Question

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

Answer 1

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

Answer 2

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

Answer 3

yes

Answer 4

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

Answer 5

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

Answer 6

yes I don't know how to figure that out

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

So ummmm

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

wait no, its A

Answer 15

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:/

Answer 16

so a is the answer?

Answer 17

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?

Answer 18

yes

Answer 19

ooh ok. Thank you for the great explanation!

Answer 20

np c: