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Pre-Calc Homework Solutions 165

# Pre-Calc Homework Solutions 165 - Section 4.4 To verify...

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To verify that this critical point represents the minimum distance, note that D 0 ( x ) 5 2 5 2 5 } ( x 2 1 a 2 a 2 ) 3/2 } 1 , which is always positive. We now know that D ( x ) is minimized when Equation 1 is true, or, equivalently, } P A R R } 5 } Q BR R } . This means that the two right triangles APR and BQR are similar, which in turn implies that the two angles must be equal. 47. } d d v x } 5 ka 2 2 kx The critical point occurs at x 5 } k 2 a k } 5 } a 2 } , which represents a maximum value because } d dx 2 v 2 } 5 2 2 k , which is negative for all x . The maximum value of v is kax 2 kx 2 5 ka 1 } a 2 } 2 2 k 1 } a 2 } 2 2 5 } k 4 a 2 } . 48. (a) v 5 cr 0 r 2 2 cr 3 } d d v r } 5 2 cr 0 r 2 3 cr 2 5 cr (2 r 0 2 3 r ) The critical point occurs at r 5 } 2 3 r 0 } . (Note that r 5 0 is not in the domain of v .) The critical point represents a maximum because } d dr 2 v 2 } 5 2 cr 0 2 6 cr 5 2 c ( r 0 2 3 r ), which is negative in the domain } r 2 0 } # r # r 0 . (b) We graph v 5 (0.5 2 r ) r 2 , and observe that the maximum indeed occurs a t v 5 1 } 2 3 } 2 0.5 5 } 1 3 } .
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