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April 11th, 2011, 03:51 AM   #1
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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
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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.



But, . So, we get:



Now, differentiate, set to 0 and solve for x.
 
April 11th, 2011, 05:57 AM   #3
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Re: need help urgently have entrance exam tomorrow

no the point is (16,1/2) that i need to measure from
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April 11th, 2011, 11:00 AM   #4
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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.
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April 11th, 2011, 04:55 PM   #5
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I got 7?17/2.
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