In Exercises 19-22, find the work done by F over the curve...

In Exercises $19-22,$ find the work done by $F$ over the curve in the direction of increasing $t .$ \begin{equation} \begin{array}{l}{\mathbf{F}=2 \mathrm{yi}+3 x \mathbf{j}+(x+y) \mathbf{k}} \\ {\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(t / 6) \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array} \end{equation}

In Exercises $19-22,$ find the work done by $F$ over the curve in the direction of increasing $t .$
\begin{equation}
\begin{array}{l}{\mathbf{F}=2 \mathrm{yi}+3 x \mathbf{j}+(x+y) \mathbf{k}} \\ {\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(t / 6) \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array}
\end{equation}

Key Concepts

Line Integral in Vector Fields

This is a method used to calculate the work done by a force field along a given curve. It involves integrating the dot product of the force vector and the differential element of the path over the curve. The line integral converts the problem into an integral over the parameter that defines the curve, representing the cumulative effect of the force along the path.

Parameterization of Curves

Parameterization is the process of expressing a curve using a single variable, typically denoted as t, which simplifies the evaluation of integrals along the curve. Instead of working with a function in terms of x, y, and z independently, the curve is described as r(t) with its components being functions of t, thus reducing the complexity of computing the line integral.

Dot Product

The dot product is a fundamental operation in vector calculus that computes the scalar projection of one vector onto another. In the context of a line integral, it is used to combine the force vector and the differential path vector, ensuring that only the component of the force that does work along the direction of the path is considered.

Definite Integral

Once the integrand is established by taking the dot product of the force field with the derivative of the parameterized path, the work done is found by evaluating a definite integral over the interval of the parameter. This process sums up the contributions of the force along the entire curve as defined by the parameter's limits.

Transcript

00:01 All right, so they want us to find the work done on this curve here.

00:08 So we can just use the formula up top there.

00:12 Now, the first thing i'm going to do is figure out what is f when we plug r into it.

00:18 So this thing right here.

00:21 So f of rt, this is going to be.

00:24 I'm going to write this in the bracket notation as opposed to the ijk notation.

00:29 I just like the bracket notation better.

00:31 That's all.

00:32 Right, so remember this is supposed to be x, y, and this is z.

00:37 So this is going to be 2 sine of t, and then 3 cosine of t.

00:46 And then over here, that's going to be a cosine t plus sine t.

00:58 And now we need to also find the derivative of our position.

01:03 So our t is going to be so that would be negative sine t then we get cosine t and then we get one -sixth here all right now we need to take the dot product of this so this would be zero to two pi and then the dot product of those so remember we just do this component -wise so this is going to be two or actually not two negative two negative two sign squared t and then plus 3 cosine squared t and then plus 1 .6 cosine t plus 1 6th sine t d t and now remember to integrate sine squared and cosine squared we can use the power reducing formulas for each of them so sine so i'll actually write these in the top corner over here so sine squared of t should be one minus cosine 2t all over 2.

02:18 And the way i remember it is when i plug in 0 here, because it's either 1 plus or 1 minus cosine.

02:25 So if i plug in 0, i should get 0 over here.

02:28 So 1 minus cosine of 0, which is 1 minus 1 to 0.

02:32 So yeah, that's that.

02:33 And then cosine squared is just the other one.

02:35 So 1 plus cosine 2t...

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