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Re: [Phys-L] f'(t) / g'(t) question



For motion of a particle confined to the xy plane, a graph of v_y vs. v_x is called a hodograph. You can think of it as the trajectory of the arrowhead of the velocity vector where the tail of the velocity vector is at the origin. For a case where the velocity is not constant, imagine the change in that velocity vector in time dt. You draw (an arrow representing) a vector from the origin to the point on the hodograph curve corresponding to the velocity at time t. Then draw another one from the origin to the point on the hodograph curve corresponding to the velocity at time t + dt. The change in velocity would be the vector represented by an arrow from the tip of the first velocity vector to the tip of the second velocity vector. In the limit as dt goes to zero, you can see that the change in velocity is tangent to the hodograph curve. This means that the acceleration is always tangent to the hodograph curve. The slope of v_y vs. v_x is thus the (trig function) tangent of the direction of the acceleration. Your intuition is right, the slope does tell you something useful. The acceleration of a particle is always directed along (tangent to) its hodograph just as the velocity of a particle is directed along (tangent to) its trajectory.

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@www.phys-l.org] On Behalf Of Lulai,
Paul
Sent: Friday, April 08, 2016 11:20 AM
To: Phys-L@phys-l.org
Subject: Re: [Phys-L] f'(t) / g'(t) question

Thanks folks. Your answers are more complete than my question.
I understand the lack of direction on the Pythagorean method.
The conversation started with her presentation of a graph of g'(t) vs f'(t).
From that visual, it /felt/ like the slope of the graph should tell me something
useful. I couldn't come up with anything. It helps to have such clear
feedback.
Have a good one.
Paul.
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