Find the domain of the vector functions

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Homework Help Overview

The discussion revolves around finding the domain of the vector function r(t) = . Participants are exploring the constraints on the variable t to ensure that all components of the vector function yield valid results.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions under which each component of the vector function is defined, questioning the implications of logarithmic and square root functions on the domain of t.

Discussion Status

There is an ongoing examination of the conditions for the domain, with some participants suggesting intervals based on their interpretations of the component functions. Guidance has been offered regarding the restrictions imposed by the logarithmic function and the square root functions.

Contextual Notes

Participants are considering the implications of t being negative, zero, or exceeding certain values, particularly in relation to the logarithmic function and the square root function. There is a focus on ensuring that all components of the vector function remain real-valued.

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Find the domain of the vector functions, r(t), listed below

a.) r(t) = <ln(6t), sqrt(t+14), 1/sqrt(16-t)>

i don't extactly know how to approach this, can someone give me a hint or two
 
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The domain of a function [tex]f(x)[/tex] basically means the set of all [tex]x[/tex] for which the function yields valid results. In your case, I presume all vector components must be real numbers. What does that mean for the values that [tex]t[/tex] is allowed to have?
 
[tex]t \geq -14[/tex]
[tex]t \leq 16[/tex]

right? so the domain should be from [-14,16]
 
You're on the right track, but what about the ln(6t) function?
 
t cannot equal to 0 for ln(6t). so should it be [-14,0) U (0, 16]
 
If ln(6t) is to yield a real number, t must be greater than 0. A negative value for t won't work! So one of the domains you posted is almost right - the one holding positive values. The reason that the interval isn't exactly correct is because t = 16 isn't allowed (take a look at the third term and see what happens when t=16). In other words: t is in (0,16) instead of (0,16].
 

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