Question

The displacement of a particle on a vibrating string is given by the equation s(t)=4+4sin(8πt), where s is measured in centimeters and t is in seconds. Find the velocity and acceleration of the particle after t seconds.

I am aware that velocity involves a first derivative and acceleration involves the second derivative.

LaTeX will be below.

Answer 1

Well.....

Answer 2

First derivative:\[s'(t)=(4+4\sin{8\pi t})'\]\[=0+4(\cos{(8\pi t)}\times8\pi)\]\[=4(\cos{(8\pi t)}\times32\pi\]

Answer 3

Oops, left out a parentheses in the final part.

How is the result? Is this correct?

Answer 4

You seem to have an extra 4 in there

Answer 5

I do? Where D:

Answer 6

Check your work in the second and third lines.

Answer 7

...I don't get it D:

Answer 8

You don't notice an 8 magically turning into a 32?

Answer 9

Well, yeah, it's \(4\times8\pi\) isn't it? o-o

Answer 10

Do you not see it? This is what you wrote earlier\[\large 4(\cos{(8\pi t)}\times8\pi)\] \[\large 4(\cos{(8\pi t)}\times32\pi)\]

Answer 11

I said I missed a parentheses in the last bit, but I meant it as \[4(\cos{(8\pi t)})\times32\pi\]

Answer 12

Math doesn't work that way.

Answer 13

\[\large a(b*c) \neq ab*ac\]

Answer 14

OH, wow. That was stupid of me.

Answer 15

\(s'(t)=32\pi\cos{(8\pi t)}\)?

Answer 16

Yes.

Answer 17

Alright, thank you!
And the derivative of that....\[s"(t)=-256\pi^{2}\sin{(8\pi t)}\]Is this?

Answer 18

LOL using quotation mark for second derivative,
that's cute XD

Answer 19

^lol

Looks about right. You can use wolfram alpha to find derivatives btw.

Answer 20

I believe I put two apostrophes :\
oh well.

alright thanks for the tip!

Answer 21

\(\large\rm "~\ne~~''\)

XD

Answer 22

I guess it must've been my auto correct then. Oops.