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OpenStudy (anonymous):

Find out the closest distance of origin to intersect line of the surfaces x+y=z and x*y=1

6 years ago
OpenStudy (owlfred):

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6 years ago
OpenStudy (dumbcow):

ok i am assuming you want the point that minimizes the distance from the origin to the intersection of the 2 surfaces. the distance of any point from the origin is: \[d = \sqrt{x^{2}+y^{2}+z^{2}}\] set 2 surfaces equal to each other \[x+y = xy -1\] \[\rightarrow y = \frac{x+1}{x-1}\] substitute this into surface equation to get z in terms of x \[z=x+y = x +\frac{x+1}{x-1}\] \[\rightarrow z =\frac{x^{2}+1}{x-1}\] \[d = \sqrt{x^{2}+\frac{(x+1)^{2}}{(x-1)^{2}}+\frac{(x^{2}+1)^{2}}{(x-1)^{2}}}\] We want to minimize d so differentiate and set equal to 0 This can get messy, be careful simplify radical first, then use quotient rule to differentiate \[d ^{\prime}= \frac{2(4x ^{3}-3x ^{2}+4x+1)(x-1) - \sqrt{2(x ^{4}-x ^{3}+2x ^{2}+x+1)}}{2(x-1)^{2}\sqrt{2(x ^{4}-x ^{3}+2x ^{2}+x+1)}}=0\] This simplifies to an 8th degree polynomial, use software to solve this I used Newtons method and found about 4 real solutions the one giving the smallest d will be the answer \[x \approx -0.292\] \[y \approx -0.548\] \[z \approx -0.84\] \[d \approx 1.045\]

6 years ago
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