In an AC circuit, the current I amperes is given by I=20sin4t, where t is the time in milliseconds.

Find the maximum value of the current and the first instant at which this maximum current occurs.

Okay, so I've actually got the answer, but I'm not sure how I got it /shot
These are my steps:

I=20sin3t

i find the maximum point first
-1

Question

In an AC circuit, the current I amperes is given by I=20sin4t, where t is the time in milliseconds.

Find the maximum value of the current and the first instant at which this maximum current occurs.

Okay, so I've actually got the answer, but I'm not sure how I got it /shot
These are my steps:

I=20sin3t

i find the maximum point first
-1<sin4t<1
-20<20sin4x<20
so the max value is positive 20

then i subbed it into the current
I=20sin4t
=20sin4(20)

then finding time:
20sin4(20) x10^-2

And i realise if i multiply that by 2, i get the answer
but why multiply by 2??
(pls explain this :3)

ganeshie8 · Answer

Your work for finding max value is correct

ganeshie8 · Answer

I = 20 is the max value you have got, plugging that in the current equation gives you :

$$\large 20 = 20\sin (4t)$$

ganeshie8 · Answer

solve $\large t$

anonymous · Answer

right, hang on

anonymous · Answer

20=20sin(4t)
1=sin(4t)
sin^-1(1)=4t
t=90/4
=45/2

ganeshie8 · Answer

you want to use radians i guess

ganeshie8 · Answer

sin^-1(1) = 4t

pi/2 = 4t

pi/8 = t

anonymous · Answer

oh uh
good point yes
no wonder

ganeshie8 · Answer

btw, that a bit cheating

anonymous · Answer

is that? why?

ganeshie8 · Answer

$$\large \sin^{-1}(1) = 4t$$

ganeshie8 · Answer

sin is periodic with a period of 2pi, so the correct way to solve it is 

$$\large \frac{\pi}{2} + 2n\pi = 4t$$

ganeshie8 · Answer

assuming t >= 0,
the first time max value occurs when n = 0

anonymous · Answer

yes

anonymous · Answer

how is simply subbing in the 20 'cheating' though?

ganeshie8 · Answer

subbing 20 is not cheating,
saying sin^-1(1) = 4t   means,  pi/2 = 4t  is bit lazy

anonymous · Answer

but but 
it's just solving an equation :o
and sin^-1(1) is a special angle, isn't it? 
so getting π/2 without working would be natural

ganeshie8 · Answer

sin^-1(1)  could be pi/2 or  5pi/2  or 9pi/2 or.. anythng
so when you solve an equation, you should never say sin^-1(1) = 4t means pi/2 = 4t

ganeshie8 · Answer

pi/2 is just one solution, there are infinitely many other solutions.

 its like saying x^2 = 1 means x = 1
but  x could be -1 also right ?

anonymous · Answer

oh yeahh there are other angles too
thanks ^^

actually sorry, if you wouldn't mind explaining one thing again, 
i don't really get the π/2+2nπ=4t working up there
is 2nπ the frequency?

ganeshie8 · Answer

nope

ganeshie8 · Answer

period of sin(x) is 2pi right ?

anonymous · Answer

yep

ganeshie8 · Answer

that means the value of sin(x) repeats after "2pi" time right ?

anonymous · Answer

and it repeats 4 times?

ganeshie8 · Answer

suppose if the value of sin(x) is 1,
then after 2pi seconds, its value will be 1 too

anonymous · Answer

wait, okay go on..

ganeshie8 · Answer

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