Answer to Question #170065 in Calculus for Keerthidas

Question

Find the domain of the vector function ?

t^5,4log(t-2),3ln(t+1)

Answer 1

$Given \ that \ the \ vector \ function \ is$ $(t^5 , 4log(t-2),3ln(t+1))\\$

$A \ vector \ function \ f(t)=(f_1 (t), f_2 (t), f_3 (t)) \ is \ said \ to \ be \ defined \ if \ the \ component \ function s\ f_1 (t)$$f_2 (t) \ and \ f_3 (t) \ are \ defined.\\ The\ set \ of \ all \ t \ \ for \ which \ f_1, f_2 \ and \ f_3 are \ defined \ is\, called \ the \ domain \ of \ f(t).$ The \ $The \ component \ function \ f_2(t) = 4log(t-2)\ is \ defined \ only \ for \ t-2>0 \\ \Rightarrow t>2 \Rightarrow The\, domain \ of \ f_2(t) is , \ the \ set \ of \ all \ t > 2.\\$ $The \ function \ f_1(t)=t^5 \ is \ defined \ for \ all\ t \isin\ R.\\$ \\

$The \ component \ function \ f_3 (t) =3ln(t+1) \ is \ defined \ if \ (t+1)>0 \Rightarrow \ t+1> 0 \\ \Rightarrow t>-1$

$The \ domain \ of \ f(t) \ is \ Domain (f_1(t))\bigcap domain \ of \ (f_2(t)) \bigcap\ Domain \ of (f_3(t) )\\$

$\Rightarrow \ {{t>0}} \bigcap {{t>2}}\bigcap \ t>-1 \\$

$\Rightarrow \ t>2$

$Therefore, \ the \ domain \ of \ the \ definition \ of \ the \ vector \ function \ is \ t>2.$