Motion in the Plane

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turophile

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I'm working on a problem related to motion in the plane, and I'm stuck for a next step. Here's the problem statement:

A particle moves so that at time t it is at P(t). It does not pass through the origin, but is nearest the origin at time t = 2. Assuming that P is differentiable, show that P(2) ∙ P'(2) = 0.

Here's my work so far:

Let P = (f(t), g(t)). Since P(t) does not pass through the origin, P(t) ≠ (0, 0). We know that the length of the segment OP is |P| = (x2 + y2)1/2 = (f(t)2 + g(t)2)1/2. We are given that OP is smallest at time t = 2.

Am I heading in the right direction? And if so, what would a useful next step be? Thanks!
 
turophile said:
I'm working on a problem related to motion in the plane, and I'm stuck for a next step. Here's the problem statement:

A particle moves so that at time t it is at P(t). It does not pass through the origin, but is nearest the origin at time t = 2. Assuming that P is differentiable, show that P(2) ∙ P'(2) = 0.

Here's my work so far:

Let P = (f(t), g(t)). Since P(t) does not pass through the origin, P(t) ≠ (0, 0). We know that the length of the segment OP is |P| = (x2 + y2)1/2 = (f(t)2 + g(t)2)1/2. We are given that OP is smallest at time t = 2.

Am I heading in the right direction? And if so, what would a useful next step be? Thanks!
Click to expand...

P(2) ∙ P'(2) = 0.

Does that mean P(2) [times] P'(2) = 0?

Are P and P' treated as scalars?

If those are scalars then your problem should be written as P(2) * P'(2) = 0.
 
Sorry if the notation wasn't clear. It comes from the textbook I'm studying. The dot notation it uses is defined to mean the scalar product (or dot product or inner product).
 
turophile said:
Sorry if the notation wasn't clear. It comes from the textbook I'm studying. The dot notation it uses is defined to mean the scalar product (or dot product or inner product).
Click to expand...

So then P(t) and P'(t) are vectors.

When the dot product of two vectors is zero - what does it say about their relative orientations (angle between the vectors)?

How can you use that information for the problem at hand?
 
That means that P(t) and P'(t) need to be perpendicular at t = 2. I'm not sure how to apply that result to the problem though.
 
turophile said:
I'm working on a problem related to motion in the plane, and I'm stuck for a next step. Here's the problem statement:

A particle moves so that at time t it is at P(t). It does not pass through the origin, but is nearest the origin at time t = 2. Assuming that P is differentiable, show that P(2) ∙ P'(2) = 0.

Here's my work so far:

Let P = (f(t), g(t)). Since P(t) does not pass through the origin, P(t) ≠ (0, 0). We know that the length of the segment OP is |P| = (x2 + y2)1/2 = (f(t)2 + g(t)2)1/2. We are given that OP is smallest at time t = 2.

Am I heading in the right direction? And if so, what would a useful next step be? Thanks!
Click to expand...
I would give the length OP the name r(t). Since r(2) is a minimum, you need to set dr/dt = 0. Differentiating implicitly,
r dr/dt =x f'(t) + y g'(t)

P' is the derivative of P with respect to time:
P'(t) = (f'(t), g'(t))

That is almost done...
What does the dot product remind you of?
 
turophile said:
That means that P(t) and P'(t) need to be perpendicular at t = 2. I'm not sure how to apply that result to the problem though.
Click to expand...

Perpendicular distance is the shortest distance between a line and a point.

Derivative at a point = slope of the tangent line to the curve at the point...
 
It's taken me a little while to get back to working on this problem, and I've been thinking about it some more. Can I let x = f(t) = cos(t) and y = g(t) = sin(t), so that OP = |P| = r(t) = (cos2t + sin2t)1/2? Then d/dt |P| = r'(t) = [(1/2)(cos2t + sin2t)-1/2][- 2 sin(t) cos (t) + 2 sin(t) cos(t)] = 0? Could I then state that since P'(t) = 0 then P(t) ∙ P'(t) = 0? That doesn't seem quite right to me, since I haven't shown anything special about t = 2.
 
Think of it this way: If you want to find a point "closest to the origin", find a vector from the point to the origin and move in that direction. With no other restrictions, that would take you directly to the origin. If you are required to stay on a specific curve (or surface in three dimensions) you might not be able to move directly toward the origin but you could move in the direction of the projection of that vector on the curve (or surface) and move in that direction. Continue to do that until you cannot- that is until the vector from the curve to the origin is perpendicular to the surface and has no projection onto it. If P(t)= <x(t), y(t)> is the "position vector" of the curve, then P'(t)= <x'(t), y'(t)> is its tangent vector and P(t) must be perpendicular to that: P(t)P'(t)= 0.
 
turophile said:
I'm working on a problem related to motion in the plane, and I'm stuck for a next step. Here's the problem statement:

A particle moves so that at time t it is at P(t). It does not pass through the origin, but is nearest the origin at time t = 2. Assuming that P is differentiable, show that P(2) ∙ P'(2) = 0.

Here's my work so far:

Let P = (f(t), g(t)). Since P(t) does not pass through the origin, P(t) ≠ (0, 0). We know that the length of the segment OP is |P| = (x2 + y2)1/2 = (f(t)2 + g(t)2)1/2. We are given that OP is smallest at time t = 2.

Am I heading in the right direction? And if so, what would a useful next step be? Thanks!
Click to expand...
I would write the formula for P(t)^2 and differentiate with respect to time. Since P is differentiable, P'(2) exists, and since P(t) is a minimum, so is P^2(t), so you can set the derivative of P^2 (evaluated at t=2) to zero.
 
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