A particle's position is r(vector) =(ct^2−2dt^3)i^+(2ct^2−dt^3)j^,
where c and d are positive constants. Find expressions for times t > 0 when the particle is moving in the x-direction.
Express your answer in terms of the variables c and d.
can anybody walk me through how to even approach this problem? I am totally lost.

- cgbork

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- anonymous

Have you taken calculus by any chance? And If so is this a calculus based class?

- anonymous

I can walk you through it but I need to think about how to tailor my approach based on your level.

- cgbork

Yes. This is calc based. I believe the derivative needs to be taken to solve this

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## More answers

- anonymous

Correct the instantaneous rate of change of position is velocity, and that can be calculated by taking the derivative. This seems function seems to be fairly tame, do you recall the derivative rule for polynomials?

- cgbork

not entirely. I am on a site for my class where it deducts 5% every time I guess wrong. I have put in d^2r/dt^2 and it said it didn't depend on dt or r

- anonymous

Sorry I didnt see that you had returned hold on...

- anonymous

Ok is this the function you gave:

- anonymous

\[ r(t)= (c*t^{2} - (2d)*t^{3})i+((2c)*t^{2} - d* t^{3})j \]

- anonymous

Note how I made sure not to run the constants together with the independent variable... * denotes typical multiplication but I added it explicitly to make sure this is what you had in the problem statement

- anonymous

Are you still there?

- cgbork

Internet failed

- anonymous

Hello

- anonymous

So if a particle is moving in the x-direction what does that mean?

- anonymous

It means the x-component of its velocity is non-zero... Please check to make sure you copied the question correctly because I suspect something is missing here.

- cgbork

I believe it is correct

- cgbork

My hint was this..
Hint 1. You will have to take a derivative with respect to time to get the velocity of the particle.

- anonymous

Correct, but what I am getting at is only requiring the x-component of the velocity v_x to be non-zero results in only one point where t>0 and v_x is zero and that is t=c/(3d).... I was expecting an interval or something more than a single point

- anonymous

So in other words any time t>0 except fot that point t=c/(3d) the particle is moving in the x direction

- cgbork

I already tried t=c/3d and it said I was missing something

- cgbork

Specifically,
Your answer either contains an incorrect numerical multiplier or is missing one.

- anonymous

No at that point there is NO motion in the x-direction. At that point v_x is zero... at all other times besides that point there is some motion in the x-direction

- cgbork

Oh crap. That is because I found it when t=0 and it was asking for when it is greater than 0. So how would I find it when it isn't 0, but instead t>0?

- anonymous

No that is the answer for when t > 0. When t=0 the answer is just v_x = 0 plain and simple but when t >0 you get c-3dt=0 or t = c/(3d)

- anonymous

Like I said I feel like this question is either ill posed or incorrect... If that isn't the answer.... all t =/= c/(3d)

- anonymous

Does it mean in the POSITIVE x direction maybe?

- cgbork

It was asking for t>0 but told me that c/3d was wrong. This computer program for this question is on crack.

- anonymous

Yes because it IS wrong.... The answer is all t NOT equal to that value

- cgbork

So I need to make an equation that excludes that point

- anonymous

Assuming any motion in the x-direction (either way) is permitted.

- anonymous

sure try 0

- cgbork

Part a is asking for the x direction and part b is asking for the y direction. So I'm hoping once I understand how to get part a that part b will be a breeze. Are you sure it is asking for all of that?

- anonymous

No Im not sure like I said I feel like this question is missing some information or poorly written or something. I mean have you tried the above answer?

- anonymous

Having only a single point seems kind of silly to me... I feel like it should be an interval or something.... Asking only about POSITIVE x-direction or something would make more sense

- anonymous

But that is just a guess

- cgbork

All of
0

- anonymous

if it were in the positive x-direction then the as long as 2ct > 6dt^2 or as long as 2c >6dt or c/(3d) > t

- anonymous

So for all times 0 < t < c/(3d)

- anonymous

After that the velocity becomes negative (moves in the -x-direction) and stays that ways forever

- anonymous

Did you try that?

- cgbork

I don't know if I can put it in like that

- cgbork

I figured it out. Yay for physics roommate! Thank you for trying!!!

- anonymous

So what did the answer end up being?

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