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Math Help - Polar Cooridinates Speed Velocity

  1. #1
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    Polar Cooridinates Speed Velocity

    The question I am trying to solve is,

    A Particle P moves in the place. The Rectangular Observer computes the magnitude of the speed and acceleration of P by observing the x- and y- coordinates of P (as functions of time) and using these formulas: speed = sqrt( (dx/dt)^2 + (dy/dt)^2 ) and acceleration = sqrt( (d^2x/dt^2)^2 + (d^2y,dt^2)^2 ). The other observer, the Polar Observer, finds it easier and more natural to measure the polar coordinates r and Theta of P(as functions of time), using herself as the origin, of course. The Polar Observer computes dr/dt, dTHETA,dt, d^2r/dt^2, etc. What formula should the Polar Observer use to compute the speed of the particle P? Deduce your answer from the formula for speed displayed above and the relationship among x, y, r, and Theta. Finally, if r and dTheta/dt are constant, derive this formula valid for uniform circular motion = ( (speed)^2 )/(r).

    So I know the relationships between x, y, r, and theta as x=rcostheta, y=rsinTheta.

    So just plugging them in I figured I would solve for r in both of these equations for x and y, and than take the derivative, but it should be with respect to t, but if it seems if I take the derivative when r = x/cosTheta, than it should be dr/dTheta = x/cosTheta. When it should be dr/dt, Can I just take the derivative and plug them in or do I need to change it to be with respect to t? If so would it looks like r(t) = x(t)cosTheta(t)?

    I was thinking if I need dr/dt, than r should be a function of time, so I'm thinking r(t) = t/cost and r(t) = t/sint and than take the derivative of each.
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  2. #2
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    Re: Polar Cooridinates Speed Velocity

    Quote Originally Posted by kevinnwhat View Post
    The question I am trying to solve is,

    A Particle P moves in the place. The Rectangular Observer computes the magnitude of the speed and acceleration of P by observing the x- and y- coordinates of P (as functions of time) and using these formulas: speed = sqrt( (dx/dt)^2 + (dy/dt)^2 ) and acceleration = sqrt( (d^2x/dt^2)^2 + (d^2y,dt^2)^2 ). The other observer, the Polar Observer, finds it easier and more natural to measure the polar coordinates r and Theta of P(as functions of time), using herself as the origin, of course. The Polar Observer computes dr/dt, dTHETA,dt, d^2r/dt^2, etc. What formula should the Polar Observer use to compute the speed of the particle P? Deduce your answer from the formula for speed displayed above and the relationship among x, y, r, and Theta. Finally, if r and dTheta/dt are constant, derive this formula valid for uniform circular motion = ( (speed)^2 )/(r).

    So I know the relationships between x, y, r, and theta as x=rcostheta, y=rsinTheta.

    So just plugging them in I figured I would solve for r in both of these equations for x and y, and than take the derivative, but it should be with respect to t, but if it seems if I take the derivative when r = x/cosTheta, than it should be dr/dTheta = x/cosTheta. When it should be dr/dt, Can I just take the derivative and plug them in or do I need to change it to be with respect to t? If so would it looks like r(t) = x(t)cosTheta(t)?

    I was thinking if I need dr/dt, than r should be a function of time, so I'm thinking r(t) = t/cost and r(t) = t/sint and than take the derivative of each.
    In rectangular coordinates, x and y are both functions of t. In polar coordinates, r and θ are both functions of t. Start with the equations x(t) = r(t)\cos\theta(t) and y(t) = r(t)\sin\theta(t). Differentiate both of them with respect to t, using the product rule and the chain rule. Then use the results to write \bigl(\tfrac{dx}{dt}\bigr)^2 + \bigl(\tfrac{dy}{dt}\bigr)^2 in terms of the polar coordinates and their derivatives.
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    Re: Polar Cooridinates Speed Velocity

    Thank you for your response, I am unsure of how to differentiate something in that form. I don't see how to differentiate "r(t)" for example, I have dx/dt = r(t)cos\theta(t) so Using the product rule I obtained dx/dt = cos\theta(t) - r(t)sin\theta(t). I know I'm doing something wrong because I am not using the chain rule, I am thinking you must use the chain rule when you differentiate r(t) and cos\theta(t)??
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  4. #4
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    Re: Polar Cooridinates Speed Velocity

    Quote Originally Posted by kevinnwhat View Post
    Thank you for your response, I am unsure of how to differentiate something in that form. I don't see how to differentiate "r(t)" for example, I have dx/dt = r(t)cos\theta(t) so Using the product rule I obtained dx/dt = cos\theta(t) - r(t)sin\theta(t). I know I'm doing something wrong because I am not using the chain rule, I am thinking you must use the chain rule when you differentiate r(t) and cos\theta(t)??
    Okay, I'll do one of them for you. If x = r\cos\theta, where x, r and θ are all functions of t, then the product rule, together with the chain rule, says that

    . . . . . . . . . . \frac{dx}{dt} = \frac{dr}{dt}\cos\theta + r(-sin\theta)\frac{d\theta}{dt}.
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