Can the Undefined Limit of a Velocity Equation be Solved for as k Approaches 0?

  • Thread starter Thread starter pr0me7heu2
  • Start date Start date
  • Tags Tags
    Limits
pr0me7heu2
Messages
13
Reaction score
2
Undefined Limits as k--> 0

Homework Statement



lim(k-->0) [ (-mg)/k + v*e^(kt/m) + (mg)/k*e^(kt/m)]

,the end result of this limit is ultimately supposed to be v -gt (or the velocity of an object at any time t neglecting air resistance).



Homework Equations



This equation comes from the differential equation dv/dt - k/m *v =g

,then using integrating factors (the equation itself is a linear ODE) I found:

v = (-mg)/k + ce^(kt/m)

,where c is found by solving for the initial condition v(0)=v0 where

v0 = (-mg)/k + c(1)

--> c = v0 + mg/k

The Attempt at a Solution



I've spent literally a few hours pouring over this, frustrated as hell that I couldn't solve a simple limit!

I tried first taking the natural log of the whole thing, then I tried using all of the rules of exponents separating the e^(x)s. Then I tried working with the differential equation for a while but ultimately was never able to find a way to fully end up with a defined answer - ie. I was unable to completely eliminate x^(-1) or ln(x) prior to taking the limit as x-->0.
 
Physics news on Phys.org
Rewrite the solution as ( I hope I haven't made a mistake)

v \left( t \right) = \exp \left( \frac{k}{m} t \right) v_0 +gm \frac{\exp \left( \frac{k}{m} t \right) - 1}{k},

and then take the limit k \rightarrow 0.

Note that if you set k = 0 in the original differential equation, the solution is v = v_0 + gt, not v = v_0 - gt.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top