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December 19th, 2018, 11:28 AM  #1 
Newbie Joined: Jun 2018 From: Jordan Posts: 24 Thanks: 0  Solving a 3 variables equation, one happens to cause an oddity
Hey, so we had this question in class: Given that the relationship between distance (m) and velocity (v) of an object is $\displaystyle v^2 = 1  m^3$ Find the acceleration of the object when $\displaystyle m=1$ By taking the derivative of each side with respect to $\displaystyle t$ $\displaystyle 2v \frac{dv}{dt} = 3m^2 \frac{dm}{dt} $ and we know that$\displaystyle \frac{dv}{dt}$ = acceleration, $\displaystyle \frac{dm}{dt} = v$, then: $\displaystyle 2v a = 3m^2 v$ and by solving for $\displaystyle a$ without dividing by $\displaystyle v$ since $\displaystyle v $ can be zero(assuming that we don't know yet, $\displaystyle a = 3/2$ But hold on... we've just substituted m with 1, this means that we have to substitute v with zero, but if we do this we would turn out with 0 = 0, without solving for a If we don't substitute v with 1, we wouldn't be able to solve for acceleration if we cancel v's with dividing, and it doesn't really make sense to keep v in there while we know that v = 0 when m = 1. My teacher clarified that by going back to the definition of derivative, that it's after all nothing but a limit and a limit is an approximation of some f(x) when x approaches some value, so a limit which equals 0 would be in real something like 0.0001 or 0.000000001 in a way we would be able to cancel v with v. But that doesn't really make sense to me. If that's true, when to use the approximate definition of limits and when to use the accurate? And what can we do for this equation? 
December 19th, 2018, 11:46 AM  #2 
Senior Member Joined: Oct 2009 Posts: 755 Thanks: 258  
December 19th, 2018, 01:03 PM  #3 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
If not divide by $\displaystyle v$ $\displaystyle v(2a+3m^2)=0$ or $\displaystyle 2a+3m^2=0$ $\displaystyle a=\frac{3m^2 }{2}$ 
December 19th, 2018, 06:45 PM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,073 Thanks: 843 Math Focus: Wibbly wobbly timeywimey stuff.  Clearly if we were to be more critical about the relationship $\displaystyle v^2 = 1  m^3$ then we should be writing $\displaystyle v^2 = p  qm^3$ where p and q are constant with the appropriate units. The we just say that p = 1 and q = 1. I've seen this kind of thing a lot when going from a Physics problem to a Calculus problem. Dan 
December 19th, 2018, 08:48 PM  #5  
Newbie Joined: Jun 2018 From: Jordan Posts: 24 Thanks: 0 
Treat it as an impartial differentiation problem along with chain rule, and taking derivatives with respect to time for now. Quote:
Also, is what my teacher said true?  
December 20th, 2018, 02:12 AM  #6 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
Let's take a simple example with velocity. If $\displaystyle s=s(t)$ then $\displaystyle vt\neq s(t)$ Once distance is given as function of time then $\displaystyle v=s’(t)$ Last edited by skipjack; December 20th, 2018 at 03:54 AM. 

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derivatives, equation, limits, oddity, solving, variables 
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