Question

A particle moves along the x-axis with position function s(t) = e^cos(x). How many times in the interval [0, 2π] is the velocity equal to 0?

1
2
3
More than 3

Answer 1

do you know how to differentiate?

Answer 2

I think so yeah

Answer 3

s(t) is the function for postition.

the derivative of s(t) with respect to time is velocity.

i'm sure the question is actually e^cos(t)

so can you find V=ds/dt ?

Answer 4

no it is s(t) = ecos(x).
So do i start with ln e^(cos x)

Answer 5

*s(t) = e^cos(x)*

Answer 6

the question makes sense only if
s(t)=e^cos(t)

Answer 7

oh but my problem has x not t....would that matter or can we continue?

Answer 8

we'll it matters. buts lets just consider it a mistake in the text book, and change the question to s(t)=e^cos(t)

Answer 9

what you should do is differentiate s(t) with respect to t

Answer 10

so I would have -e^cos(x) sin(x)

Answer 11

yes. except its not x its t.

now we know that the whole expression is zero when ever sin(t) is zero.

do you know when sin( ) is zero in the intervall [0,2pi] ?

Answer 12

does this have to do with the unit circle

Answer 13

yes

Answer 14

i don't know ;/

Answer 15

\(sin(n\pi) \) =0 where n is any integer

so in the interval [0,2pi]

we have

sin(0), sin(pi) and sin (2pi) (n=0,1 and)

3 times (at t=0 , pi and 2pi)

Answer 16

ooohh okay i see. Thank you :D