MHB How to Integrate a Function to Solve Air Resistance Equations?

[Us477]

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

 

[chisigma]

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 $\chi$ $\sigma$

 

[Us477]

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]

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

$\chi$ $\sigma$

 

[Us477]

Yeah, I tried that but couldn't come up with your result. For your information, I use a N-spire calculator.

 

[chisigma]

...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]

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.

 

[Us477]

Thank you very very much you have been a great help!