Question

I have a vector field F(x,y) and a hyperbola; I'm supposed to calculate the work from the field on a particle traveling on the right curve of the hyperbola; how do you go about deciding a good parametrization?

Answer 1

For example: 1/4*x^2-1/9*y^2=1 from (2*sqrt(5),-6) to (2*sqrt(2),3)

Answer 2

lets try...
i did line integrals very long back.... is that a real example problem ? if so, where is the vector field ?

Answer 3

The vector field is (1/4*x^2,1/6*x^2)

Answer 4

Thoughts? :D

Answer 5

isnt the work is the derevative ?

Answer 6

\[\int\limits_{\alpha}^{\beta}(P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t))dt\]
Is the work of the field F=(P,Q) along some curve (on differential form).

Answer 7

sorry i dont know , still learning these stuff :)

Answer 8

a dumb way wud be to parameterize :-
x = 2cosht
y = 3sinht

Answer 9

Why dumb?

Answer 10

uhmm lets try it

Answer 11

Oh, you mean it's a hassle? :d

Answer 12

x= 2 sec(t)
y= 3 tan(t)

Answer 13

Why?

Answer 14

x=2cosh(t),y=3sinh(t) gives a crazy t interval, no? :c

Answer 15

logs take over :o

Answer 16

try sec and tan..

Answer 17

What about the interval for t?

Answer 18

This shouldn't be so hard. :(

Answer 19

yup! i think logs wud make hyperbolics easy... finish the hyperbolic parametrization first maybe...

Answer 20

log(sqrt(5)+sqrt((sqrt(5)-1) (1+sqrt(5))))<=t<=log(sqrt(2)+sqrt((sqrt(2)-1) (1+sqrt(2))))
This does not feel right.

Answer 21

\(-\log (2+\sqrt{5}) \le t \le -\log (\sqrt{2} - 1) \)

Answer 22

http://www.wolframalpha.com/input/?i=%5Cint_%7B-%5Clog%282%2Bsqrt%285%29%29%7D%5E%7B-%5Clog%28%5Csqrt%282%29+-+1%29%7D++cosh%5E2%28t%29+2sinht++%2B+cosh%5E2%28t%29+3cosh%28t%29+dt

Answer 23

Supposedly not right. :/

Answer 24

This problem is driving me nuts!

Answer 25

wat do u mean not right ?

Answer 26

it doesnt look that pleasant, but it does look right to meh

Answer 27

Apparently not :C

Answer 28

The answer is supposedly gnarly though.

Answer 29

hmm-

Answer 30

Is your field
\[
\left ( \frac {x^2}4, \frac {x^2}6 \right)
\]
There is no y?

Answer 31

There is no y.

Answer 32

If the second component is in terms of y, then it would be very easy. That is why I am asking.

Answer 33

I think that your teacher should not ask such a question that yields to complicated integrals.

Answer 34

I tend to agree, the problem is generated through RNG though.

Answer 35

What is RNG?

Answer 36

Random number generator

Answer 37

With restrictions of course.

Answer 38

There is for sure one trick to make this easy.

Answer 39

It should be known, that between all integrals that you can generate, very few can be done by hand. The others might need numerical approximations.

Answer 40

It's generated to be hand-solved with relative ease.

Answer 41

The trick that I was looking for is was that the field is conservative, but it is not.

Answer 42

If you find an easy solution, please let us know.

Answer 43

Here is an idea
\[
\left ( \frac {x^2}4, \frac {x^2}6 \right)= \left ( \frac {x^2}4, 0 \right)+\left ( 0, \frac {x^2}6 \right)
\]

Answer 44

The first one is conservative and the work for it depends of the end points. You need to complete the work done by the second.

Answer 45

You need to compute
\[
\int_{t_1}^{t_2} \frac 1 6 x(t)^2 y'(t) dt
\]

Answer 46

Try it