The point B(a,b) is on the curve f(x)=x² such that B is the point which is closest to A(6,0). Calculate the value of a.
a point on the curve \(y=x^2\) looks like \((x,x^2)\) and distance between \((x,x^2)\) and \((0,6)\) is by the distance formula \[\sqrt{(x-0)^2+(x^2-6)^2}\] easier to work with the square of the distance, so minimize \[x^2+(x^2-6)^2\]
i am guessing this is calc, so multiply out to get \[d=x^4-11 x^2+36\] then maybe take the derivative to find the critical points good from there?
Oh, shoot, I'm sorry the question said A(6,0) not (0,6).
should be easy now since \[d'=4x^3-22x=2x(x^2-11)\] zeros are easy and they will give you a minimum
But it's the same, right? Just swap the numbers in the distance formula.
in that case it is even easier
yeah it is the same, just swap
\[d^2=(x-6)^2+x^4=x^4+x^2-12 x+36\]
oh actually now it is not easier. you have to find the zeros of a cubic function. i liked the last one more
we can do it a different way, but you get exactly the same equation to solve, so it will be no easier
Well, I am allowed to use a calculator. So I think the cubic equation won't be a problem. :)
well then we can go right to the answer http://www.wolframalpha.com/input/?i=%28x-6%29^2%2Bx^4
So since local min at x=1.33, is that the answer?
yeah looks like that
Ok thanks! :)
yw
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