Mechanics question derivations terminal velocity (quadratic case)

AI Thread Summary
The discussion focuses on the derivation of the quadratic case of terminal velocity, specifically questioning the signs and equations presented in a textbook. A participant points out a potential error in the sign of an equation and seeks clarification on the derivation of certain terms, particularly how 1/2 * dv²/dy is obtained. The conversation also addresses the relationship between velocity and its derivatives, explaining how dv²/dt relates to v and dv/dt. The clarification provided confirms the correct interpretation of the equations and their derivations. Overall, the discussion emphasizes the importance of understanding the mathematical relationships in the context of terminal velocity.
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hi so basically my question is derivation of derivation of the quadratic case of the terminal velocity


I have attached the pictures in picture 1 I don't know I think the book has error in his derivation @ equation 2.4.13 shouldn't it be + instead of negative ??

& in second picture their is something I don't understand two stuff acctually first is in equation 2.4.15
how did he come up with 1/2 * dv^2 / dy? and my second question is
du/dy = -1/vt^2 * dv^2/dy how did he come up with dv^2/dy ??

thanks alot.
 

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Yes the sign is wrong.
And dv/dt = dv/dy dy/dt = dv/dy v
since v = dy/dt .
And since dv²/dt = 2 v dv/dt, the last result follows.
 
could u explain this line a little more ->dv²/dt = 2 v dv/dt, the last result follows?
 
read:

dv²/dy = 2 v dv/dy

which implies

v dv/dy = 1/2 dv²/dy
 
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