Consider the parabola y = x^2. Find the point on the parabola that is closest to the point (3, 1).
this is the problem of finding the minima closest to point, means MINIMUM distance from that point
let the point be (x,y) distance between x,y and 3,1 will be given by ?
\[1=3^2\]
??
use the distance formula Distance between points (x1,y1) and (x2,y2) is \(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
uuu.....I'll have to put it in that order right?
order doesn't matter x1 - x2 or x2 - x1 both are same
here, x1 = x y1 = y = x^2 x2 = 3 y2 = 1
So..what's next?
plug those in the distance formula
Ok...Give me a sec:)
\[d=\sqrt{(x-3)^2}+(x^2-1)\]
Right??
to simplify things, we can also minimize the square of distance = d^2 so that we could ignore the square root sign, and your function just becomes \(f(x) = (x-3)^2+ (x^2-1)^2\)
So, that's my final answer..right?
did you cancel out square root and square for x^2-1 ?? \(\sqrt{a^2+b^2} \ne \sqrt {(a^2)} +b \)
Yes. I did!:)
that was not correct thing to do .... and do you know how to minimize the function ?
My bad.. and no..I don't know..
whats the topic of this question ? derivatives ?
What? I don't know what exactly you're asking for...
pre-calc, right ? any sub-topic? any similar solved question ? so that i know which method to use to explain...
Hang on..I'm loosing connection real bad..As you can see....I'm replying late...so I'll just get back a little later.When it's better...
no problem :)
Ok. @hartnn Yes. It is Pre-calculus. No subtopic and I don't have any similar solved question yet..This is my first one...
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