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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 ) = \left( 1 + \frac { 9 } { 4 } z ^ { 1 / 3 } \right) ^ { 1 / 4 } ; \quad \mathbf { r } ( t ) = ( \cos 2 t ) \mathbf { i } + ( \sin 2 t ) \mathbf { j } + } \\ { t ^ { 5 / 2 } \mathbf { k } , \quad 0 \leq t \leq 2 \pi } \end{array}

$$

A. $d s=\sqrt{4+\frac{25}{4} t^{3}} d t$

B. $f(g(t), h(t), k(t))|v(t)|=\left(1+\frac{9}{4}\left(t^{\frac{5}{6}}\right)\right)^{\frac{1}{4}}\left(\sqrt{4+\frac{25}{4} t^{3}}\right)$

C. $\int_{C} f \cdot d s \approx 172.441$

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okay. What we want to go ahead and do is we're given a function of X, y and Z, which is equal to one plus 9/4 Z to the 1/3 all raised to the 1/4 and R T is equal to co sign of to t I plus sign of to t j plus t to the five half's Kay and T is going to go from 0 to 2 pi inclusive. And so what we want to eventually do is we want to write the integral of the function X, y and Z with respect to d s all in terms of tea. And we also want to go to her calculator and integrate that. And so we know that in order find d s, we've got to find this function vey of tea, which is a derivative of our tea. So this is gonna be native to sign of to t I plus to co sign of to t J plus five over to T to three halves. Okay. And so, um, the magnitude of e of t is the square root of native to signing to tea, and we're going to square that plus to co sign a two tea and we're being square that, um plus, um, 25/4, um, T Cube, because we square that that z component as well. And so this becomes the square root of four. Class 25/4 t cubed S O. D s is that square root which is square root of four plus 25 over fourty cube TT. So now we're integrate from 0 to 2 pi, and now we're going to change our function in terms of tea as well. And so this will be one plus nine force. And then we know that Z is to you the 5/2. So this will be t to the 56 that is all raised to the 1/4 times the square root of four plus 25/4 t cubed. And we are going to integrate that with respect to tea. So now let's go ahead and go to our calculator. Um, and I have a 84 um, online emulator. And so how I use this is, um I get in my math button, I go down to function in it, which sends for function integration, and I still like that. So we're gonna go from zero thio to pie, and then the function is and I have to insert some additional parentheses in here. Um, one plus we do, uh, one plus, um, that nine divided by four. Um, And then it's gonna be we're gonna integrate with respect to X X rays to the 56 home. And then that, of course, is raised to the 14 Um And so instead of parentheses, that's gonna be raised an additional 1/4 power, and then we come down, Um, and then that is gonna be multiplied by the square root of that four plus 25 divided by four X to the third power. Um, we get out of our square root, and that is integrating with respect to X. And so hopefully it will come back. I know for my emulators, it takes it a tad bit for it to come back with the answer. And so 1 72.44 eso so that integration is equal to 1 72.44

University of Central Arkansas