My Math Forum need help urgently have entrance exam tomorrow

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 April 11th, 2011, 03:51 AM #1 Newbie   Joined: Apr 2011 Posts: 8 Thanks: 0 need help urgently have entrance exam tomorrow the problem is this What is the (closest) distance from the curve y = x^2 to the point (x, y) = (16,1/2) ? thanks
April 11th, 2011, 05:13 AM   #2
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Re: need help urgently have entrance exam tomorrow

Quote:
 (x, y) = (16,1 2) ?
What is this?. (16,12)?.

If so, it helps when minimizing a distance to use the observation that the distance and square of the distance occur at the same points.

So, we do not need the radical in the distance formula.

$S=(x-16)^{2}+(y-12)^{2}$

But, $y=x^{2}$. So, we get:

$S=(x-16)^{2}+(x^{2}-12)^{2}=x^{4}-23x^{2}-32x+400$

Now, differentiate, set to 0 and solve for x.

 April 11th, 2011, 05:57 AM #3 Newbie   Joined: Apr 2011 Posts: 8 Thanks: 0 Re: need help urgently have entrance exam tomorrow no the point is (16,1/2) that i need to measure from
 April 11th, 2011, 11:00 AM #4 Senior Member   Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0 Re: need help urgently have entrance exam tomorrow Just change the 12 to 1/2 in galactus's post and redo the algebra. Then set to 0 and solve for x as he said. If you have more trouble show us where you're stuck and we'll help you through it.
 April 11th, 2011, 04:55 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 I got 7?17/2.

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