Vector valued function

[JJ007]

Find the domain of the vector valued functions
$\displaystyle r(t)=3ti- \sqrt{1-t^2}j+4k$
\(\displaystyle $-1\leq$ t $\leq$1?\)

$\displaystyle r(t)=\frac{1}{t}i+ln(-t)j+t^2k$
$\displaystyle (-\infty,o)?$

Thanks

 

[Deleted member 4993]

Find the domain of the vector valued functions
\(\displaystyle r(t)=3ti- sqrt{1-t^}j+4k\)

I think you are missing a set of parenthesis. Is your function:

r(t)=3ti - sqrt(1-t^)j + 4k <<< What's the intention of the carret sign over 't'?


\(\displaystyle $-1\leq$ t $\leq$1?\)

$\displaystyle r(t)=\frac{1}{t}i+ln(-t)j+t^2k$
$\displaystyle (-\infty,o)?$

Thanks

 

[JJ007]

Find the domain of the vector valued functions
\(\displaystyle r(t)=3ti- sqrt{1-t^}j+4k\)

I think you are missing a set of parenthesis. Is your function:

$\displaystyle r(t)=3ti - \sqrt{(1-t^2)}j + 4k$ <<< What's the intention of the carret sign over 't'?


\(\displaystyle $-1\leq$ t $\leq$1?\)

$\displaystyle r(t)=\frac{1}{t}i+ln(-t)j+t^2k$
$\displaystyle (-\infty,o)?$

Thanks

Sorry, fixed it

 

[BigGlenntheHeavy]

$\displaystyle Now, \ unless \ stated \ otherwise, \ we \ consider \ the \ domain \ of \ the \ vector-valued \ function$

$\displaystyle r \ to \ be \ the \ intersection \ of \ the \ domain \ of \ the \ component \ functions \ f, \ g, \ and \ h.$

$\displaystyle Hence, \ r(t) \ = \ 3ti-\sqrt{1-t^2}j+4k.$

$\displaystyle f(t) \ = \ 3t, \ domain \ = \ all \ reals.$

$\displaystyle g(t) \ = \ -\sqrt{1-t^2}, \ domain \ = \ [-1,1].$

$\displaystyle h(t) \ = \ 4, \ domain \ = \ all \ reals.$

$\displaystyle Therefore, \ the \ domain \ of \ r(t) \ = \ [-1,1].$

$\displaystyle What \ is \ the \ range \ of \ r(t)?$