Answer to Question #170065 in Calculus for Keerthidas

Question #170065

Find the domain of the vector function ?

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


1
Expert's answer
2021-03-10T07:06:35-0500

Given that the vector function isGiven \ that \ the \ vector \ function \ is (t5,4log(t2),3ln(t+1))(t^5 , 4log(t-2),3ln(t+1))\\

A vector function f(t)=(f1(t),f2(t),f3(t)) is said to be defined if the component functions f1(t)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)f2(t) and f3(t) are defined.The set of all t  for which f1,f2 and f3are defined iscalled the domain of f(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 f2(t)=4log(t2) is defined only for t2>0t>2Thedomain of f2(t)is, the set of all t>2.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 f1(t)=t5 is defined for all t R.The \ function \ f_1(t)=t^5 \ is \ defined \ for \ all\ t \isin\ R.\\ \\

The component function f3(t)=3ln(t+1) is defined if (t+1)>0 t+1>0t>1The \ 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(f1(t))domain of (f2(t)) Domain of(f3(t))The \ domain \ of \ f(t) \ is \ Domain (f_1(t))\bigcap domain \ of \ (f_2(t)) \bigcap\ Domain \ of (f_3(t) )\\

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

 t>2\Rightarrow \ t>2

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment