Question

A particle is moving with the given data.
Find the position of the particle.
a(t)=10sin(t)+3cos(t)
s(0)=0
s(2pi)=12

Answer 1

So you're getting into anti-differentiation now i see? :)

They gave us a function for the particles ACCELERATION.
Acceleration is the derivative of Velocity.
Velocity is the derivative of Position.

So we'll need to anti-differentiate 2 times.

If we anti-differentiate a(t) we'll get,\[\huge v(t)=-10\cos(t)+3\sin(t)+c\]Understand how the trig functions changed?
Also, whenever we anti-differentiate an unknown constant value will be tacked onto the end of the problem. If you need an explanation as to why the +c is there, let me know.

Ok where you stuck at? :)

Answer 2

Okay, I understand that (:

Answer 3

Now, do I need to find the anti derivative of that anti derivative..?

Answer 4

Yes c:

Answer 5

hmm, can I hire you to teach me? (; you're so much easier to understand then my chinese professor hahaha!

Answer 6

chinese? oh boy that sounds rough :3 lol

Answer 7

hahah oh it is rough. She's insane in the membrane.

Answer 8

heh :D do you understand how to find the next anti-derivative? it's a little trickier than the first, since something will happen with our c term.

Answer 9

oh no! Let me try (:

Answer 10

I got...
-10sin(t)-3cos(t)....
now im stuck?

Answer 11

When we anti-differentiate a second time, this is what we should end up with,
\[\huge s(t)=-10\sin(t)-3\cos(t)+cx+d\]

Answer 12

uhm, why cx+d?

Answer 13

So think about this a second.
When you take a DERIVATIVE, every term is DECREASING in power by 1 degree yes? (at least for the powers of x).

So if we had something like \[\huge \frac{d}{dx}3x^1 \qquad \rightarrow \qquad 3x^0 \qquad \rightarrow \qquad 3\]
Now let's think about that in reverse for a moment...
If I start with a constant, I can think of it as a constant times x^0. (Because x^0 is just 1). From there, we want to INCREASE the power of X, since we're anti-differentiating.

Answer 14

ohh... I'm getting ya...

Answer 15

\[3 \qquad \rightarrow \qquad 3x^0\]Anti-differentiating gives us,\[3x^1\]

Answer 16

In our case, it's not a 3, it's a C, but the principle still applies.
As for the d, that is just a NEW constant term that appeared!
I could have called it C2 or something perhaps, but D seemed fine...

Remember we get a constant EACH TIME we anti-differentiate.

Answer 17

Okay so now.. what are s(0)=0 and s(2pi)=12 for?

Answer 18

Those are specific values of s(t) that they provided us with.
What we have come up with so far, is a GENERAL SOLUTION for s(t).
It includes a couple of UNKNOWN constants that we can't solve for unless we're given initial conditions.

Well, we were! :) So we'll use those values to solve for c and d.

Answer 19

\[\large s(0)=0 \qquad \rightarrow \qquad 0=-10\sin(0)-3\cos(0)+3(0)+d\]

Answer 20

Understand how we did that? We set it equal to 0, before s(0) is equal to zero.
And then plug in 0 for all of our x values.

We should be able to easily solve for d at this point.

Answer 21

d=3?

Answer 22

k sounds good c:

Answer 23

Then use the other given value of s(t) to find c \:D/

Answer 24

s(2pi)=12... 12=-10sin(2pi)-3cos(2pi)+3c+3?

Answer 25

did i set that up wrong?

Answer 26

all looks good except the c part.
We have already solved for D, so D is 3.

But +cx should change to C(2pi) right?

Answer 27

i dunno why i changed from lowercase c to uppercase, that wasn't intentional lol

Answer 28

ohhh okay (:

Answer 29

I simplified it down to...
2pi*c=12

Answer 30

6/pi=c
yay! I got it (:

Answer 31

yay team \c:/

Answer 32

hha thanks so much~!