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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}

$$

A. $d s=\sqrt{1+\frac{4 t^{2}}{9}+\frac{1}{4 t}} d t$

B. $f(g(t), h(t), k(t))|v(t)|=\left(\sqrt{1+t^{3}+\frac{5}{27} t^{6}}\right)\left(\sqrt{1+\frac{4 t^{2}}{9}+\frac{1}{4 t}}\right)$

C. $\int_{C} f \cdot d s \approx 5.8080$

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okay. What we want to do is we're given a function f of X y Z, which is equal to the square root of one plus X cube plus five white cube R t is equal to t I plus 1 30 square J plus the square root, or T, to the 1/2 K and T is going to go between zero and two inclusive. And so we want to, um, find the of tea, or we actually want to find the magnitude of V of tea. And before I find the magnitude of e of t, I've got to find V of tea, which is a derivative of our and so this is gonna be I plus 2/3 t j plus one have tea to the negative 1/2 K, and so the magnitude is gonna be one squared plus that 2/3 squared 2/3 t squared, plus the 1/2 t to the negative 1/2 of the most square. That, and so that becomes the square root of one plus four nights T squared, plus 1/4 t. And so D s is defined to be that magnitude. Um, which is the square root of one plus four ninth T squared plus one over fourty t t. And now we want to set up our integral and then and then we're gonna evaluate are integral using a calculator. So that inner goal, um, goes from 0 to 2. It's gonna be or function, though transferred into terms of tea. So that is gonna be the square root of one plus and X is t So that's gonna be a T cubed, um, plus five times Why cubed and why is 1 30 squared to this becomes 1/27 t to the sixth, and then we're gonna multiply that by V of tea, which is one plus four nights t squared, plus 1/4 tea. And we're gonna integrate with respect to t. So now what we're gonna do is go ahead and put this in our calculator or online calculator system. And so what we're gonna do is I'm gonna go ahead and switch. Um, I have a t I 84 emulator fun. And so what I need to do is on this, um, I hit the math button. I go down to where it says function into which is function integration, and I now compote in my lower limit and then my upper limit of two. And then now I can actually put in, um, my into grand, and that is gonna be the square root of one. Plus an unfortunate we're gonna have to go back to excess here. So eggs raised to the third power, and then that is gonna be, um plus, um, five divided by 27. And then this is gonna be times and X raised to the sixth. And so now we need to get out of, um, that first square root, and that is gonna be multiplied by a second square root, which is that one plus that four nights times x squared. And then we also have, um, plus that one divided by that four X. And I had to go ahead and put, um, that four x in parentheses. Also, it might be easier if you have ah, a nicer emulator system. And so we're integrating with respect to x him But C minus still working, and that is continuing to work. There we go. There we go to took mine a little bit longer. So that is 5.808 Um, didn't like the fact that it took so long, but that's okay. So 5.8 08 now, if you have a T I 89 or a T I inspire, that might work a little bit better, or some of the other online options might work a little bit better as well.

University of Central Arkansas