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

1. amy0799

$\frac{ d ^{2}y }{ dx ^{2} }>0$ a. y<0 b. y>0 c. x>0

2. zepdrix

Have you tried to find your y''? :)

3. amy0799

yes

4. amy0799

$y''=-\frac{ x ^{2}+y ^{2} }{ y ^{3}}$

5. zepdrix

Ooo ok ya :O derivative looks good! So we have to figure out some conditions huh? Hmm

6. amy0799

yes I don't know how to figure that out

7. zepdrix

Notice that x^2 is ALWAYS positive. So y is the only thing dictating the sign of y''.

8. zepdrix

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

$\large\rm y''=-\frac{c}{y^3}$ Where c is a positive number. y'' is positive when $$\large\rm -\frac{c}{y^3}\gt0$$

10. zepdrix

So ummmm

11. zepdrix

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

There's probably a more algebraic way to do this, but I can't think of it :) lol

13. amy0799

so b is the answer if I'm understanding correctly?

14. amy0799

wait no, its A

15. zepdrix

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

17. zepdrix

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

yes

19. amy0799

ooh ok. Thank you for the great explanation!

20. zepdrix

np c: