- #1
Saladsamurai
- 3,020
- 7
In fluid mechanics velocity is given in the form
[tex]\textbf{V}=u\textbf{i}+v\textbf{j}+w\textbf{k}[/tex]
A two-dimensional velocity field is given by
[tex]\textbf{V}=(x^2-y^2+x)\textbf{i}+(-2xy-y)\textbf{j}[/tex]
At [itex](x_o,y_o)[/itex] compute the accelerations [itex]a_x\text{ and }a_y[/itex]
I am having trouble with the books definition of [itex]a_x\text{ and }a_y[/itex]
Now I can see that the w term is zero. They define ax to be
[tex]a_x=\frac{\partial{u}}{\partial{t}}=u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}}[/tex]
I know that the chain rule has been used to arrive at this, but I seem to be getting lost along the way. So I am attempting to do it out here. So when we say
[itex]\frac{\partial{u}}{\partial{t}}[/itex] we are really saying
[tex]\frac{\partial{}}{\partial{t}}[u(x(t),y(t),z(t))][/tex]
right?
I am just confused as to how to evaluate this
[tex]\textbf{V}=u\textbf{i}+v\textbf{j}+w\textbf{k}[/tex]
Homework Statement
A two-dimensional velocity field is given by
[tex]\textbf{V}=(x^2-y^2+x)\textbf{i}+(-2xy-y)\textbf{j}[/tex]
At [itex](x_o,y_o)[/itex] compute the accelerations [itex]a_x\text{ and }a_y[/itex]
I am having trouble with the books definition of [itex]a_x\text{ and }a_y[/itex]
Now I can see that the w term is zero. They define ax to be
[tex]a_x=\frac{\partial{u}}{\partial{t}}=u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}}[/tex]
I know that the chain rule has been used to arrive at this, but I seem to be getting lost along the way. So I am attempting to do it out here. So when we say
[itex]\frac{\partial{u}}{\partial{t}}[/itex] we are really saying
[tex]\frac{\partial{}}{\partial{t}}[u(x(t),y(t),z(t))][/tex]
right?
I am just confused as to how to evaluate this