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A point moves along the curve y=2x^2+1

  1. Nov 11, 2014 #1

    dnt

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    • Thread was started in a non-homework section, so is missing the template
    A point moves along the curve y=2x^2+1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x=(3/2)?


    Ok here's what I've done...

    Dy/dt=(4x)(dx/dt)

    -2 = (4*3/2)(dx/dt)

    So dx/dt=-1/3 (ie decreasing by 1/3 units per second)

    First, is that right? Second when I look at the graph and y is decreasing it's the left side of the parabola however as I go down the graph the x increasing from left to right. So either my work is wrong above or I am misinterpreting the graph. What have I done wrong? Thanks.
     
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  3. Nov 11, 2014 #2

    pasmith

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    Yes.

    Looking at a graph of [itex]y[/itex] against [itex]x[/itex] only tells you how [itex]y[/itex] varies with [itex]x[/itex]. That graph tells you nothing about how [itex]y[/itex] varies with [itex]t[/itex] unless you know how [itex]x[/itex] varies with [itex]t[/itex]. If [itex]y[/itex] increases with [itex]x[/itex] but [itex]x[/itex] itself is decreasing with [itex]t[/itex], then [itex]y[/itex] will decrease with [itex]t[/itex].
     
  4. Nov 11, 2014 #3

    PeroK

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    In which direction do you think the particle is moving? (You've not done anything wrong.)
     
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