Question

Trig question! Please I can't figure this out, have been trying for a couple of weeks not wanting help but I do give up, please help! an electric generator produces a 30-cycle alternating current described by I=40sin(60pi(t-7/72)) where I is the current at t seconds. find the smallest positive value of t for which the current is 20 amps

Answer 1

\[20 = 40\sin(60\pi (t - \frac{ 7 }{ 72 }))\]

\[\frac{ 1 }{ 2 } = \sin(60\pi(t - \frac{ 7 }{ 72 }))\]

the opposite operation of sin is arcsin (the sin^(-1) button on your calculator)

\[\sin^{-1}(\frac{ 1 }{ 2 }) = \sin^{-1}(\sin(60\pi(t - \frac{ 7 }{ 72 }))\]

\[\frac{ \pi }{ 6 } = 60\pi(t - \frac{ 7 }{ 72 })\]

solving for t from here is trivial

Answer 2

I'm looking at your answer to see if I get it...btw thanks for replying

Answer 3

let me know if you have follow-up questions :)

Answer 4

yes i do, thanks for helping. at the first step, you divided both sides by 20 correct?

Answer 5

no, by 40

Answer 6

oh yes, that's how you get the 1/2?

Answer 7

yup

Answer 8

and what do you mean solving the rest is trivial?

Answer 9

well. im pretty sure you can do the algebra to solve to t

the complication was in the trig

Answer 10

hehehe. i understand. that's my problem, I am so saturated with the trig that I often get confused on how to keep going. ugh. I will try this. Thanks a lot!

Answer 11

t = 1/10 or 0.1

Answer 12

well u know, I chose that answer in my last test and I got it wrong, or should say it was marked wrong, and that's why I couldnt figure out this problem

Answer 13

\[\frac{ \pi }{ 6 } = 60\pi(t - \frac{ 7 }{ 72 })\]

divide both sides by 60pi

\[\frac{ 1 }{ 360 } = t - \frac{ 7 }{72 }\]

Answer 14

0.1 was marked wrong?

Answer 15

no, i marked 1/10 and the teacher said I got it wrong.

Answer 16

choices were: 1/10, 1/90, 1/9, 2/15

Answer 17

i see

Answer 18

answer is 1/9

Answer 19

I thought so but only by intuition, not because I know the answer

Answer 20

both of these are equal

Answer 21

but.... 1/9 > 1/10 so it shouldn't be right

Answer 22

oooooooooooo! i see!

Answer 23

my drawing is terrible since 5pi/6 should be the whole angle from 0 to there: