This is from Applications of the Derivative under Curvilinear Motion:
So a rocket follows a path given by y=x-(1/90)x^3. Distance in km.
If the horizontal velocity is Vx=x, find the magnitude and direction of the velocity when the rocket hits the ground, if time is in minutes.
So I have so far = I do the product rule
dx/dy=1-(1/90)x^3 + x-(1/30)x^2(dx/dt)
So then I multiply by 90 to get rid of the 90 and 30.
So I have now
Vy=-x^3+90X-0.1x^3
Gives me now
Vy = -1.1x^3+90x
I am stuck at this part and I don't know if I did this correct??
How do I solve with Vx=x???
Can someone direct me in the right direction.
Thanks
Joanne
OK I kinda somewhat understand. what I am not getting is that you have dy/dt is that not Vy?? Why sub in the V, is what I am lost on now?
Sorry to keep asking, something is not clear to me here.
Ok looking at it and going over it. the -18.9734 us tge Vy component of it.
V is the resultant in a way then? correct?
I think my confusion is this:
dy/dx is Vy and your saying dy/dt is the same as Vy. So dx/dy is Vx and dx/dt is the same correct??
But regarding to time, it's the dy or dx over dt correct?? to still solve Vy or Vx? See the teacher never really covered this in class.
My book does not explain it that well, so I think this is where my confusion is. Trying to find info online is not that helpful either.
Jo
I do appreciate the help very much!