Finding flow lines (vector calc problem)

[jaejoon89]

Homework Statement

F = (x^2 / y) i + y j + k

a) Use parametric equations to determine the equation for the flow line for the function F which passes thru the point (1,1,0)
b) Show that the flow line also passes thru the point (e,e,1)

Homework Equations

F = F1 i + F2 j + F3 k
dx/F1 = dy/F2 = dz/F3

The Attempt at a Solution

I didn't think parametric equations were actually needed here, but I think we're supposed to use them...

Somebody told me
x'[t] = x[t]2/y[t], y'[t] = y[t], z'[t] = 1 with the initial conditons that x[0] = 1, y[0] = 1, and z[0] = 0.
The solution is: x[t] = et, y[t] = et, z[t] = t
Setting t = 1 proves part b of the problem.

But I get
dx/(x^2 / y) = dt
dy/y = dt
dz = dt

Thus,
-y/x = t + c1 => x = -y / (t + c1)
ln|y| = t + c2 => y = c3e^t
z = t + c4

where c1, c2, c3, c4 are constants
What am I doing wrong? Also, why is t = 0?

 

[tiny-tim]

F = (x^2 / y) i + y j + k

a) Use parametric equations to determine the equation for the flow line for the function F which passes thru the point (1,1,0)
b) Show that the flow line also passes thru the point (e,e,1)
…
I get
dx/(x^2 / y) = dt
dy/y = dt
dz = dt

Thus,
-y/x = t + c1 => x = -y / (t + c1)
… What am I doing wrong? Also, why is t = 0?

Hi !

You can't integrate dx/(x^2 / y) = dt to get -y/x = t + c1 …

you have to eliminate y first, by expressing it as a function of t (or of x, I suppose).

And you're given (1,1,0), so if you're using t = z as a parameter, then obviously (1,1,0) corresponds to t = 0

 

[jaejoon89]

Thanks but I'm still not getting x = e^t, y = e^t. How do you eliminate y?

 

[jaejoon89]

Hi !

You can't integrate dx/(x^2 / y) = dt to get -y/x = t + c1 …

you have to eliminate y first, by expressing it as a function of t (or of x, I suppose).

And you're given (1,1,0), so if you're using t = z as a parameter, then obviously (1,1,0) corresponds to t = 0

You can't just take the antiderivative of x and treat y as a constant? (Thus getting -y/x)? When I try eliminating y I get

dx / (x^2 / y) = dy/y
y*dx/x^2 = dy/y
dx/x^2 = dy/y^2
-1/x = -1/y => x = (1/y + c5)^-1

If y = c3e^t as calculated earlier,
x = [(1/c3e^t)+c5]^-1 = c6e^t
Is it ok to eliminate y in this manner (sort of after the fact since I still take dx/(x^2/y)? How else would you do it?

 

[tiny-tim]

Hi jaejoon89!

(try using the X² and X₂ tags just above the Reply box )

You can't just take the antiderivative of x and treat y as a constant?
(Thus getting -y/x)?

No!

When I try eliminating y I get
dx / (x^2 / y) = dy/y
y*dx/x^2 = dy/y
dx/x^2 = dy/y^2
-1/x = -1/y => x = (1/y + c5)^-1

If y = c3e^t as calculated earlier,
x = [(1/c3e^t)+c5]^-1 = c6e^t
Is it ok to eliminate y in this manner (sort of after the fact since I still take dx/(x^2/y)? How else would you do it?

yes, that's fine, except it would have been a lot simpler to leave dt as it was, and just substitute y = c₃e^t, so as to give:
dx /x² = c₃e^-t dt