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.

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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.

Physics
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Have you taken calculus by any chance? And If so is this a calculus based class?
I can walk you through it but I need to think about how to tailor my approach based on your level.
Yes. This is calc based. I believe the derivative needs to be taken to solve this

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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?
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
Sorry I didnt see that you had returned hold on...
Ok is this the function you gave:
\[ r(t)= (c*t^{2} - (2d)*t^{3})i+((2c)*t^{2} - d* t^{3})j \]
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
Are you still there?
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Hello
So if a particle is moving in the x-direction what does that mean?
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.
I believe it is correct
My hint was this.. Hint 1. You will have to take a derivative with respect to time to get the velocity of the particle.
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
So in other words any time t>0 except fot that point t=c/(3d) the particle is moving in the x direction
I already tried t=c/3d and it said I was missing something
Specifically, Your answer either contains an incorrect numerical multiplier or is missing one.
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
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?
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)
Like I said I feel like this question is either ill posed or incorrect... If that isn't the answer.... all t =/= c/(3d)
Does it mean in the POSITIVE x direction maybe?
It was asking for t>0 but told me that c/3d was wrong. This computer program for this question is on crack.
Yes because it IS wrong.... The answer is all t NOT equal to that value
So I need to make an equation that excludes that point
Assuming any motion in the x-direction (either way) is permitted.
sure try 0
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?
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?
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
But that is just a guess
All of 0
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
So for all times 0 < t < c/(3d)
After that the velocity becomes negative (moves in the -x-direction) and stays that ways forever
Did you try that?
I don't know if I can put it in like that
I figured it out. Yay for physics roommate! Thank you for trying!!!
So what did the answer end up being?

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