MHB Why Should Constants Be Substituted in Integration?

[harryt]

This is quite a basic question as my understanding of integration is not that good.

I have an equation concerning the time taken for an object to fall.

$m\frac{dv}{dt} = mg-R{v}^{2}$

I need to get an expression for time as a function of velocity and I have been told to integrate with the substitution

${u}^{2}=R/mg$

I don't understand this though, as the substitution is not in terms of either v or t. You can't do integration by substitution as you get du/dv (or du/dt) = 0. Why would I be substituting constants?

I know I'm missing something here... I'm very keen to get to the bottom of this myself but if anyone could give me a hint, I'd be most appreciative!

Thank you,
Harry

 

[Fantini]

That substitution will not help you. However, consider dividing by $m$ and writing $\frac{R}{m} = k$. You get

$$\frac{dv}{dt} = g -kv^2,$$

therefore

$$\frac{dv}{g - kv^2} = dt.$$

Can you take it from here? :)

 

[topsquark]

This is quite a basic question as my understanding of integration is not that good.

I have an equation concerning the time taken for an object to fall.

$m\frac{dv}{dt} = mg-R{v}^{2}$

I need to get an expression for time as a function of velocity and I have been told to integrate with the substitution

${u}^{2}=R/mg$

I don't understand this though, as the substitution is not in terms of either v or t. You can't do integration by substitution as you get du/dv (or du/dt) = 0. Why would I be substituting constants?

I know I'm missing something here... I'm very keen to get to the bottom of this myself but if anyone could give me a hint, I'd be most appreciative!

Thank you,
Harry

The substitution merely gets rid of some of the extra manipulation of constants, nothing more. In fact, in many cases the constants are manipulated such that the remaining quantities, such as v and t, are unitless. I'm not sure if there is a Mathematical reason behind this but I find unitless quantities somewhat easier to work with.

-Dan

 

[harryt]

That substitution will not help you.

I didn't think it would!

Thank you for your help; it certainly clarified things.

 

[harryt]

The substitution merely gets rid of some of the extra manipulation of constants, nothing more. In fact, in many cases the constants are manipulated such that the remaining quantities, such as v and t, are unitless. I'm not sure if there is a Mathematical reason behind this but I find unitless quantities somewhat easier to work with.

-Dan

Substituting the constants has definitely helped me!