HIgher Derivatives and Applications

Question

Which point on the curve y-x^2 is nearest to the point (3,0)?

Answer 1

A point on the curve y = x² can be expressed as (x, x²).

 

We want to find x so that the distance from (x,x²) to (3,0) is as small as possible.

 

By the Distance Formula, distance = √[(x-3)²+(x²-0)²]

 

                                                = √[x⁴+x²-6x+9]

 

It's enough to minimize f(x) = x⁴+x²-6x+9.  We can assume that x>0 because when x≤0, points of the form (x,x²) are clearly further away from (3,0) than points on the graph of y = x² that lie in quadrant 1. 

 

  f'(x) = 4x³+2x-6 = 0

 

  By inspection, we see that 1 is a root.  Divide synthetically by x-1:

 

       1⌋      4        0       2      -6

           _________4____4____6

                4        4       6       0

 

  f'(x) = (x-1)(4x²+4x+6)

 

   4x²+4x+6 has imaginary roots, so the only critical number is 1

 

   If 0<x<1, f'(x) < 0 , so f(x) is decreasing

 

   If x>1, f'(x) > 0, so f(x) is increasing

 

   Therefore, the minimum distance occurs when x = 1.  That is, the          point (1,1) lies on the curve y = x² and is closest to (0,3).  

 

 

Answer 2

I assume that the curve is  y =  x²   (a parabola)

 

This is an optimization problem.   The task is to minimize the distance, d, between a point on the parabola and the point (3,0).     It practice, it is somewhat easier to minimize d²  .  This is permissible because d² is monotonic with d and d will not be negative.  

 

The setup is the distance formula.   

 

d²  = (x -3)²  + (y- 0)²   .   The relation y = x²  can be viewed as a constraint equation.  

Using it to substitute for y in the d²  equation gives

 

d² =  (x -3)² + x⁴   .          This is the objective function that must be minimized.  

 

Proceeding as usual,   compute the derivative of d²  with respect to x and set to zero

 

 0 =     2(x -3) + 4 x³  .    The solution to this equation is x =1   (very easy to check this). 

  Since y = x²       y =   1

 

Thus the final answer is the point  ( 1, 1)