Welcome to our community

Be a part of something great, join today!

[SOLVED] Air resistance equation

U477

New member
Feb 11, 2013
4
I have been unable to solve this problem featured in "Elementary Differential Equations" by Edwards and Penney. The problem is this: I do not know how to integrate the function, and get the desired result. I would very much appreciated any help and instructions on how to get the two air resistance equations by integrating the function.

Thanks

Søren, Denmark

equation.jpg
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have been unable to solve this problem featured in "Elementary Differential Equations" by Edwards and Penney. The problem is this: I do not know how to integrate the function, and get the desired result. I would very much appreciated any help and instructions on how to get the two air resistance equations by integrating the function.

Thanks

Søren, Denmark

View attachment 615
The ODE can be written with the variables v and t separated as...

$\displaystyle \frac{d v}{1 - \frac{k}{m g}\ v^{2}} = g\ d t$ (1)

... and remembering that is...

$\displaystyle \int \frac{d u}{1-a\ u^{2}} = \frac{\tanh^{-1} (\sqrt{a}\ u)}{\sqrt{a}} + c$ (2)

... You obtain...


$\displaystyle \tanh^{-1} (\sqrt{\frac{k}{m\ g}}\ v) = \sqrt{\frac{k\ g}{m}}\ t + c $ (3)

... and from (3)...

$\displaystyle v = \sqrt{\frac{m\ g}{k}} \tanh (\sqrt{\frac{k\ g}{m}}\ t + c)$ (4)

Now You can proceed if You are able to find the constant c and that is possible if the initial speed v(0) is given...


Kind regards


$\chi$ $\sigma$
 

U477

New member
Feb 11, 2013
4
Thanks for your help.

I have a few questions for your solution -

1) Why are the position of the constants changed inside the radical in (3) – I am thinking they should be the same.

2) When I try to isolate ”v” I end up with a fraction - how did you go from (3) to (4)

I'm still new to all this thanks for helping me out
 

chisigma

Well-known member
Feb 13, 2012
1,704
Thanks for your help.

I have a few questions for your solution -

1) Why are the position of the constants changed inside the radical in (3) – I am thinking they should be the same.

2) When I try to isolate ”v” I end up with a fraction - how did you go from (3) to (4)

I'm still new to all this thanks for helping me out
I apologize to have omitted some intermediate stage and I suggest You, as useful exercise, to try Yourself to complete my work...

Kind regards

$\chi$ $\sigma$
 

U477

New member
Feb 11, 2013
4
Yeah, I tried that but couldn't come up with your result. For your information, I use a N-spire calculator.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
...for your information, I use a N-spire calculator...
Unfortunately I don't have available such a powerful math tool...

Kind regards

$\chi$ $\sigma$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
So you have
$$\frac{1}{g} \, \frac{dv}{dt}=1- \frac{k}{mg}\,v^{2}$$
$$\int \frac{dv}{1- \frac{k}{mg} \,v^{2}}=g \int dt$$
$$ \frac{ \tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)}{ \sqrt{ \frac{k}{mg}} }=gt+c$$
$$\tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)= \sqrt{ \frac{k}{mg}}\, (gt+c)$$
$$\tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)= \sqrt{ \frac{k g^{2}}{mg}}\,t+\sqrt{ \frac{k}{mg}} \, c= \sqrt{ \frac{kg}{m}}\,t+c'.$$
Can you continue from here? Note that a constant times an arbitrary constant is just another arbitrary constant - in DE's, you often absorb known constants into arbitrary constants, and relabel as the original arbitrary constant. So the $c'$ above becomes $c$ again.
 
Last edited:

U477

New member
Feb 11, 2013
4
Thank you very very much you have been a great help!!