Air resistance equation derivation

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Homework Help Overview

The discussion revolves around the derivation of the air resistance equation, with the original poster expressing a desire for a detailed breakdown of the derivation process. The context involves mathematical concepts related to calculus and differential equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessary mathematical background for the derivation, with some questioning the original poster's familiarity with integral calculus and differential equations. There are attempts to clarify specific equations and steps involved in the derivation process, including the use of hyperbolic functions.

Discussion Status

The discussion is ongoing, with participants providing insights into the mathematical requirements and suggesting resources for further study. The original poster has indicated progress in understanding, but no consensus or complete solution has been reached yet.

Contextual Notes

There is mention of the original poster's lack of recent study in differential equations, which may impact their ability to follow the derivation. The discussion also highlights the complexity of the problem and the need for foundational knowledge in calculus.

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I would like to derive the air resistance equation showed in the attached photo. A little while ago, I tried asking at a different forum, however, the users were very experienced and omitted several stages of the derivation, and thereby made the interpretation difficult as I am new to math, but extremely eager to learn. I hope someone can help me illuminate each step of the derivation process, this would be a great help for my future studies, which are independent - I am not attending school at the moment. I have spent much time with this problem and I appreciate all help

Thanks from Denmark :)
 

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When you say you are new to math, how new is that? Have you studied integral calculus and differential equations? Without knowledge of some advanced math, this forum is not suitable for detailed explanation of the derivation you seek.
 
I think that that second-to-last equation where you finally get v(t) to the left hand side follows from:

<br /> <br /> tanh ( arctanh( \sqrt{ \frac{\rho}{g} }v ) ) = \sqrt{ \frac{\rho}{g} } v = tanh( \sqrt{ \frac{\rho}{g} } (gt + c) )<br /> <br />
Which follows from the given fact that \rho = \frac{k}{m}.

From there it's a matter of rearraning, and using what I assume will be some given physical constraint which tells us that c is 0 or may be taken to be 0.

To do the final step, it's not immediately clear to me how to integrate that, I would consult an integral table for integrands containing tanh(z) .

Hope that helps somewhat.

Note also that tanh^{-1}(z) denotes the inverse hyperbolic tangent, or arctanh(z), and not the reciprocal of tanh(z). That would be coth(z).

EDIT: Yes, I would also agree that you should make yourself comfortable with basic calculus and differential equations. There are a number of free reading courses online for both, and you should be able to find good references using Google or the search function. There are many skills and tricks in Calculus that are very important to being able to manipulate physical equations.
 
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Yes, I have studied calculus and differential equations, and solved many different types of problems, but I am unable to solve this specific problem. When I wrote "new" I meant that I had not studied differential equations for several years

Thanks Ocifer, now I am one step further, will update the picture I uploaded later today.
 
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