Question

find the domain of the vector function r(t) =

Answer 1

find the domain of each part then take the intersection

Answer 2

yea i get that

Answer 3

i'm having trouble with the actual construction

Answer 4

the intersection

Answer 5

ln(4t) has no restrictions so can be (-inf, inf)......sqrt(t+16) has restrictions at -16, and 1/sqrt(12-t) has restrictions at 12

Answer 6

no

Answer 7

ln(4t)

t>0

Answer 8

okay..so the only restriction then is 0

Answer 9

\[r(t)=(\ln 4t, \sqrt{t+16}, 1/\sqrt{12-t})\]
for the i component the domain is x>0
for j, x >or=-16
for k x<12

Answer 10

and don't put negative numbers under the radical

Answer 11

so the domain is (-16,12)

Answer 12

12>x>0

Answer 13

i have to do it in interval notation

Answer 14

so
(0,12)

Answer 15

(0,12)

Answer 16

awesome..thanks so much

Answer 17

i should prolly figure this out before the test

Answer 18

notice that the restriction \[x \ge -16\]doesn'e matter because we already have that x>0, so the intersection will not include x=-16

Answer 19

ohhhhh

Answer 20

that's what u meant earlier by don't use the negative numbers under the radical

Answer 21

well, the x≥−16 is because if x<-16 we get an imaginary number for the j-component, but after looking at the other restrictions we see that because the i-component has x>0 anyway, so in this case it doesn't matter.

Answer 22

feel like helping me with another one?

Answer 23

I'm eating, give me 5 min, then ok

Answer 24

alright...i'll create another post so i can give medals

Answer 25

thx