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

Mathematics
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\[\frac{ d ^{2}y }{ dx ^{2} }>0\] a. y<0 b. y>0 c. x>0
Have you tried to find your y''? :)
yes

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\[y''=-\frac{ x ^{2}+y ^{2} }{ y ^{3}}\]
Ooo ok ya :O derivative looks good! So we have to figure out some conditions huh? Hmm
yes I don't know how to figure that out
Notice that x^2 is ALWAYS positive. So y is the only thing dictating the sign of y''.
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
\[\large\rm y''=-\frac{c}{y^3}\] Where c is a positive number. y'' is positive when \(\large\rm -\frac{c}{y^3}\gt0\)
So ummmm
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
There's probably a more algebraic way to do this, but I can't think of it :) lol
so b is the answer if I'm understanding correctly?
wait no, its A
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:/
so a is the answer?
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?
yes
ooh ok. Thank you for the great explanation!
np c:

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