SOLUTION: 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?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: 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?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1202345: 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?
Answer by ikleyn(52886) About Me  (Show Source):
You can put this solution on YOUR website!
.
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?
~~~~~~~~~~~~~~~~~~~~ 

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 %28dx%29%2F%28dt%29 = 3 units/sec.

        They want you find  %28%28dy%29%2F%28dt%29%29  at x = 1.


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

    2y%2A%28%28dy%29%2F%28dt%29%29 = 16%2A%28%28dx%29%2F%28dt%29%29,    (2)


At x= 1, y^2 = 16, from equation (1).  So, substitute in (2) y = +/- sqrt%2816%29 = +/- 4,  %28%28dx%29%2F%28dt%29%29 = 3, as it is given.  You will get

    2*(+/-4)*%28%28dy%29%2F%28dt%29%29 = 16%2A3,

which gives you

    %28%28dy%29%2F%28dt%29%29 = +/- 6.



ANSWER.  Under given conditions,  %28%28dy%29%2F%28dt%29%29 = +/- 6.

         It means that when the point (x,y) moves on the upper branch y =  4%2Asqrt%28x%29,  %28%28dy%29%2F%28dt%29%29 = 6 at x= 1.
                       When the point (x,y) moves on the lower branch y = -4%2Asqrt%28x%29, %28%28dy%29%2F%28dt%29%29 = -6 at x= 1.

Solved.