Question 1183948
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If {{{x = sqrt(y)}}}, then {{{1 = (dy/dx)/(2*sqrt(y))}}}, or {{{dy/dx = 2sqrt(y) = 2x}}}

If we let point P be ({{{x[0]}}}, {{{y[0]}}}), then the tangent line is 

{{{y - y[0] = (dy(x[0])/dx)*(x-x[0])}}}, whose x-intercept is obtained by letting y = 0:

{{{-y[0] = 2sqrt(y[0])(x-x[0])}}} ===> {{{-sqrt(y[0])/2 = x-x[0]}}} <===> {{{-x[0]/2 = x-x[0]}}} ===> {{{x[i] = x[0]/2}}}.

===> With respect to time,  {{{dx[i]/dt = (1/2)(dx[0]/dt)}}}, or {{{2(dx[i]/dt) = dx[0]/dt}}}.

Now the arc length "s" from x = 0 to x = {{{x[0]}}} is given by 

{{{s = int(sqrt(1+(dy/dx)^2), dx, 0, x[0]) = int(sqrt(1+ 4x^2), dx, 0, x[0])}}}.


===> {{{ds/dt = sqrt(1+ 4x[0]^2)*(dx[0]/dt))}}},   by a direct application of the fundamental theorem of calculus.


===> {{{ds/dt = sqrt(1+ 4x[0]^2)*2*(dx[i]/dt))}}}.


Since at the instant when P=(2,4) {{{ds/dt = 2}}}, we get 

{{{2= sqrt(1+ 4x[0]^2)*2*(dx[i]/dt))}}}, so that {{{dx[i]/dt = 1/ sqrt(1+ 4*2^2) = highlight(1/sqrt(17))}}} unit per second,


the rate of change of the x-intercept Q.