Question 1202345
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a point moves on the parabola y^2=16x in such a way that the rate of change of the abscissa 
is always 3 units/sec. how fast is the ordinate changing when the abscissa is 1 unit?
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The point  (x,y)  moves  x = x(t),  y = y(t)  along the parabola 

    y^2 = 16x      (1)

in such a way that the rate of change of the abscissa  always is {{{(dx)/(dt)}}} = 3 units/sec.

        They want you find  {{{((dy)/(dt))}}}  at x = 1.


It is easy.  Take the derivatives respective the time "t" of the equation (1).  You will get

    {{{2y*((dy)/(dt))}}} = {{{16*((dx)/(dt))}}},    (2)


At x= 1, y^2 = 16, from equation (1).  So, substitute in (2) y = +/- {{{sqrt(16)}}} = +/- 4,  {{{((dx)/(dt))}}} = 3, as it is given.  You will get

    2*(+/-4)*{{{((dy)/(dt))}}} = {{{16*3}}},

which gives you

    {{{((dy)/(dt))}}} = +/- 6.



<U>ANSWER</U>.  Under given conditions,  {{{((dy)/(dt))}}} = +/- 6.

         It means that when the point (x,y) moves on the upper branch y =  {{{4*sqrt(x)}}},  {{{((dy)/(dt))}}} = 6 at x= 1.
                       When the point (x,y) moves on the lower branch y = {{{-4*sqrt(x)}}}, {{{((dy)/(dt))}}} = -6 at x= 1.
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Solved.