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Math Help - calc 1 question

  1. #1
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    calc 1 question

    point moves along a curve y=2x^2+1 in such a way that the y value is decreasing at the rate of 2 units per sec. at what rate is x changing when x=3/2.

    y' = 4x so F'(3/2)= m = 6
    so slope=6 and f(3/2) = y= 11/2

    point slope formula = y-(11/2) = 6(x-(3/2)) = y= 12x/2 -7/2

    so do I set x=0 to find rate of change for x?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by weezie23 View Post
    point moves along a curve y=2x^2+1 in such a way that the y value is decreasing at the rate of 2 units per sec. at what rate is x changing when x=3/2.

    y' = 4x so F'(3/2)= m = 6
    so slope=6 and f(3/2) = y= 11/2

    point slope formula = y-(11/2) = 6(x-(3/2)) = y= 12x/2 -7/2

    so do I set x=0 to find rate of change for x?
    nope. implicit differentiation is what is needed here. what you are doing is for a whole other class of problem. are you sure you're getting what's going on in class?


    y = 2x^2 + 1 ..............differentiate with respect to time

    \Rightarrow \frac {dy}{dt} = 4x \frac {dx}{dt}

    now, you are told \frac {dy}{dt} = -2, \mbox{ and } x = \frac 32, you are asked for \frac {dx}{dt}, just solve for it
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