In Exercises 43 - 46 , use a CAS to perform the following...

In Exercises $43 - 46 ,$ use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array} { l } { \text { a. Find } d s = | \mathbf { v } ( t ) | d t \text { for the path } \mathbf { r } ( t ) = g ( t ) \mathbf { i } + h ( t ) \mathbf { j } + k ( t ) \mathbf { k } \text { . } } \\ { \text { b. Express the integrand } f ( g ( t ) , h ( t ) , k ( t ) ) | \mathbf { v } ( t ) | \text { as a function of the parameter } t . } \\ { \text { c. Evaluate } \int _ { C } f d s \text { using Equation } ( 2 ) \text { in the text. } } \end{array} $$ $$ \begin{array} { l } { f ( x , y , z ) = \sqrt { 1 + x ^ { 3 } + 5 y ^ { 3 } } ; \quad \mathbf { r } ( t ) = t \mathbf { i } + \frac { 1 } { 3 } t ^ { 2 } \mathbf { j } + \sqrt { t } \mathbf { k } } \\ { 0 \leq t \leq 2 } \end{array} $$

In Exercises $43 - 46 ,$ use a CAS to perform the following steps to evaluate the line integrals.
$$
\begin{array} { l } { \text { a. Find } d s = | \mathbf { v } ( t ) | d t \text { for the path } \mathbf { r } ( t ) = g ( t ) \mathbf { i } + h ( t ) \mathbf { j } + k ( t ) \mathbf { k } \text { . } } \\ { \text { b. Express the integrand } f ( g ( t ) , h ( t ) , k ( t ) ) | \mathbf { v } ( t ) | \text { as a function of the parameter } t . } \\ { \text { c. Evaluate } \int _ { C } f d s \text { using Equation } ( 2 ) \text { in the text. } } \end{array}
$$
$$
\begin{array} { l } { f ( x , y , z ) = \sqrt { 1 + x ^ { 3 } + 5 y ^ { 3 } } ; \quad \mathbf { r } ( t ) = t \mathbf { i } + \frac { 1 } { 3 } t ^ { 2 } \mathbf { j } + \sqrt { t } \mathbf { k } } \\ { 0 \leq t \leq 2 } \end{array}
$$

Key Concepts

Tangent Vectors and Speed

The tangent vector, obtained by differentiating the position vector r(t), provides the direction of the curve at any given point. Its magnitude, or speed, |v(t)|, is used to compute the arc length differential. This concept links the geometry of the curve with the analytical process of integration.

Arc Length Differential (ds)

The arc length differential, expressed as ds = |v(t)| dt where v(t) is the derivative of the vector function, represents an infinitesimally small segment of the curve. This concept is crucial because it quantifies how the parameter t scales with actual distance along the curve, which is necessary for correctly weighting the function being integrated.

Substitution into the Integral

This key idea involves replacing the function and differential in the line integral with their expressions in terms of the parameter t. By substituting f(x, y, z) with f(r(t)) and ds with |r'(t)| dt, the line integral is transformed into a standard integral with respect to t, making it manageable with conventional integration techniques or computer algebra systems.

Line Integrals of Scalar Functions

This concept involves integrating a scalar function along a curve in space, usually by summing up contributions of the function weighted by small lengths (ds) along the path. It is used in various applications such as calculating mass or work when density or force is distributed along a curve.

Parameterization of Curves

Parameterization refers to representing a curve using a vector-valued function of a single parameter, typically denoted as r(t). This allows one to express the coordinates of points on the curve in terms of t, facilitating the conversion of geometric integrals into parameter-space integrals.

Transcript

00:01 Okay, what we want to do is we are given a function f of xyz, which is equal to the square root of 1 plus x, cube, plus 5y cubed.

00:16 R of t is equal to t i plus 1 third t square j plus a square root or t to 1 half k.

00:28 And t is going to go between zero and two inclusive.

00:33 And so we want to find v of t, or we actually want to find the magnitude of v of t.

00:44 And before i find the magnitude of v of t, i've got to find v of t, which is the derivative of r.

00:53 And so this is going to be i plus two thirds t, j, plus one -half t to the negative one -half k.

01:02 And so the magnitude is going to be one squared plus that two -thirds squared, two -thirds t squared, plus the one -half t to the negative one -half, and i'm going to square that.

01:19 And so that becomes the square root of 1 plus 4 9th t squared plus 1 over 4 t.

01:35 And so ds is defined to be that magnitude, which is the square root of 1 plus 4 nt squared plus 1 over 4 t, dt.

01:53 And now we want to set up our integral.

01:56 And then we're going to evaluate our integral using a calculator.

02:01 So that integral goes from zero to two.

02:05 It's going to be our function, though transferred into terms of t.

02:11 So that is going to be the square root of 1 plus, and x is t.

02:20 So that's going to be a t -cubed.

02:25 Plus five times y cubed and y is one third t squared.

02:32 So this becomes one over 27 t to the six.

02:37 And then we're going to multiply that by v of t, which is one plus four ninth t squared plus one over four t...