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Homework Help: Related Rates Elipse Question

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle is moving along the ellipse x2/16 + y2/4 = 1. At each time t its x and y coordinates are given by x = 4cost, y = 2sint. At what rate is the particle's distance from the origin changing at time t? At what rate is the distance from the origin changing when t = pi/4?

    2. Relevant equations
    x2/16 + y2/4 = 1
    x = 4cost
    y = 2sint
    3. The attempt at a solution
    I am fairly sure i can do this question, the only problem is that i am not so sure what it is asking for. I'm assuming that i am to find dy/dx.
    dy/dx = (dy/dt)(dt/dx)
    dy/dt = 2cost
    dx/dt = -4sint

    x2/16 + y2/4 = 1
    2x(dx/dt) (1/16) + 2y(dy/dt)(1/4) = 0
    (x/8) (-4sint) + (y/2)(2cost) = 0
    (-xsint)/2 + ycost = 0
    At t = pi/4
    2ycos(pi/4) = xsin(pi/4)
    2y(1/√2) = x(1/√2)
    x = 2y

    dy/dx = (dy/dt) (dt/dx)
    = (2cost)(-1/4sint)
    = (-1/2) (cost/sint)
    = (-1/2)(cott)
    at t = pi/4
    dy/dx = (-1/2)(1/1)
    = -1/2

    Does this mean that the answer is -1/2 because then i'm not sure why they gave us the equation for the ellipse because it doesnt play any effect in finding dy/dx. Thank you.
    Last edited by a moderator: Nov 3, 2011
  2. jcsd
  3. Nov 3, 2011 #2


    Staff: Mentor

    No, they're not asking for dy/dx. You wrote what the problem is asking for in your problem statement:
    What expression represents the particle's distance from the origin?
  4. Nov 3, 2011 #3
    the rate of change of x with respect to t plus the rate of change of y with respect to t:
    dx/dt + dy/dt ?
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