# Math Help - Domain of the vector function

1. ## 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

2. 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).

3. Originally Posted by matt3D
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).

4. Originally Posted by HallsofIvy
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?

5. Right. You have already removed any negative number from consideration.