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I've got a stream function:

$$u_\infty y + \frac{Q}{2\pi} \operatorname{arctg}\frac{y}{x} = 0 $$

How do I solve it for y? I know the solution, just don't know how to get there step by step.

EDIT: The solution given in the book is: $$ x = -y \operatorname{ctg} \left( \frac{2 \pi u_\infty}{Q} \right) $$

I'm sorry, I actually didn't realize the given solution is for x, not y.

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So, what's the solution that you know? –  Kaster Feb 17 '13 at 10:21
    
@Kaster see above, edited my question –  mmm Feb 17 '13 at 10:26
    
Actually, that's not right; there should be a $y$ in the $\cot$, see my answer below. –  Ron Gordon Feb 17 '13 at 10:26
    
@rlgordonma yes, you're right, there's an (now) obvious error in my materials. –  mmm Feb 17 '13 at 10:32
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1 Answer 1

up vote 2 down vote accepted

You essentially have an equation for $y$ of the form

$$x = -\frac{y}{\tan{a y}}$$

where $a=2 \pi u_{\infty}/Q$. There is no analytical solution for $y$ in terms of $x$ and $a$, but there are ways to approximate the solution depending on the domain of $x$.

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what is $\mu_{\infty}$ and $Q$ mean? –  yiyi Feb 17 '13 at 10:17
    
See the problem statement. –  Ron Gordon Feb 17 '13 at 10:18
    
I have [read this][1] and the wikipedia page, but I still don't get the infinity subscript for mu. Also the Q I don't see mentioned, that is why I asked. I don't know about this and was just trying to understand it better, I see the pattern of your answer, nice by the way; however, don't understand those terms in the context of the problem. [1]: see.ed.ac.uk/~johnc/teaching/fluidmechanics4/2003-04/fluids2/… –  yiyi Feb 17 '13 at 10:24
    
Thank you @rlgordonma. I didn't see before the solution given in the book is for x, not y. Silly me :). –  mmm Feb 17 '13 at 10:26
    
$u_\infty$ stands for flow velocity in the x-direction, you could skip the subscript, it's just used to indicate undisturbed flow in the materials I'm studying i.e. "long before flow reaches the obstacle". Q is spring discharge. I've never studied fluid mechanics before and don't know english nomenclature, hope I didn't confuse anything. –  mmm Feb 17 '13 at 10:43
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