# Math Help - Question on Curvilinear Motion

1. ## Question on Curvilinear Motion

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

2. 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
When the rocket hits the ground,

$\displaystyle y=0$

$\displaystyle x-\frac{x^3}{90}=0$

Solve for $\displaystyle x$.

------------------------------------------------------------------------------------------------

$\displaystyle y=x-\frac{x^3}{90}$

Differentiating both sides wrt $\displaystyle t$,

$\displaystyle \frac{dy}{dt}=(1-\frac{x^2}{30})\frac{dx}{dt}$

Using $\displaystyle \frac{dx}{dt}=x$, we get

$\displaystyle \frac{dy}{dt}=(1-\frac{x^2}{30})x$

Now, substitute the value of $\displaystyle x$ you previously found to find the velocity vector $\displaystyle \vec{v}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}$.

3. So I get 90x-x^3=0
For some stupid reason I forgot this part.
is it 90=x^3/x ???
so I have 90=x^2
So x =9.4868??

In my question when it said Vx=x, how is that used??
I am a bit lost on this one.

4. Originally Posted by bradycat
So I get 90x-x^3=0
For some stupid reason I forgot this part.
is it 90=x^3/x ???
so I have 90=x^2
So x =9.4868??
Yes.

Originally Posted by bradycat
In my question when it said Vx=x, how is that used??
I am a bit lost on this one.
$\displaystyle v_x=\frac{dx}{dt}=x$

5. So Find the derivative of the original equation and sub in the x value??
Sorry I am not getting this for some reason. Brain dead tonight I think.

6. Originally Posted by bradycat
So Find the derivative of the original equation and sub in the x value??
Sorry I am not getting this for some reason. Brain dead tonight I think.
$\displaystyle \frac{dx}{dt}=9.4868$

$\displaystyle \frac{dy}{dt}=(1-\frac{9.4868^2}{30})9.4868=-18.9734$

$\displaystyle \vec{v}=9.4868\hat{i}-18.9734\hat{j}$

$\displaystyle |\vec{v}|=\sqrt{9.4868^2+18.9734^2}=21.2130$

7. 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?

8. Originally Posted by bradycat
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?
$\displaystyle \frac{dy}{dt}$ and $\displaystyle v_y$ mean the same thing. I have found $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$ in terms of $\displaystyle x$ and finally substituted $\displaystyle x=9.4868$.

$\displaystyle \vec{v}$ is the resultant.

Go over all my posts again and tell me if there's something you don't follow.

9. 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!

10. Originally Posted by bradycat
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!
$\displaystyle \frac{dy}{dx}$ is the slope of the curve. $\displaystyle v_y$ is the y-component of the rocket's velocity.

$\displaystyle \frac{dy}{dx}\neq v_y$ and $\displaystyle \frac{dx}{dy}\neq v_x$.

$\displaystyle \frac{dx}{dt}$ and $\displaystyle v_x$ mean the same thing, as do $\displaystyle \frac{dy}{dt}$ and $\displaystyle v_y$.

11. AWESOME, thank you, that helps ALOT let me tell you!!!!!!!!!!!!!!!!!!!!!!!!