March 1, 2015

Homework Help: Calculus

Posted by Tracy on Friday, November 17, 2006 at 4:51pm.

Hey there. I have a quick question. Ive done questions like these before so I dont know whats stoping me. Anways the question is this:
Find a point on the graph of
y = x^2

with x>0 that lies closest to (0, 3). What is the x-coordinate of such a point?

So I got 3^2 + y^2 = x^2 and solved for x but after that I'm lost. Normally thats what we did, but our other equations was normally like xy=8 so now I'm kinda lost. Any help would be apperciated. Thanks

The distance d from apoint with coordinates (x,y) to the point (0,3) is given by:

d^2 = x^2 + (y-3)^2

You have to minimize this while satisfying the constraint that he point (x,y) be on te parabola. So the contraint is:

y = x^2

This means that:

d^2 = y + (y-3)^2

You have to minimize this function.

Okay, so would you mind checking this? I've lost so many points already I don't wanna enter any more wrong answers.
If we have d^2 = y + (y-3)^2 then d =(y + (y-3)^2) ^1/2 (aka the root is what im getting at) So then if I take the derivative and then make it equal zero I get y = 3, so then x would equal the root of 3?

Hmm well I guess I went wrong somwheres because I put that it and it says its not right.

Sorry again. I missed something when I did the derivative and while adding something together I forgot something out. So now I have that y=5/2 and x=sqrt5/2

This is correct. Note that you can just differentiate d^2 instead of d. The square of a positive function is minimal (maximal) if and only if the function itself is minimal (maximal).

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