text: # a particle is moving.. w/ equation.
  
  • Feb 27th 2007, 05:22 PM
  rcmango
  a particle is moving.. w/ equation.
  A particle is moving so that the position at time t is given by:
  r(t) = e^t(cost i - sint j)
  
  Find the velocity vector at time t:
  
  find the acceleration vector at time t:
  
  find the speed at time t:
  
  i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?
  
  i wasn't sure on the derivatives of a natural log.
  
  thanks for any help.
  • Feb 27th 2007, 06:07 PM
  Jhevon
  Quote:
  
  Originally Posted by rcmango
  A particle is moving so that the position at time t is given by:
  r(t) = e^t(cost i - sint j)
  
  Find the velocity vector at time t:
  
  find the acceleration vector at time t:
  
  find the speed at time t:
  
  i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?
  
  i wasn't sure on the derivatives of a natural log.
  
  thanks for any help.
  
  you are right about first and second derivatives for velocity and acceleration. speed is just velocity without direction, so i believe you just remove the i and j. derivative of lnx is 1/x, of course if x is a function you have to multiply by it's derivative
  • Feb 27th 2007, 07:49 PM
  topsquark
  Quote:
  
  Originally Posted by Jhevon
  speed is just velocity without direction, so i believe you just remove the i and j.
  
  Yes, and no.
  
  Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
  v = ai + bj
  then the speed would be:
  |v| = sqrt(a^2 + b^2)
  
  -Dan
  • Feb 27th 2007, 07:51 PM
  Jhevon
  Quote:
  
  Originally Posted by topsquark
  Yes, and no.
  
  Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
  v = ai + bj
  then the speed would be:
  |v| = sqrt(a^2 + b^2)
  
  -Dan
  
  A yes, now i remember. speed = |v|. There i go running my mouth off again without knowing the facts
  • Feb 27th 2007, 09:39 PM
  rcmango
  might need some help fixing my differentiation skills..
  
  for the first derivative my guess would be to use the chain rule to differentiate?
  
  so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)
  
  and for the 2nd derivative i guessed to use the chain and product rule in combination: so i got something like:- 1/x^2(cost i - sin j)(sint i - cost j)[(sint i - cost j)(sint i - cost j) + (cost i - sint j)(cost i + sint i)]
  
  how am i doing.
  
  thanks.
  • Feb 28th 2007, 09:00 AM
  topsquark
  Quote:
  
  Originally Posted by rcmango
  A particle is moving so that the position at time t is given by:
  r(t) = e^t(cost i - sint j)
  
  Find the velocity vector at time t:
  
  find the acceleration vector at time t:
  
  find the speed at time t:
  
  Quote:
  
  Originally Posted by rcmango
  might need some help fixing my differentiation skills..
  
  for the first derivative my guess would be to use the chain rule to differentiate?
  
  so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)
  
  What is x??
  
  I think you are making this too hard. Given h(x) = f(x)*g(x)
  h'(x) = f'(x)*g(x) + f(x)*g'(x)
  
  So:
  r(t) = e^t*(cos(t)i - sin(t)j)
  r'(t) = e^t*(cos(t)i - sin(t)j) + e^t*(-sin(t)i - cos(t)j)
  r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)
  
  r''(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j) + e^t*([-sin(t) - cos(t)]i + [-cos(t) + sin(t)]j)
  r''(t) = e^t*(-2sin(t)i + -2cos(t)j)
  r''(t) = -2e^t*(sin(t)i + cos(t)j)
  
  For the speed:
  r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)
  
  So
  |r'(t)| = sqrt{e^(2t)[cos(t) - sin(t)]^2 + e^(2t)[-sin(t) - cos(t)]^2}
  
  |r'(t)| = e^t*sqrt{cos^2(t) + sin^2(t) - 2sin(t)cos(t) + sin^2(t) + cos^2(t) + 2sin(t)cos(t)}
  
  |r'(t)| = e^t*sqrt{1 + 1}
  
  |r'(t)| = e^t*sqrt{2}
  
  -Dan
explanation: The document is a math question from a forum post, where a user is asking for help with a calculus problem involving a particle's motion. The user provides an equation for the particle's position and asks for the velocity, acceleration, and speed at time \( t \). The user also provides some initial thoughts and attempts at solving the problem.
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  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # a particle is moving.. w/ equation.
  
  • Feb 27th 2007, 05:22 PM
  rcmango
  a particle is moving.. w/ equation.
  A particle is moving so that the position at time t is given by:
  r(t) = e^t(cost i - sint j)
  
  Find the velocity vector at time t:
  
  find the acceleration vector at time t:
  
  find the speed at time t:
  
  i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?
  
  i wasn't sure on the derivatives of a natural log.
  
  thanks for any help.
  • Feb 27th 2007, 06:07 PM
  Jhevon
  Quote:
  
  Originally Posted by rcmango
  A particle is moving so that the position at time t is given by:
  r(t) = e^t(cost i - sint j)
  
  Find the velocity vector at time t:
  
  find the acceleration vector at time t:
  
  find the speed at time t:
  
  i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?
  
  i wasn't sure on the derivatives of a natural log.
  
  thanks for any help.
  
  you are right about first and second derivatives for velocity and acceleration. speed is just velocity without direction, so i believe you just remove the i and j. derivative of lnx is 1/x, of course if x is a function you have to multiply by it's derivative
  • Feb 27th 2007, 07:49 PM
  topsquark
  Quote:
  
  Originally Posted by Jhevon
  speed is just velocity without direction, so i believe you just remove the i and j.
  
  Yes, and no.
  
  Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
  v = ai + bj
  then the speed would be:
  |v| = sqrt(a^2 + b^2)
  
  -Dan
  • Feb 27th 2007, 07:51 PM
  Jhevon
  Quote:
  
  Originally Posted by topsquark
  Yes, and no.
  
  Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
  v = ai + bj
  then the speed would be:
  |v| = sqrt(a^2 + b^2)
  
  -Dan
  
  A yes, now i remember. speed = |v|. There i go running my mouth off again without knowing the facts
  • Feb 27th 2007, 09:39 PM
  rcmango
  might need some help fixing my differentiation skills..
  
  for the first derivative my guess would be to use the chain rule to differentiate?
  
  so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)
  
  and for the 2nd derivative i guessed to use the chain and product rule in combination: so i got something like:- 1/x^2(cost i - sin j)(sint i - cost j)[(sint i - cost j)(sint i - cost j) + (cost i - sint j)(cost i + sint i)]
  
  how am i doing.
  
  thanks.
  • Feb 28th 2007, 09:00 AM
  topsquark
  Quote:
  
  Originally Posted by rcmango
  A particle is moving so that the position at time t is given by:
  r(t) = e^t(cost i - sint j)
  
  Find the velocity vector at time t:
  
  find the acceleration vector at time t:
  
  find the speed at time t:
  
  Quote:
  
  Originally Posted by rcmango
  might need some help fixing my differentiation skills..
  
  for the first derivative my guess would be to use the chain rule to differentiate?
  
  so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)
  
  What is x??
  
  I think you are making this too hard. Given h(x) = f(x)*g(x)
  h'(x) = f'(x)*g(x) + f(x)*g'(x)
  
  So:
  r(t) = e^t*(cos(t)i - sin(t)j)
  r'(t) = e^t*(cos(t)i - sin(t)j) + e^t*(-sin(t)i - cos(t)j)
  r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)
  
  r''(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j) + e^t*([-sin(t) - cos(t)]i + [-cos(t) + sin(t)]j)
  r''(t) = e^t*(-2sin(t)i + -2cos(t)j)
  r''(t) = -2e^t*(sin(t)i + cos(t)j)
  
  For the speed:
  r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)
  
  So
  |r'(t)| = sqrt{e^(2t)[cos(t) - sin(t)]^2 + e^(2t)[-sin(t) - cos(t)]^2}
  
  |r'(t)| = e^t*sqrt{cos^2(t) + sin^2(t) - 2sin(t)cos(t) + sin^2(t) + cos^2(t) + 2sin(t)cos(t)}
  
  |r'(t)| = e^t*sqrt{1 + 1}
  
  |r'(t)| = e^t*sqrt{2}
  
  -Dan
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a math question from a forum post, where a user is asking for help with a calculus problem involving a particle's motion. The user provides an equation for the particle's position and asks for the velocity, acceleration, and speed at time \( t \). The user also provides some initial thoughts and attempts at solving the problem.
  
  Label: question-and-answer