Thread: Parametric Curve and Travelling Object, help

1. Parametric Curve and Travelling Object, help

Hi everyone, I received this problem in my Calculus 2 class today as a bonus problem and I don't know how to proceed on it:

---

A bead is sliding on a wire having position x(t) = (2-3t)^2, y(t) = 2-3t. If the bead slides off the wire at time t=1, where is it when t=3?

---

So far I've found that at t = 1, the point on the wire is (1,-1). Eliminating t in the parametric equation leaves the equation x = y^2 which is a parabola opening to the right. The first derivative should be 2y, which means the velocity at t = 1 should be 2. What I don't know is what path the bead will take after it slides off the wire and what to do after that.

Any help will be greatly appreciated, thanks

2. Originally Posted by Semloh
Hi everyone, I received this problem in my Calculus 2 class today as a bonus problem and I don't know how to proceed on it:

---

A bead is sliding on a wire having position x(t) = (2-3t)^2, y(t) = 2-3t. If the bead slides off the wire at time t=1, where is it when t=3?

---

So far I've found that at t = 1, the point on the wire is (1,-1). Eliminating t in the parametric equation leaves the equation x = y^2 which is a parabola opening to the right. The first derivative should be 2y, which means the velocity at t = 1 should be 2. What I don't know is what path the bead will take after it slides off the wire and what to do after that.

Any help will be greatly appreciated, thanks
The velocity at any time the bead is on the wire is the vector:

$
\frac{d}{dt}\left((2-3t)^2, 2-3t)\right)=\left(18t-12, -3\right)
$

So at $t=1$ the velocity is $(6, -3)$, and as you know its position at that time you can calculate its subsequent path (either
assuming constant velocity or constant acceleration of $(0,-g)$)

RonL

3. Ahhhh, thanks!

If I plug in t=1 in the velocity formula, wouldn't I get (6,-3) though?

So assuming constant velocity from the moment it drops off the wire at point (1,-1), I can just add the displacement to the points which would be velocity multiplied by time. So that means at time t=3 the bead is at (13,-7). Is this right?

Thanks again, I'm just wondering if it's safe to assume that velocity is constant with the given information. What if velocity is changing due to the force of gravity?

4. Originally Posted by Semloh
Ahhhh, thanks!

If I plug in t=1 in the velocity formula, wouldn't I get (6,-3) though?
Opps, yes.

So assuming constant velocity from the moment it drops off the wire at point (1,-1), I can just add the displacement to the points which would be velocity multiplied by time. So that means at time t=3 the bead is at (13,-7). Is this right?
Yes

Thanks again, I'm just wondering if it's safe to assume that velocity is constant with the given information. What if velocity is changing due to the force of gravity?
You need to determine if you are supposed to treat this as being in a
problem where you treat the trajectory as being in a vertical plane and
subject to gravity, of being in a horizontal plane on a low friction table say.

There should be some hint somewhere in the question paper or the context
of the course at the point that it was set.

If gravity is to be allowed for you ave at time t=1, position (1,-1), velocity
(6, -3) and acceleration (0,-g), and you integrate the differential equations
for horizontal and vertical components seperatly to get the trajectory

RonL