Problem 43 - Thomas Calculus

Question

In Exercises \(43 - 46 ,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array} { l } { \text { a. Find } d s = | \mathbf { v } ( t ) | d t \text { for the path } \mathbf { r } ( t ) = g ( t ) \mathbf { i } + h ( t ) \mathbf { j } + k ( t ) \mathbf { k } \text { . } } \\ { \text { b. Express the integrand } f ( g ( t ) , h ( t ) , k ( t ) ) | \mathbf { v } ( t ) | \text { as a function of the parameter } t . } \\ { \text { c. Evaluate } \int _ { C } f d s \text { using Equation } ( 2 ) \text { in the text. } } \end{array} $$ $$ \begin{array} { l } { f ( x , y , z ) = \sqrt { 1 + 30 x ^ { 2 } + 10 y } ; \quad \mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + 3 t ^ { 2 } \mathbf { k } } \\ { 0 \leq t \leq 2 } \end{array} $$

Step-by-Step Solution

Answer

The value of the line integral is \( \frac{326}{3} \).

Step 1: Compute the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is given by the derivative of \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 3t^2 \mathbf{k} \) with respect to \( t \). Compute the derivative: \[ \mathbf{v}(t) = \frac{d}{dt} (t \mathbf{i} + t^2 \mathbf{j} + 3t^2 \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + 6t \mathbf{k}. \]

Step 2: Compute the Magnitude of the Velocity Vector

The magnitude \( |\mathbf{v}(t)| \) is found by taking the square root of the sum of the squares of its components: \[ |\mathbf{v}(t)| = \sqrt{1^2 + (2t)^2 + (6t)^2} = \sqrt{1 + 4t^2 + 36t^2} = \sqrt{1 + 40t^2}. \]

Step 3: Find the Element of Arc Length, ds

The element of arc length \( ds \) is given by \( ds = |\mathbf{v}(t)| \, dt \). Substitute the expression from Step 2: \[ ds = \sqrt{1 + 40t^2} \, dt. \]

Step 4: Express the Integrand as a Function of t

The function \( f(g(t), h(t), k(t)) = \sqrt{1 + 30x^2 + 10y} \) where \( x = t, y = t^2, z = 3t^2 \). Substitute these into \( f \): \[ f(t, t^2, 3t^2) = \sqrt{1 + 30t^2 + 10t^2} = \sqrt{1 + 40t^2}. \]

Step 5: Set Up the Line Integral

Substitute the expressions for \( f(g(t), h(t), k(t)) \) and \( ds \) into the line integral \( \int_C f \, ds \). This results in: \[ \int_0^2 \sqrt{1 + 40t^2} \times \sqrt{1 + 40t^2} \, dt = \int_0^2 (1 + 40t^2) \, dt. \]

Step 6: Evaluate the Line Integral

Evaluate the integral \( \int_0^2 (1 + 40t^2) \, dt \). First, compute the antiderivative: \[ \int (1 + 40t^2) \, dt = t + \frac{40}{3}t^3 + C. \] Evaluate from 0 to 2: \[ \left[ t + \frac{40}{3}t^3 \right]_0^2 = \left(2 + \frac{40}{3}(2^3)\right) - \left(0 + \frac{40}{3}(0^3)\right) = 2 + \frac{320}{3}. \] Calculate the result: \[ 2 + \frac{320}{3} = \frac{326}{3}. \]

Key Concepts

Velocity Vector

When dealing with line integrals, understanding the velocity vector is crucial. The velocity vector, often denoted as \( \mathbf{v}(t) \), plays a key role as it represents the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). This differentiation provides the instantaneous rate of change of the position vector, essentially giving us the direction and speed at each point along the curve.

For example, in the problem, the position vector is given by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 3t^2 \mathbf{k} \). By differentiating this with respect to \( t \), we obtain the velocity vector:
\[ \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} + 6t \mathbf{k}. \]
This vector tells us how each of the components along the \( i, j, \) and \( k \) directions changes as \( t \) varies.

• The component \( \mathbf{i} \) remains constant.
• The \( \mathbf{j} \)-component increases linearly with \( 2t \).
• The \( \mathbf{k} \)-component rises steeper, as \( 6t \), indicating quicker change in that direction.

Arc Length

Arc length is the measure of the distance along a curve, and in line integrals, an element of arc length \( ds \) is a small segment of this curve. It serves as the differential over which the function and path are integrated.

The calculation of \( ds \) usually involves finding the magnitude of the velocity vector \( |\mathbf{v}(t)| \), because this magnitude gives the 'speed' at which we traverse the curve.

For our specific example, the magnitude is:
\[ |\mathbf{v}(t)| = \sqrt{1 + 4t^2 + 36t^2} = \sqrt{1 + 40t^2}. \]
The expression for \( ds \) thus becomes \( ds = |\mathbf{v}(t)| dt = \sqrt{1 + 40t^2} dt \). Evaluating this across the desired interval helps in finding the total length over which integration is performed. This formulation ensures that the contribution of each part of the curve to the integral can be calculated accurately by considering the curve's geometry.

Parameterization

Parameterization involves expressing a curve or geometric figure by assigning a parameter, typically \( t \). This process is mostly about breaking complex shapes into simpler parts that depend on a singular variable.

In the context of line integrals, parameterization simplifies how we deal with paths by converting spatial coordinates \( x, y, z \) into functions of \( t \). Here, our original position vector is:
\[ \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 3t^2 \mathbf{k}. \]
This parameterization of \( x = t, y = t^2, \) and \( z = 3t^2 \) lets us replace these variables when setting up integrals. It allows the transformation of the function we wish to integrate over the path into a function of \( t \).

• This conversion simplifies the function \( f(x, y, z) = \sqrt{1 + 30x^2 + 10y} \) to \( f(t, t^2, 3t^2) = \sqrt{1 + 40t^2} \).
• Once parameterized, the integral becomes much more manageable as it compacts the scenario into a single-variable calculus problem.