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amy0799

  • one year ago

if x^2+y^2=a where a is a non-zero constant, which of the following conditions are necessary for

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  1. amy0799
    • one year ago
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    \[\frac{ d ^{2}y }{ dx ^{2} }>0\] a. y<0 b. y>0 c. x>0

  2. zepdrix
    • one year ago
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    Have you tried to find your y''? :)

  3. amy0799
    • one year ago
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    yes

  4. amy0799
    • one year ago
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    \[y''=-\frac{ x ^{2}+y ^{2} }{ y ^{3}}\]

  5. zepdrix
    • one year ago
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    Ooo ok ya :O derivative looks good! So we have to figure out some conditions huh? Hmm

  6. amy0799
    • one year ago
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    yes I don't know how to figure that out

  7. zepdrix
    • one year ago
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    Notice that x^2 is ALWAYS positive. So y is the only thing dictating the sign of y''.

  8. zepdrix
    • one year ago
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    And further than that, y^2 doesn't affect our sign either. So we need only to pay attention to the negative in front, and the y^3

  9. zepdrix
    • one year ago
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    \[\large\rm y''=-\frac{c}{y^3}\] Where c is a positive number. y'' is positive when \(\large\rm -\frac{c}{y^3}\gt0\)

  10. zepdrix
    • one year ago
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    So ummmm

  11. zepdrix
    • one year ago
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    When you cube a negative, you get an extra negative sign popping out, ya? So I guess we just need to make sure that y is negative, in order for the two negatives to cancel out. Make sense? :o

  12. zepdrix
    • one year ago
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    There's probably a more algebraic way to do this, but I can't think of it :) lol

  13. amy0799
    • one year ago
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    so b is the answer if I'm understanding correctly?

  14. amy0799
    • one year ago
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    wait no, its A

  15. zepdrix
    • one year ago
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    It's certainly not c, the sign of x has no effect on our second derivative. A? When y is less than 0? So y has to be negative for the entire expression to be positive? Yay good job \c:/

  16. amy0799
    • one year ago
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    so a is the answer?

  17. zepdrix
    • one year ago
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    Here is a way you can visualize it. Recall that the second derivative tells us about concavity. So a positive second derivative should represent a shape of `concave up`. When y is negative, we're dealing with the BOTTOM HALF OF THE CIRCLE. The shape of which is concave up! :) ya?

  18. zepdrix
    • one year ago
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    yes

  19. amy0799
    • one year ago
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    ooh ok. Thank you for the great explanation!

  20. zepdrix
    • one year ago
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    np c:

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