OpenStudy (yanasidlinskiy):

Consider the parabola y = x^2. Find the point on the parabola that is closest to the point (3, 1).

4 years ago
hartnn (hartnn):

this is the problem of finding the minima closest to point, means MINIMUM distance from that point

4 years ago
hartnn (hartnn):

let the point be (x,y) distance between x,y and 3,1 will be given by ?

4 years ago
OpenStudy (yanasidlinskiy):

$1=3^2$

4 years ago
OpenStudy (yanasidlinskiy):

??

4 years ago
hartnn (hartnn):

use the distance formula Distance between points (x1,y1) and (x2,y2) is $$\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$

4 years ago
OpenStudy (yanasidlinskiy):

uuu.....I'll have to put it in that order right?

4 years ago
hartnn (hartnn):

order doesn't matter x1 - x2 or x2 - x1 both are same

4 years ago
hartnn (hartnn):

here, x1 = x y1 = y = x^2 x2 = 3 y2 = 1

4 years ago
OpenStudy (yanasidlinskiy):

So..what's next?

4 years ago
hartnn (hartnn):

plug those in the distance formula

4 years ago
OpenStudy (yanasidlinskiy):

Ok...Give me a sec:)

4 years ago
OpenStudy (yanasidlinskiy):

$d=\sqrt{(x-3)^2}+(x^2-1)$

4 years ago
OpenStudy (yanasidlinskiy):

Right??

4 years ago
hartnn (hartnn):

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 $$f(x) = (x-3)^2+ (x^2-1)^2$$

4 years ago
OpenStudy (yanasidlinskiy):

4 years ago
hartnn (hartnn):

did you cancel out square root and square for x^2-1 ?? $$\sqrt{a^2+b^2} \ne \sqrt {(a^2)} +b$$

4 years ago
OpenStudy (yanasidlinskiy):

Yes. I did!:)

4 years ago
hartnn (hartnn):

that was not correct thing to do .... and do you know how to minimize the function ?

4 years ago
OpenStudy (yanasidlinskiy):

My bad.. and no..I don't know..

4 years ago
hartnn (hartnn):

whats the topic of this question ? derivatives ?

4 years ago
OpenStudy (yanasidlinskiy):

What? I don't know what exactly you're asking for...

4 years ago
hartnn (hartnn):

pre-calc, right ? any sub-topic? any similar solved question ? so that i know which method to use to explain...

4 years ago
OpenStudy (yanasidlinskiy):

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...

4 years ago
hartnn (hartnn):

no problem :)

4 years ago
OpenStudy (yanasidlinskiy):

Ok. @hartnn Yes. It is Pre-calculus. No subtopic and I don't have any similar solved question yet..This is my first one...

4 years ago