Question

c

Answer 1

Any suggestions?! am I doing this totally wrong?

Answer 2

Consider a change of coordinate system.

Answer 3

Or check if it is a gradient field.

Answer 4

It looks like it might have a potential function, which would make it a lot easier.

Answer 5

it does

Answer 6

The potential function was a question asked previously before and I got f(x,y,z)=cos(xy)+sinz+C

Answer 7

soo....how would I do that? it thought you had to use the vector field? What would the integral look like? Would it be similar to the work integral?

Answer 8

You need to find the endpoints of the parametrizaton

Answer 9

....I'm not exactly sure how to do that. Wasn't covered at all during out lectures

Answer 10

The initial endpoint \(r_0\) is just the parametrization at the initial time \(t_0\).
So \[
r_0 = \mathbf r(t_0)
\]

Answer 11

ok uh so that would just be (1, 0,0)?

Answer 12

Yes, and then you need the final endpoint

Answer 13

and that would be (0,1,1)?

Answer 14

\[
\cos(2\pi)=0
\]???

Answer 15

hahah oh my b that's a retarded mistake so (1,0,1)

Answer 16

is my potential function equation right though?

Answer 17

Okay so the result of the line integral is the potential function at the endpoint minus the potential function at the initial end point

Answer 18

I got f(x,y,z) = cos(xy)+sinz+C

Answer 19

check by finding the gradient of it.

Answer 20

I see a similar example in my book so all I do is f(1,0,0)-f(1,0,1) into the potential function right?

Answer 21

Wrong order

Answer 22

I mean yeah the opposite. How do I find the gradient?

Answer 23

cant I just integrate with respect to dx, dy, and dx seperately to check if it fits the vector field?

Answer 24

Just brush up on gradients?

Answer 25

oh yeah that's just the partial derivative of fx, fy, fz...okay yeah thanks.All these terms are confusing!lol