You are here:

Advanced Math/Parametric equations

Advertisement

Expert: - 5/25/2008


Question
Hi Paul,
It is my understanding that speed is the absolute value of velocity. But then how do I find the functions that gives me the speed. Confused .. pls help - Thank you

Consider the graph defined parametrically by  x=t^2 and  y=t^3 -3t for  -5<=t<=5.

  1. Find a function s(t) which gives the speed of the particle.
  2. Find the time(s) that the particle reaches its slowest speed.
  3. Find the perimeter of the “loop” in this graph.
  4. Find the average speed of the particle while it is traveling on the “loop.”

Answer

Parametric equations
Questioner:   sami
Category:  Advanced Math
Private:  No
 
Subject:  parametric function
Question:  Hi Paul,
It is my understanding that speed is the absolute value of velocity. But then how do I find the function that gives me the speed. Confused .. pls help - Thank you

Consider the graph defined parametrically by  x=t^2 and  y=t^3 -3t for  -5<=t<=5.

 1. Find a function v(t) which gives the speed of the particle.
 2. Find the time(s) that the particle reaches its slowest speed.
 3. Find the perimeter of the “loop” in this graph.
 4. Find the average speed of the particle while it is traveling on the “loop.”
..............................................
Hi, Sami,

I have changed your notation a bit -- I made it v(t) for your speed, BECAUSE it is customary in this case to write  s(t) for arc length.  As your 'particle' travels, and its coordinates (x,y) are given by your p.e.'s, the distance traveled is given by s(t), and the velocity is

v(t) = ds/dt = sqrt(dx/dt^2 + dy/dt^2)


Now for your questions:
 1. Find a function v(t) which gives the speed of the particle.

If:
x = t^2, dx/dt = 2t

y = t^3 - 3t, dy/dt = 3t^2 - 3

v(t) = sqrt(4t^2 + (3t^2 - 3)^2)

v(t) = sqrt(4t^2 + 9(t^4 - 2t^2 + 1))

v(t) = sqrt(4t^2 + 9t^4 - 18t^2 + 9)

v(t) = sqrt(9t^4 - 14t^2 + 9)

[No problem with negative numbers here -- sqrt() is >= 0.]
....................................
 2. Find the time(s) that the particle reaches its slowest speed.

This would mean to minimize v(t).  Usual stuff -- find a(t) = v'(t) and see when it is zero.  Might as well use v^2 and not bother with roots.

v^2 = 9t^4 - 14t^2 + 9

Dv^2 = 36t^3 - 28t = 0

t(36t^2 - 28) = 0

t1 = 0 and t2 = +- sqrt(7)/3

v^2(t2) = 9(49/81) - 14(7/9) + 9

v^2(t2) = 49/9 - 98/9 + 9

v^2(t2) = - 49/9 + 81/9 = 32/9  << that's your min. (of the square, that is)

v^2(0) = 9. = max

Except that  v^2(+-5) = much more.

 3. Find the perimeter of the “loop” in this graph.

I am not sure what you mean.  Does the 'loop' mean the entire interval [-5,5] or are you referring to the little piece that you see in the picture?

In any event, the 'perimeter' is the distance traveled, so you must do this integral:

{t2
|  sqrt(9t^4 - 14t^2 + 9) dt
}t1

-- not easy.  But once you find it, divide by (t2 - t1) to get the next:

 4. Find the average speed of the particle while it is traveling on the “loop.”

Paul Klarreich

Expertise

I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience

I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

Education/Credentials
-----------

©2012 About.com, a part of The New York Times Company. All rights reserved.