Material Derivative and Implicitly Given Variables for Velocity Calculation

[member 428835]

Homework Statement

Show $DF/Dt=0$. $F = x-a-e^b\sin(a+t)$ and $a$ is given implicitly as $y=b-e^b\cos(a+t)$ where $a=f(y,t)$ and $b$ is a constant. Also, velocity is $$u=e^b\cos(a+t)\\v=e^b\sin(a+t)$$

Homework Equations

$DF/Dt=F_t+v\cdot\nabla F$

The Attempt at a Solution

$F_t = -e^b\cos(a+t)$
$v\cdot \nabla F = e^b\cos(a+t) \cdot 1 + e^b\sin(a+t) \cdot 0 = e^b\cos(a+t)$.
Then $DF/Dt = -e^b\cos(a+t)+e^b\cos(a+t)=0$. Is this correct? It feels too easy.

Thanks!

 

[Orodruin]

What happened to Da/Dt?

 

[member 428835]

So you're saying $F_t=a'(t)-e^b\cos (a+t)\cdot (a'(t)+1)$. But then the convective term would also have the $y$ component, namely $v\cdot (a'(y)-e^b\cos (a+t)\cdot a'(y)$?

Also, $x$ velocity is given by $u=e^b\cos(a+t)$. Then $a$ was not differentiated for $u=x'(t)$.

Sorry if this looks weird, I'm using the app for the first time and I can't see Tex output.

 

[member 428835]

Orodruin, do you still think I should differentiate $a$ since they did not for the velocity term?

 

[Orodruin]

You make it sound as if the velocity was not given on the form you have quoted. My understanding of what you posted is that you were given the velocity.

 

[member 428835]

You make it sound as if the velocity was not given on the form you have quoted. My understanding of what you posted is that you were given the velocity.

I apologize for the ambiguity. So the particle's position is given by: $$x=a+e^b\sin(a+t)\\y=b-e^b\cos(a+t)$$ Then the velocity as in the first post, which is given in the question stem. Does this clarify my question?
Thanks for your patience!

 

[Orodruin]

Could you quote the problem statement verbatim?

 

[member 428835]

Could you quote the problem statement verbatim?

Definitely. It follows:
A particle's flow path is described as
$$x=a+e^b\sin(a+t)\\y=b-e^b\cos(a+t)$$
Thus the spatial velocity is
$$u=e^b\cos(a+t)\\v=e^b\sin(a+t)$$ Show that the kinematic boundary condition is satisfied along the curve derived from above by setting $b=const$ and $a$ is a parameter.
Hint: One could consider this curve to be $$x-a(y,t)-e^b\sin(a(y,t)+t)=0$$.

The kinematic boundary condition is $DF/Dt=0$ where $F$ is a curve of the boundary, presumably the hint's curve.

 

[member 428835]

But $u=\partial_tx$, and they did not implicitly differentiate $a$, but treated it as a constant. If this is true for the time derivative, since $a$ is a functino of $t$ and $y$, then $D a/D t = 0$. Do you agree?