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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
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 Originally Posted by Integraluser Hey, so we had this question in class: Given that the relationship between distance (m) and velocity (v) of an object is $\displaystyle v^2 = 1 - m^3$
Meaningless question since the units don't match. December 19th, 2018, 01:03 PM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 540 Thanks: 82 If not divide by $\displaystyle v$ $\displaystyle v(2a+3m^2)=0$ or $\displaystyle 2a+3m^2=0$ $\displaystyle a=\frac{-3m^2 }{2}$ Thanks from topsquark December 19th, 2018, 06:45 PM   #4
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 Originally Posted by Micrm@ss Meaningless question since the units don't match.
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
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Treat it as an impartial differentiation problem along with chain rule, and taking derivatives with respect to time for now.

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 Originally Posted by idontknow If not divide by $\displaystyle v$ $\displaystyle v(2a+3m^2)=0$ or $\displaystyle 2a+3m^2=0$ $\displaystyle a=\frac{-3m^2 }{2}$
Yes I did mention this implicitly in my post, but why can't we substitute for $\displaystyle m$ directly in the equation rather than simplifying it to this form?

Also, is what my teacher said true? December 20th, 2018, 02:12 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 540 Thanks: 82 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. Tags derivatives, equation, limits, oddity, solving, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post joswhale New Users 4 June 20th, 2017 12:57 AM lawlsh Algebra 4 January 15th, 2017 01:50 PM iman.sayah Applied Math 5 January 10th, 2014 08:29 PM tom.lavy Algebra 6 March 18th, 2012 05:32 PM truecb Algebra 9 August 19th, 2009 12:00 PM

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