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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  
Guest Joined: Posts: n/a Thanks:  Re: need help urgently have entrance exam tomorrow Quote:
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. But, . So, we get: 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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