Asked by Son
                Find the point on the graph of the function closest to the given point.  Function  f(x) = xsquared  Point (2, 1/2)
            
            
        Answers
                    Answered by
            Steve
            
    using distance from (2,1/2) to y=x^2,
d^2 = (x-2)^2 + (y-1/2)^2
= (x-2)^2 + (x^2 - 1/2)^2
= x^2 - 4x + 4 + x^4 - x^2 + 1/4
= x^4 - 4x + 17/4
d = sqrt(x^4 - 4x + 17/4)
d' = 2(x^3 - 1)/sqrt(x^4 - 4x + 17/4)
d' = 0 when x = 1
so, (2,1/2) is closest to (1,1)
______________________
using normal line, we want the line from (2,1/2) normal to the curve. The distance from P to the curve will be minimum along the line normal to the curve.
At (x,y) the slope of x^2 = 2x
so, the normal has slope -1/2x
So, the equation of the normal line from (2,1/2) is
(y - 1/2)/(x-2) = -1/2x
2x(x^2 - 1/2) = -(x-2)
2x^3 - x = -x + 2
2x^3 = 2
x = 1
So, the normal line from (2,1/2) is the line to (1,1)
    
d^2 = (x-2)^2 + (y-1/2)^2
= (x-2)^2 + (x^2 - 1/2)^2
= x^2 - 4x + 4 + x^4 - x^2 + 1/4
= x^4 - 4x + 17/4
d = sqrt(x^4 - 4x + 17/4)
d' = 2(x^3 - 1)/sqrt(x^4 - 4x + 17/4)
d' = 0 when x = 1
so, (2,1/2) is closest to (1,1)
______________________
using normal line, we want the line from (2,1/2) normal to the curve. The distance from P to the curve will be minimum along the line normal to the curve.
At (x,y) the slope of x^2 = 2x
so, the normal has slope -1/2x
So, the equation of the normal line from (2,1/2) is
(y - 1/2)/(x-2) = -1/2x
2x(x^2 - 1/2) = -(x-2)
2x^3 - x = -x + 2
2x^3 = 2
x = 1
So, the normal line from (2,1/2) is the line to (1,1)
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.