Question

looking for help with line integrals and Independence of path questions

Answer 1

If it has a potential function, it is path independent.

Answer 2

ah okay i see im not on path indepenance ones yet im trying to get a of line integral stuff first

Answer 3

get a grasp*

Answer 4

look at this question \[\int\limits_{?}^{?}xy^2 dx x=4\cos t, y - 4\sin t , 0< t<\pi/2\]

Answer 5

The point of a line integral is that instead of integrating over a simple interval, like \(0\leq x \leq 10\), you integrate over some parametrized curve, like \((x,y) = (t,2t), 0 \leq t \leq 4\).

Answer 6

its just dx not dxx

Answer 7

what does that mean visually on a graph

Answer 8

oh okay

Answer 9

so lets say therse a curve on x and y axis
if i do the line integral with respect to x instead of ds what wud look like

Answer 10

How would you do it with respect to x?

Answer 11

You supposed to do it with respect to \(t\)

Answer 12

Okay, I think I understand what you mean now...

Answer 13

With respect to \(t\):\[
\int_C fds = \int_{t_0}^{t_1}f(x(t),y(t))\cdot \sqrt{[x'(t)]^2+[y'(t)]^2}dt
\]With respect to \(x\): \[
\int_C fds = \int_{t_0}^{t_1}f(x,y(x))\cdot \sqrt{1+[y'(x)]^2}dx
\]

Answer 14

ya like just putting only dx in terms of dt where as ds is Sqrt (dx/dt)^2 (dy/dt)^2 dt

Answer 15

Is this what you meant?

Answer 16

If the curve is expressed as a function of \(x\) then you can use the second formula. If it is a parametrized curve, you use the first formula.

Answer 17

\[\int\limits_{}^{?}xy^2 dx = \int\limits_{?}^{?}(4cost 16) (\sin^2t) (-4 \sin t dt) ; dx=- 4\sin dt\]

Answer 18

Ummm I'm not sure what you're doing.

Answer 19

(4 cos t) (16 sin^2 t)*

Answer 20

x = 4 cos t and y = 4 sin t
so i just got dx from there and then my function is xy^2 so i subbed it in
it wants me to do Integral xy^2 dx and the next question is integral xy^2 dy and then xy^2 ds so i wanted to know what exactly doesn it mean on a graph if im only taking the line integral with respect to x is that just like a normal integral not a curvy integral anymore

Answer 21

First of all, what did you get for \[
\sqrt{[x'(t)]^2+[y'(t)]^2}
\]

Answer 22

sqrt16cos^2t + sin^2t dt

Answer 23

so 4

Answer 24

4 dt

Answer 25

Alright and what do you get for \(xy^2\) in terms of \(t\)?

Answer 26

Lastly, we know our limits of \(t\) are \(0\) to \(\pi/2\)

Answer 27

with everything together i got
256 Integral sin^2 t cos t dt = Integral xy^2 ds

Answer 28

ya i get 256/3

Answer 29

Alright, there you go.

Answer 30

i see what u mean im reading something about
ds = sqrt (1+[f'(x)]^2)

Answer 31

for explicit functions where dy = f'(x) dx