves in vector calculus.
With parametric curves, distinct equations define the components of the vector independently based on a parameter. In the exercise, \( \mathbf{r}(t) = \sqrt{6} t^{2} \mathbf{i} + \frac{2}{3} t^{3} \mathbf{j} + 6 t \mathbf{k} \) gives positions on the curve as \( t \) changes. The curve is expressed in terms of \( t \) so that each value of \( t \) corresponds to a specific point.
Advantages of using parametric equations include:
  • Handling more complex curves that cannot be represented by \( y = f(x) \).
  • Providing a natural approach to 3D representation.
  • Flexibility to trace curves more accurately with separate control over changes in each axis.

In this exercise, by calculating the derivatives and using them in integrals, we find precise measurements of curves.
Integral Calculus
Integral calculus is a branch that focuses on integration, which is the calculation of areas, volumes, and other quantities through sums of infinitely small data points. In our exercise, integral calculus finds application in measuring the arc length of a curve defined by vectors.
The formula for arc length \( L \) associated with a parametric curve is: \[ L = \int_a^b \| \mathbf{r}'(t) \| \, dt \] This integral, based on the magnitude of the derivative of the vector function over an interval \([a, b]\), calculates precise arc lengths.
Main uses in the context of this problem include:
  • Finding total path length a particle travels in space.
  • Summing up infinitesimal segments of a parameterized curve to obtain overall curvature.
  • Offering an approach to calculate physical properties like work and center of mass along curves.

The step-by-step solution demonstrates the process, from differentiation, simplifying under the square root, and evaluating definite integrals to reach the final answer.
Vector Functions
Vector functions are functions where the domain consists of one-dimensional input, typically time \( t \), and the range is multidimensional vectors. These functions are used to describe curves and motions in space, aiding in the mathematical description of dynamics.
A vector function \( \mathbf{r}(t) \) is typically written as an expression for the \( i, j, \) and \( k \) components, as seen in the exercise: \[ \mathbf{r}(t) = \sqrt{6} t^{2} \mathbf{i} + \frac{2}{3} t^{3} \mathbf{j} + 6 t \mathbf{k} \] This notation clearly designates each spatial dimension separately.
The properties and usage of vector functions include:
  • Depicting paths or trajectories of points moving through space over time.
  • Allowing the representation of quantities that have magnitude and direction.
  • Facilitating operations such as differentiation and integration, essential for problems like arc length and surface calculations.

In the given solution, integration of the magnitude of the derivative \( \mathbf{r}'(t) \) demonstrates how vector functions facilitate the computation of distances and geometric properties of paths in physics and engineering fields.

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