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Math Help - Domain of the vector function

  1. #1
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    Question Domain of the vector function

    Hello, I am trying to find the domain of the vector function:
    I know that for \ln(7t) we have t\ge 0, \sqrt{t+12} \Rightarrow t\ge-12,\frac{1}{\sqrt{19-t}} \Rightarrow t\le18

    So I thought the domain was from [-12,0,]U[0,18] but it isn't correct.
    What am I doing wrong?

    Thanks,
    Matthew
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Well your first condition (which should be t>0) would exclude [-12,0] from the domain.

    Moreover the third condition should be t<19, so the domain would be (0,19).
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  3. #3
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    Quote Originally Posted by matt3D View Post
    Hello, I am trying to find the domain of the vector function:
    I know that for \ln(7t) we have t\ge 0, \sqrt{t+12} \Rightarrow t\ge-12,\frac{1}{\sqrt{19-t}} \Rightarrow t\le18

    So I thought the domain was from [-12,0,]U[0,18] but it isn't correct.
    What am I doing wrong?

    Thanks,
    Matthew
    In order that "x" be in the domain, all of those functions must be defined so the domain is the intersection of the domains of those individual functions, NOT the union. The domain of ln(7t) is t> 0 (NOT " t\ge 0"- ln(0) is not defined). The domain of \sqrt{t+ 12} requires that t\ge -12 and the intersection of those is still t> 0. The domain of \frac{1}{\sqrt{19-t}} is t< 19 (NOT " t\le 18! t= 18.999999 is perfectly good.) and the intersection of t> 0 and t< 19 is (0, 19).
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    In order that "x" be in the domain, all of those functions must be defined so the domain is the intersection of the domains of those individual functions, NOT the union. The domain of ln(7t) is t> 0 (NOT " t\ge 0"- ln(0) is not defined). The domain of \sqrt{t+ 12} requires that t\ge -12 and the intersection of those is still t> 0. The domain of \frac{1}{\sqrt{19-t}} is t< 19 (NOT " t\le 18! t= 18.999999 is perfectly good.) and the intersection of t> 0 and t< 19 is (0, 19).
    Okay, but since t>0 for ln(7t), I don't have to worry about -12 for the y component?
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  5. #5
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    Right. You have already removed any negative number from consideration.
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