Unit
Polynomial and Rational Functions
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I want to do another optimization problem. This time I need to give a short definition first. The distance between a curve and a point b is the minimum of distances between any point a on the curve and point b. So let me give you an example; find the distance between the curve y equals x² and the point 0,1.
This is going to be our point b 0,1 so I have graph here of y equals x² and point b. To find the distance between the point and the curve, I have to find the minimum of all distances between points on the curve and this point.
So how we get started here? Well the first thing I want to do is express this distance using the distance formula. Remember the distance formula involves a big square root and then the difference of the x coordinates, x minus 0² and the difference of the y coordinates y minus 1². And I can simplify this, I get x² plus y² minus 2y plus 1, just expanding these binomials. Now if you look back at the curve, the curve we are dealing with is y equals x² so I can make a substitution for x² I can call it y.
Y plus y² minus 2 y plus 1 and that simplifies further to y² minus y plus 1. This is going to be my function d of y. So I want to find the minimum value of this function. Now let’s think about how to do that because this is not a polynomial function because of the square root but it makes sense that if I can figure out when the inside part is a minimum, I’ll have the minimum distance because the way the square root works, the smaller the input the smaller the output.
So let’s find where y² minus y plus 1 is a minimum. Now recall for a quadratic function f of x is equals a x² plus b x² plus c. You know when you are graphing quadratics that the vertex occurs at x equals -b over 2a. Now all quadratic is f of y equals y² minus y plus 1. So a is 1, b is -1 and c is 1. So our vertex happens at y equals negative minus 1 that’s b over 2 times a which is 1. So one half so the y value that minimizes this quantity is at 1/2 and when this quantity is a minimum, this distance is a minimum.
So find the distance between the curve y equals x² and the point. The minimum distance is going to be given by d of 1/2 so that’s the square root of remember it's y², 1/2 squared minus y 1/2 plus 1. 1/2 squared is a 1/4 minus 1/2 plus 1. So we have now minus 1/2 plus 1 is 1/2 plus a 1/4 is 3/4 so I have the square root of 3/4 which is root 3 over 2. That’s the minimum distance root 3 over 2 which is about .87.
Now let’s see does this make sense going back to our problem. It does kind of make since because if you think about y equals x², you might thing the vertex down here is the closest point 0,0. But it’s a whole unit away from 0,1 and of course there are points at 1,1 and -1,1 that are also one unit away. But somewhere in here there is a point that's root 3 over 2 away from point b that’s the closest point and that point defines the distance between the curve and the point b.
So the problem didn’t ask us to find what point was closest just what’s the distance between the curve and the point and it's root 3 over 2.