AP Calc for Reiny
posted by Sanya on .
You did just fine and your second derivative is correct, if you meant (6y^2  4x^2)/(9y^3)
except they took it a bit further.
notice your numerator is
4x^2 + 6y^2
from the original
2x^2  3y^2 = 4 , then
4x^2  6y^2 = 8 , and
4x^2 + 6y^2 = 8
to get their 8/9y^3
If you were marked "wrong", then that is bad,
when differentiating implicitly, there are often multiple variations of the same answer.
Here is a way to check if two possible answers are equivalent:
Pick any point which satisfies the original equation, sub that point into the two variations of the derivatives, you should get the same answer if they are equivalent.
e.g.
the point (√8,2) is on the original curve
y''  your answer  = (24  32)/72 = 8/72 = 1/9
y''  their answer = 8/72 = 1/9
ok, then !!
When you wrote
2x^2  3y^2 = 4 , then
4x^2  6y^2 = 8 , and
4x^2 + 6y^2 = 8
Did you just run a ratio for the first two equations? And, then for te third since it was negative, then you just took the negative number for it 8 to negative 8?
And, I was not marked wrong for it it was a hw problem and the answer in the back had 8 but had gotten the 4x^2 + 6y^2 expression.
Also, how were you able to get the point (√8,2)as a point on the curve? Did you just solve for y? Thanks so much for your help.

Come on. You asked that, and it's just simple algebra, which should be no problem for someone taking calculus.
As for the point, he probably just picked an easy value, y=2 and solved for x:
2x^23y^2=4
2x^2  12 = 4
x^2 = 8
x = √8
I would have not marked the answer wrong, but would have noted that it could be made simpler. After all, if you do a problem and end up with 15/20, don't you try to reduce it?
In this kind of exercise, when you get an answer, always check back to the beginning or intermediate steps to see whether som expression can be made simpler.