How do you find a point on curve y = x^2 that is closet to the point (16, 1/2) ?
First we will draw the graph of given curve and point To find the closest point on curve we draw a perpendicular from point P(16, 1/2) to the curve y = x^2. Let the point Q(t, t^2) be the point on curve that is foot of perpendicular from point P on the curve. Now, Slope of line PQ = (t^2 - 1/2)/(t - 16) Slope of tangent to the curve at Q is value of dy/dx at (t, t^2). So, dy/dx = 2x dy/dx at (t, t^2) = 2t Now, as tangent at Q and line PQ are perpendicular. So, (Slope of PQ)*(Slope of TANGENT at Q) = -1 => ((t^2 - 1/2)/(t - 16))*2t = -1 => 2t^3 - t = -t + 16 =>2t^3 = 16 => t^3 = 8 => t = 2 So, Q(2 , 4) is the point.