Originally Posted by

**bradycat** 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