Differential equation of gradually varied flow

[fonseh]

Homework Statement

I have no idea how the third formula of dy/dx is derived ...

Homework Equations

The Attempt at a Solution

I know that the Q = (1/n)(A)(R^2/3) [(s)(^0.5)] ,
Q = K [(s)(^0.5)]
, so , K= (1/n)(A)(R^2/3)

i know that for very wide channel , y = R
A = by
K= (1/n)(A)(R^2/3)
= (1/n)(by)(y^2/3)
= (1/n)(b)(y^5/3)
Thus , (K^2) = [[ (1/n)(b)]^2 ](y^10/3)
but , it seems that the author just got (y^10/3) ,

So , is the author wrong ?

 

[BvU]

On first sight: no. Consider ${K_0\over K}$ instead of just $K$.
There is no second sight because your use of the template is utterly unenlightening...

 

[fonseh]

On first sight: no. Consider ${K_0\over K}$ instead of just $K$.
There is no second sight because your use of the template is utterly unenlightening...

sorry , i mean $$K_0$$ corresponds to $$y_0$$
and $$K$$ corresponds to $$y$$

 

[BvU]

Yes, I understand. Don't the n's and the b's cancel ?

 

[fonseh]

Yes, I understand. Don't the n's and the b's cancel ?

in my working , i can't cancel the n and b ... Is there anything wrong with my working ?

 

[BvU]

Not from what I can see because what I can see is nothing ...(refer to post #2). I have no idea what this is about.

I see you write $K^2 = (b/n)^2\; y^{10/3}$ and I add $K_0^2 = (b/n) ^2\; y_0^{10/3}$ leading to $\left ( {K_0\over K}\right ) ^2 = \left ({y_0\over y}\right )^{10/3}$. That simple.