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Calculus gradient at a specific time

  1. Oct 16, 2011 #1
    The path of a particle is given by r(t) = tsin(t) * i - tcos(t) * j where t≥0. The particle leaves the origin at t = 0 and then spirals outwards.



    Let θ be the acute angle at which the path of the particle crosses the x-axis.
    Find tan(θ) when t = 3pi/2.




    I was able to figure out a few things:
    At t = 3pi/2, the particle crosses the x-axis for a second time (first being at t = pi/2).
    At t = 3pi/2, the speed of the particle is (1/2)sqrt(9*pi^2 + 4).
    Tan(θ) = gradient = dy/dx
    Other than that, I don't know where to start or what to do.
     
  2. jcsd
  3. Oct 16, 2011 #2

    HallsofIvy

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    Why are you talking about the second time the particle crosses the x-axis? The problem you give says nothing about "second time". Also I see no mention of "speed" so I don't see why you are calculating that.

    dy/dx= (dy/dt)/(dx/dt). Can you calculate that? Remember that when the particle crosses the x-axis, cos(t)= 0.
     
  4. Oct 16, 2011 #3
    I calculated that:
    v(t) = (sin(t) + tcos(t))*i + (tsin(t) - cos(t))*j
    But now I am completely lost. This is the graph of velocity vs time, not position vs time. If I substituted in t = 3pi/2, I get:
    v(3pi/2) = -1*i + (-3pi/2)*j
    Which doesn't make any sense to me because that is a velocity, not an angle.
     
  5. Oct 16, 2011 #4

    HallsofIvy

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    I told you exactly what to do in my first post. Read it.
     
  6. Oct 16, 2011 #5
    So if I let x = i and y = j, then:
    dx/dt
    = i component of r'(t)
    = sin(t) + tcos(t)
    dy/dt
    = j component of r'(t)
    = tsin(t) - cos(t)
    (dy/dt)/(dx/dt)
    = (tsin(t) - cos(t)) / (sin(t) + tcos(t))
    Let cos(t) = 0
    = tsin(t) / sin(t)
    = t
    since t = 3pi/2
    tan(θ) = 3pi/2

    Is that the answer?
     
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