Finding the limit of a velocity vector

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Homework Help Overview

The problem involves finding the limit of the derivative of a velocity vector as time approaches infinity for a particle moving in the XY plane, where the position is defined by exponential functions of time.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of L'Hôpital's rule due to an indeterminate form encountered when substituting infinity into the derivative. There is also a consideration of neglecting terms that approach zero as t increases.

Discussion Status

Some participants have provided insights regarding the simplification of the limit, noting that certain terms can be disregarded. However, there is still some confusion about the implications of the indeterminate form and the resulting limit.

Contextual Notes

Participants are navigating the complexities of limits involving exponential functions and the nuances of indeterminate forms in calculus.

jcook735
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Homework Statement


A particle moves in the XY plane so that at any time t is greater than or equal to 0 its position (x,y) is given by x = e^t + e^-t and y = e^t - e^-t.

Find the limit as t approaches infinity of dy/dx.


Homework Equations



I found dy over dx to be e^t + e^-t over e^t - e^-t

The Attempt at a Solution



I plugged in infinity for t but it comes out as infinity over infinity, and then I tried L'Hôpital's rule and then multiplied by the reciprocal, but still nothing. I don't know where to go from here.
 
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As t goes to infinity, e^(-t) goes to zero, so you can neglect those terms.
 
Yeah I figured that, but that leaves me with e^t over e^t which ends up as infinity
 
The limit of e^t/e^t is just 1. Yes, "infinity over infinity" is an indeterminate form, but that just means that, for arbitrary limits that both go to infinity, you can't tell what the limit of their quotient is. If you know what the limits are, then you often can get an answer.

Look, e^t/e^t=1 for any finite t. How could it have any other limit than 1 at infinity (or anywhere else)? To make it rigorous, you could put a bound on the difference between your original quotient and this one, and show that the difference goes to 0 as t goes to infinity.
 
oh yeah. hahahahaha whoops. thanks man
 

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