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frutty [35]
3 years ago
11

Consider F and C below. F(x, y, z) = y2 sin(z) i + 2xy sin(z) j + xy2 cos(z) k C: r(t) = t2 i + sin(t) j + t k, 0 ≤ t ≤ π (a) Fi

nd a function f such that F = ∇f. f(x, y, z) = (b) Use part (a) to evaluate C ∇f · dr along the given curve C.
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
5 0

Answer:

a) f (x,y,z)= xy^2\sin(z)

b) \int_C F \cdot dr =0

Step-by-step explanation:

Recall that given a function f(x,y,z) then \nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}). To find f, we will assume it exists and then we will find its form by integration.

First assume that F = \nabla f. This implies that

\frac{\partial f}{\partial x} = y^2\sin(z) if we integrate with respect to x we get that

f(x,y,z) = xy^2\sin(z) + g(y,z) for some function g(y,z). If we take the derivative of this equation with respect to y, we get

\frac{\partial f}{\partial y} = 2xy\sin(z) + \frac{\partial g}{\partial y}

This must be equal to the second component of F. Then

2xy\sin(z) + \frac{\partial g}{\partial y}=2xy\sin(z)

This implies that \frac{\partial g}{\partial y}=0, which means that g depends on z only. So f(x,y,z) = xy^2\sin(z) + g(z)

Taking the derivative with respect to z and making it equal to the third component of F, we get

xy^2\cos(z)+\frac{dg}{dz} = xy^2\cos(z)

which implies that \frac{dg}{dz}=0 which means that g(z) = K, where K is a constant. So

f (x,y,z)= xy^2\sin(z)

b) To evaluate \int_C F \cdot dr we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.

\int_C F \cdot dr = f(r(\pi))-f(r(0))

Recall that r(\pi) = (\pi^2, 0, \pi) so f(r(\pi)) = \pi^2\cdot 0 \cdot \sin(\pi) = 0

Also r(0) = (0, 0, 0) so f(r(0)) = 0^2\cdot 0 \cdot \sin(0) = 0

So \int_C F \cdot dr =0

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If you're using the app, try seeing this answer through your browser:  brainly.com/question/3000586

——————————

The answer is option D)  r < 5  or  r > – 1.

I'm going to graph each inequality below on a number line.


A)  r > 5  or  r > – 1.

\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B)  r > 5  or  r < – 1.

\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r

Numbers between – 1  and  5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C)  r < 5  or  r < – 1.

\large\begin{array}{cl} \mathsf{r

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D)  r < 5  or  r > – 1.

\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}

Now all real numbers are included in the union. So this is the right choice.


Answer:  option D)  r < 5  or  r > – 1.


I hope this helps. =)

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3 years ago
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