Velocity Dependent Forces

[ollienor]

I Need to integrate the following but I am having problems with the completing the square and substituting u and du method because I keep ending up with a negative constant in the denominator.

\(\displaystyle \displaystyle{ \int_{V_0}^{V_2}\, }\) \(\displaystyle \dfrac{dv}{-0.02231V^2\, -\, 0.05358V\, +\, 1.60858}\, =\, \dfrac{1}{m}\,\) \(\displaystyle \displaystyle{ \int_{t_0}^{t_2}\, dt}\)

Any help much appreciated. (obviously integrating with respect to time on the RHS is easy I just end up with t/m since t0 will always be 0, I will also be happy if V0 is also 0, I don't think I need to analyse it from a starting speed but whatever). Please help!

 

[Ishuda]

I Need to integrate the following but I am having problems with the completing the square and substituting u and du method because I keep ending up with a negative constant in the denominator.

\(\displaystyle \displaystyle{ \int_{V_0}^{V_2}\, }\)\(\displaystyle \dfrac{dv}{-0.02231V^2\, -\, 0.05358V\, +\, 1.60858}\, =\, \dfrac{1}{m}\,\) \(\displaystyle \displaystyle{ \int_{t_0}^{t_2}\, dt}\)

Any help much appreciated. (obviously integrating with respect to time on the RHS is easy I just end up with t/m since t0 will always be 0, I will also be happy if V0 is also 0, I don't think I need to analyse it from a starting speed but whatever). Please help!

After completing the square you should end up with something like

\(\displaystyle a\, \)\(\displaystyle \displaystyle{ \int_{v_0}^{v_2}\, }\) \(\displaystyle \dfrac{1}{(x\, +\, b)^2\, -\, c} \,dv\)

where c is positive. To put this in 'standard form', let

\(\displaystyle u\, =\, \dfrac{x\, +\, b}{\sqrt{c}}\)

If you don't recognize that integral, use partial fractions.

 

[ollienor]

Thanks for the tip, but I think I'm just getting lost even further.

\(\displaystyle u\, =\, \dfrac{x\, +\, b}{\sqrt{c\,}},\, u^2\, =\, \dfrac{(x\, +\, b)^2}{c},\, cu^2\, =\, (x\, +\, b)^2\)

\(\displaystyle a\,\)\(\displaystyle \displaystyle{ \int_{V_0}^{V_2}\, }\)\(\displaystyle \dfrac{1}{(x\, +\, b)^2\, -\, c}\, dv\, =\, \dfrac{a}{c}\,\)\(\displaystyle \displaystyle{ \int_{V_0}^{V_2}\, }\)\(\displaystyle \dfrac{1}{u^2\, -\, 1}\)

\(\displaystyle \mbox{if }\, u\, =\, \dfrac{x\, +\, b}{\sqrt{c\,}}\, =\, \dfrac{x}{\sqrt{c\,}}\, +\, \dfrac{b}{\sqrt{c\,}}\, =\, c^{-\frac{1}{2}}x\, +\, bc^{-\frac{1}{2}},\)

\(\displaystyle \mbox{then }\, \dfrac{d}{dx}\, \dfrac{x}{\sqrt{c\,}}\, =\, \sqrt{c\,}\, \mbox{ and }\, \dfrac{d}{dx}\, bc^{-\frac{1}{2}}\, =\, 0.\)

\(\displaystyle \therefore\, du\, =\, {\sqrt{c\,}}.\)

I've mucked about with what you said but I have a hunch I'm just doing it wrong.

 

[Ishuda]

Thanks for the tip, but I think I'm just getting lost even further.

\(\displaystyle u\, =\, \dfrac{x\, +\, b}{\sqrt{c\,}},\, u^2\, =\, \dfrac{(x\, +\, b)^2}{c},\, cu^2\, =\, (x\, +\, b)^2\)

\(\displaystyle a\,\)\(\displaystyle \displaystyle{ \int_{V_0}^{V_2}\, }\)\(\displaystyle \dfrac{1}{(x\, +\, b)^2\, -\, c}\, dv\, =\, \dfrac{a}{c}\,\)\(\displaystyle \displaystyle{ \int_{V_0}^{V_2}\, }\)\(\displaystyle \dfrac{1}{u^2\, -\, 1}\)

\(\displaystyle \mbox{if }\, u\, =\, \dfrac{x\, +\, b}{\sqrt{c\,}}\, =\, \dfrac{x}{\sqrt{c\,}}\, +\, \dfrac{b}{\sqrt{c\,}}\, =\, c^{-\frac{1}{2}}x\, +\, bc^{-\frac{1}{2}},\)

\(\displaystyle \mbox{then }\, \dfrac{d}{dx}\, \dfrac{x}{\sqrt{c\,}}\, =\, \sqrt{c\,}\, \mbox{ and }\, \dfrac{d}{dx}\, bc^{-\frac{1}{2}}\, =\, 0.\)

\(\displaystyle \therefore\, du\, =\, {\sqrt{c\,}}.\)

I've mucked about with what you said but I have a hunch I'm just doing it wrong.

You (almost) have the right form,

\(\displaystyle \dfrac{du}{dv} = \dfrac{1}{\sqrt{c\,}}\, \)

\(\displaystyle \mbox{NOT }\, du\, =\, \sqrt{c\,}\, \mbox{ so that } \, dv\, =\, \sqrt{c\,}\, du:\)

\(\displaystyle \int\, \dfrac{1}{u^2\, -\, 1}\, du\)

but you need to get your constants straight.

To solve that integral write

\(\displaystyle \cfrac{1}{u^2\, -\, 1} = \dfrac{1}{2}\left[\, \dfrac{1}{u-1}\, -\, \dfrac{1}{u+1}\,\right]\)

 

[ollienor]

but isn't

\(\displaystyle \displaystyle{ \int \,}\) \(\displaystyle \dfrac{1}{u^2\, -\, 1}\, =\, -\tanh^{-1}(u)\)

?

OK I may seem like I'm following, but I do feel quite lost I'll be honest.

What do you mean by getting my constants straight?

 

[Ishuda]

but isn't

\(\displaystyle \displaystyle{ \int \,}\) \(\displaystyle \dfrac{1}{u^2\, -\, 1}\, =\, -\tanh^{-1}(u)\)

?

OK I may seem like I'm following, but I do feel quite lost I'll be honest.

What do you mean by getting my constants straight?

I'm not sure about the sign but, yes, it is the hyperbolic tangent which can also be written as the log functions you would get from the partial fraction expansion. They are equivalent.

As far as getting the constants straight, if you have the integral

\(\displaystyle a\, \int\, \dfrac{dv}{(v\, +\, b)^2\, -\, c}\)

and make the substitutions c u² = (v+b)² [that is \(\displaystyle \sqrt{c\,}\) u = (v + b)] and \(\displaystyle dv\, =\, \sqrt{c\,}\, du,\) what is(are) the proper constant(s) in front of the integral

\(\displaystyle \int \dfrac{du}{u^2\,-\,1}\)

 

[ollienor]

Awesome thank you very much for your help, I managed to get through it based on your hints and advice, cheers Ishuda!!

Incase you were wondering, it was a formula for dynamic thrust and drag. I can now calculate how fast my quad-copter will be flying at time t.

I tested it for time 5000s and it equals my calculation for terminal velocity (quadratic equation) so I'm pretty sure i've done it correctly.